113
\$\begingroup\$

Write the shortest program in your favourite language to interpret a brainfuck program. The program is read from a file. Input and output are standard input and standard output.

  1. Cell size: 8bit unsigned. Overflow is undefined.
  2. Array size: 30000 bytes (not circled)
  3. Bad commands are not part of the input
  4. Comments begin with # and extend to the end of line Comments are everything not in +-.,[]<>
  5. no EOF symbol

A very good test can be found here. It reads a number and then prints the prime numbers up to that number. To prevent link rot, here is a copy of the code:

compute prime numbers
to use type the max number then push Alt 1 0
===================================================================
======================== OUTPUT STRING ============================
===================================================================
>++++++++[<++++++++>-]<++++++++++++++++.[-]
>++++++++++[<++++++++++>-]<++++++++++++++.[-]
>++++++++++[<++++++++++>-]<+++++.[-]
>++++++++++[<++++++++++>-]<+++++++++.[-]
>++++++++++[<++++++++++>-]<+.[-]
>++++++++++[<++++++++++>-]<+++++++++++++++.[-]
>+++++[<+++++>-]<+++++++.[-]
>++++++++++[<++++++++++>-]<+++++++++++++++++.[-]
>++++++++++[<++++++++++>-]<++++++++++++.[-]
>+++++[<+++++>-]<+++++++.[-]
>++++++++++[<++++++++++>-]<++++++++++++++++.[-]
>++++++++++[<++++++++++>-]<+++++++++++.[-]
>+++++++[<+++++++>-]<+++++++++.[-]
>+++++[<+++++>-]<+++++++.[-]

===================================================================
======================== INPUT NUMBER  ============================
===================================================================
+                          cont=1
[
 -                         cont=0
 >,
 ======SUB10======
 ----------

 [                         not 10
  <+>                      cont=1
  =====SUB38======
  ----------
  ----------
  ----------
  --------

  >
  =====MUL10=======
  [>+>+<<-]>>[<<+>>-]<     dup

  >>>+++++++++
  [
   <<<
   [>+>+<<-]>>[<<+>>-]<    dup
   [<<+>>-]
   >>-
  ]
  <<<[-]<
  ======RMOVE1======
  <
  [>+<-]
 ]
 <
]
>>[<<+>>-]<<

===================================================================
======================= PROCESS NUMBER  ===========================
===================================================================

==== ==== ==== ====
numd numu teid teiu
==== ==== ==== ====

>+<-
[
 >+
 ======DUP======
 [>+>+<<-]>>[<<+>>-]<

 >+<--

 >>>>>>>>+<<<<<<<<   isprime=1

 [
  >+

  <-

  =====DUP3=====
  <[>>>+>+<<<<-]>>>>[<<<<+>>>>-]<<<

  =====DUP2=====
  >[>>+>+<<<-]>>>[<<<+>>>-]<<< <


  >>>


  ====DIVIDES=======
  [>+>+<<-]>>[<<+>>-]<   DUP i=div

  <<
  [
    >>>>>+               bool=1
    <<<
    [>+>+<<-]>>[<<+>>-]< DUP
    [>>[-]<<-]           IF i THEN bool=0
    >>
    [                    IF i=0
      <<<<
      [>+>+<<-]>>[<<+>>-]< i=div
      >>>
      -                  bool=0
    ]
    <<<
    -                    DEC i
    <<
    -
  ]

  +>>[<<[-]>>-]<<          
  >[-]<                  CLR div
  =====END DIVIDES====


  [>>>>>>[-]<<<<<<-]     if divides then isprime=0


  <<

  >>[-]>[-]<<<
 ]

 >>>>>>>>
 [
  -
  <<<<<<<[-]<<

  [>>+>+<<<-]>>>[<<<+>>>-]<<<

  >>




  ===================================================================
  ======================== OUTPUT NUMBER  ===========================
  ===================================================================
  [>+<-]>

  [
   ======DUP======
   [>+>+<<-]>>[<<+>>-]<


   ======MOD10====
   >+++++++++<
   [
    >>>+<<              bool= 1
    [>+>[-]<<-]         bool= ten==0
    >[<+>-]             ten = tmp
    >[<<++++++++++>>-]  if ten=0 ten=10
    <<-                 dec ten     
    <-                  dec num
   ]
   +++++++++            num=9
   >[<->-]<             dec num by ten

   =======RROT======
      [>+<-]
   <  [>+<-]
   <  [>+<-]
   >>>[<<<+>>>-]
   <

   =======DIV10========
   >+++++++++<
   [
    >>>+<<                bool= 1
    [>+>[-]<<-]           bool= ten==0
    >[<+>-]               ten = tmp
    >[<<++++++++++>>>+<-] if ten=0 ten=10  inc div
    <<-                   dec ten     
    <-                    dec num
   ]
   >>>>[<<<<+>>>>-]<<<<   copy div to num
   >[-]<                  clear ten

   =======INC1=========
   <+>
  ]

  <
  [
   =======MOVER=========
   [>+<-]

   =======ADD48========
   +++++++[<+++++++>-]<->

   =======PUTC=======
   <.[-]>

   ======MOVEL2========
   >[<<+>>-]<

   <-
  ]

  >++++[<++++++++>-]<.[-]

  ===================================================================
  =========================== END FOR ===============================
  ===================================================================


  >>>>>>>
 ]
 <<<<<<<<



 >[-]<
  [-]
 <<-
]

======LF========

++++++++++.[-]
@

Example run:

$ python2 bf.py PRIME.BF 
Primes up to: 100
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 
\$\endgroup\$
  • 5
    \$\begingroup\$ You should clarify about 1) size of memory 2) is memory circled 4) maybe any other details \$\endgroup\$ – Nakilon Jan 28 '11 at 1:37
  • 3
    \$\begingroup\$ I wonder if there should be two categories: Those programs that use eval (or shell out to compile) -- and those that don't. \$\endgroup\$ – MtnViewMark Feb 15 '11 at 7:52
  • 34
    \$\begingroup\$ I'd love to see someone answer this in brainfuck. \$\endgroup\$ – Hannesh Mar 14 '11 at 19:15
  • 3
    \$\begingroup\$ What does "no EOF symbol" mean? That the cell value remains unchanged when trying , on EOF? Or that it's up to us to choose a value when trying , on EOF? Or is EOF undefined behaviour altogether? \$\endgroup\$ – Martin Ender Apr 1 '16 at 14:07
  • 3
    \$\begingroup\$ Likewise, what should happen when someone tries to leave the 30k cells to either side? Should the tape head remain in place or is this undefined behaviour? \$\endgroup\$ – Martin Ender Apr 1 '16 at 14:09

52 Answers 52

46
\$\begingroup\$

Perl, 120 138

%c=qw(> $p++ < $p-- + D++ - D-- [ while(D){ ] } . print+chrD , D=ord(getc));
$/=$,;$_=<>;s/./$c{$&};/g;s[D]'$b[$p]'g;eval

This runs hello.bf and primes.bf flawlessly:

$ perl bf.pl hello.bf
Hello World!
$ perl bf.pl prime.bf
Primes up to: 100
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

Initialization: The opcode to Perl translation table is stored in %c. The readable form looks like this:

%c=(
  '>' => '$p++',
  '<' => '$p--',
  '+' => '$b[$p]++',
  '-' => '$b[$p]--',
  '[' => 'while($b[$p]){',
  ']' => '}',
  '.' => 'print chr$b[$p]',
  ',' => '$b[$p]=ord(getc)',
);

Step 1: Slurp program input to $_ and transform it to Perl code using the translation table. Comments are automatically stripped (replaced with undef) in this step.

Step 2: Uncompress all $b[$p] occurrences

Step 3: Launch the program using eval.

\$\endgroup\$
  • \$\begingroup\$ Just use Perl's qw syntax to define %c directly -- good for 7 fewer chars (you'll have to say print+chr$b[$p] and ord(getc), though) \$\endgroup\$ – mob Sep 11 '12 at 23:36
  • \$\begingroup\$ I count 18 saved… thanks! (updating in a minute) \$\endgroup\$ – J B Sep 12 '12 at 8:19
  • 1
    \$\begingroup\$ @olivecoder What on earth are you talking about? \$\endgroup\$ – J B Sep 19 '12 at 7:12
  • \$\begingroup\$ The %c table is declared and defined in the first line; its characters are perfectly accounted for. \$\endgroup\$ – J B Sep 19 '12 at 13:46
  • \$\begingroup\$ @JB hey, I accidentally pressed down vote on your answer and it is locked in, can you edit this so I can reverse the down Vote? \$\endgroup\$ – Teoc Sep 9 '15 at 0:38
69
\$\begingroup\$

Python (no eval), 317 bytes

from sys import*
def f(u,c,k):
 while(c[1]>=k)*u:
  j,u='[]<>+-,.'.find(u[0]),u[1:];b=(j>=0)*(1-j%2*2);c[1]+=b*(j<2)
  while b*c[c[0]]and j<1:f(u,c,k+1);c[1]+=1
  b*=c[1]==k;c[[0,c[0],2][j/2-1]]+=b
  if(j==6)*b:c[c[0]]=ord(stdin.read(1))
  if(j>6)*b:stdout.write(chr(c[c[0]]))
f(open(argv[1]).read(),[-1]+[0]*30003,0)
\$\endgroup\$
  • 70
    \$\begingroup\$ +1 for the f(u,c,k) \$\endgroup\$ – Joel Cornett Aug 4 '12 at 21:48
  • 10
    \$\begingroup\$ That is one beautiful piece of noise, sir \$\endgroup\$ – globby Dec 6 '14 at 1:44
  • \$\begingroup\$ -1 byte if you replace while b*c[c[0]]and j<1 with while b*c[c[0]]*(j<1) \$\endgroup\$ – Daniil Tutubalin Jun 24 at 7:58
50
\$\begingroup\$

16 bit 8086 machine code: 168 bytes

Here's the base64 encoded version, convert and save as 'bf.com' and run from Windows command prompt: 'bf progname'

gMYQUoDGEFKzgI1XAgIfiEcBtD3NIR8HcmOL2LQ/i88z0s0hcleL2DPA86sz/zP2/sU783NHrL0I
AGgyAU14DTqGmAF194qOoAH/4UfDJv4Fwyb+DcO0AiaKFc0hw7QBzSGqT8MmODV1+jPtO/NzDaw8
W3UBRTxddfJNee/DJjg1dPoz7U509YpE/zxddQFFPFt18U157sM+PCstLixbXUxjTlJWXmV+

EDIT

Here's some assembler (A86 style) to create the executable (I had to reverse engineer this as I'd misplaced the original source!)

    add dh,10h                              
    push dx                                 
    add dh,10h                              
    push dx                                 
    mov bl,80h                              
    lea dx,[bx+2]                         
    add bl,[bx]                            
    mov [bx+1],al                         
    mov ah,3dh                              
    int 21h                                 
    pop ds                                 
    pop es                                 
    jb ret                               
    mov bx,ax                              
    mov ah,3fh                              
    mov cx,di                              
    xor dx,dx                              
    int 21h                                 
    jb ret                               
    mov bx,ax                              
    xor ax,ax                              
    repz stosw                                     
    xor di,di                              
    xor si,si                              
    inc ch                                 
program_loop:
    cmp si,bx                              
    jnb ret                               
    lodsb                                    
    mov bp,8                            
    push program_loop
symbol_search:                       
    dec bp                                 
    js ret
    cmp al,[bp+symbols]
    jnz symbol_search
    mov cl,[bp+instructions]
    jmp cx                                 
forward:
    inc di                                 
    ret                                    
increment:
    inc b es:[di]                      
    ret                                    
decrement:
    dec b es:[di]                      
    ret                                    
output:
    mov ah,2                              
    mov dl,es:[di]                            
    int 21h                                 
    ret                                    
input:
    mov ah,1                              
    int 21h                                 
    stosb                                    
backward:
    dec di                                 
    ret                                    
jumpforwardifzero:
    cmp es:[di],dh                            
    jnz ret                               
    xor bp,bp
l1: cmp si,bx                              
    jnb ret
    lodsb                                    
    cmp al,'['                              
    jnz l2
    inc bp
l2: cmp al,']'                              
    jnz l1
    dec bp                                 
    jns l1
    ret                                    
jumpbackwardifnotzero:
    cmp es:[di],dh                            
    jz  ret
    xor bp,bp
l3: dec si                                 
    jz  ret
    mov al,[si-1]                         
    cmp al,']'
    jnz l4
    inc bp  
l4: cmp al,'['                              
    jnz l3
    dec bp                                 
    jns l3
    ret                                    
symbols:
    db '><+-.,[]'
instructions:
    db forward and 255
    db backward and 255
    db increment and 255
    db decrement and 255
    db output and 255
    db input and 255
    db jumpforwardifzero and 255
    db jumpbackwardifnotzero and 255
\$\endgroup\$
  • \$\begingroup\$ I've added a source code version of the program. I've just noticed that non-bf characters cause the program to exit rather than be ignored. Easy to fix that and I'll leave it as an exercise for people to do that themselves. \$\endgroup\$ – Skizz Aug 14 '12 at 13:18
  • \$\begingroup\$ I remember I got the Linux ELF version 166 bytes, 10 years ago, here muppetlabs.com/~breadbox/software/tiny \$\endgroup\$ – Emmanuel Jul 28 '14 at 17:34
39
\$\begingroup\$

brainfuck, 843 691 bytes

Edit: decided to revisit this and found a surprising number of ways to golf off bytes

>>>,[>++++[-<-------->]<-[>+<<]>[----------[>]>[<+<+>>>>]<<<-[>]>[<+<+>>>>]<<<-[>]>[<+<+>>>>]<<<-[>]>[<-<+++>>>>]<<<--------------[>]>[<++<+>>>>]<<<--[>]>[<-<+++++++>>+>>]<<++++[-<------>]+<+[>]>[<++<+>>>>]<<<--[>]>[<<+>>>>]<<-<[+]<[>]>,>]<]<-[<]>[-[<<]>[<+[>]>>[<+[<<[<]<<-[>>]<[>>>>[>]>+<<[<]<]<-[>>]<[>>>>[>]>-<<[<]<]<++[->>+<<]>>[>]>]]<<<[<]>-<]>-[<<]>[<++[>]>+>[<-]<[<<[<]>[-<<+>>]>--[<<]>[[>]>+<<[<]<]>+[<<]>[[>]>-<<[<]<]>+[>]>]<<[<]>--<]>-[<<]>[[>]>>.<<<[<]<]>-[<<]>[[>]>>-<<<[<]<]>-[<<]>[[>]>>,<<<[<]<]>-[<<]>[[>]>>+<<<[<]<]>-[<<]>[[>]>>>[>>]>[<<<[<<]<+>>>[>>]>-]>[-]<<+[<[->>+<<]<]<[->>+<<]<[<]<]>-[<<]>[[>]>-[+>[-<<+>>]>]+<<[-]+[-<<]<[->>>[>>]>+<<<[<<]<]<[<]<]<++++++++>>[+<<->>]>]

This takes input in the form code!input where the !input is optional. It also simulates negative cells without using negative cells itself and can store up to (30000-(length of code+6))/2 cells.

Try it online!

\$\endgroup\$
  • \$\begingroup\$ Just to make sure I got this right, if I ran this program with this program I could nest it 5 levels deep and still handling code-inputs of length 262. \$\endgroup\$ – Draco18s Oct 1 at 21:46
  • \$\begingroup\$ @Draco18s I suspect you'd run out the 30000 cells before that, since the size of each nested interpreter increases exponentially. I think you'd get 2, maybe 3 levels deep \$\endgroup\$ – Jo King Oct 1 at 23:03
  • \$\begingroup\$ Even 3 deep would be hilariously silly. \$\endgroup\$ – Draco18s Oct 2 at 2:34
27
\$\begingroup\$

Ruby 1.8.7, 188 185 149 147 characters

eval"a=[i=0]*3e4;"+$<.bytes.map{|b|{?.,"putc a[i]",?,,"a[i]=getc",?[,"while a[i]>0",?],"end",?<,"i-=1",?>,"i+=1",?+,"a[i]+=1",?-,"a[i]-=1"}[b]}*";"

Somewhat readable version:

code = "a = [0] * 3e4; i = 0;"
more_code ARGF.bytes.map {|b|
  replacements = {
    ?. => "putc a[i]",
    ?, => "a[i] = getc",
    ?[ => "while a[i] > 0 do",
    ?] => "end",
    ?< => "i -= 1",
    ?> => "i += 1",
    ?+ =>"a[i]+=1",
    ?- =>"a[i]-=1"
  }
  replacements[b]
}.join(";")
eval code+more_code

As you see I shamelessly stole your idea of translating to the host language and then using eval to run it.

\$\endgroup\$
  • \$\begingroup\$ You can shave off a byte byte comparing to zero >0 rather than testing equality: !=0. The specs say unsigned, and overflow is undefined. \$\endgroup\$ – anonymous coward Feb 6 '11 at 5:32
  • \$\begingroup\$ 3e4 will also work as opposed to 30000 \$\endgroup\$ – anonymous coward Feb 6 '11 at 5:36
  • \$\begingroup\$ @Charlie: Thanks. Though to be fair it didn't say "unsigned" when I wrote the code. I honestly didn't know that you could write 3e4 though. That's a very good point and good to know. \$\endgroup\$ – sepp2k Feb 6 '11 at 5:40
  • \$\begingroup\$ File.read($*.pop).bytes -> $<.bytes should work too \$\endgroup\$ – Arnaud Le Blanc Feb 12 '11 at 22:31
  • 1
    \$\begingroup\$ Ruby 1.8.7 has an even shorter syntax to build a literal hash: {?a,"foo"}, which is equivalent to {?a=>"foo"}. And testing here shows that you actually can replace File.read($*.pop).bytes with $< without any problems. Also inlining everything to something like eval"a[0]..."+$<.bytes.map{?.,"putc a[i]",...}*";" shortens the solution by another few characters. \$\endgroup\$ – Ventero Feb 14 '11 at 15:58
26
\$\begingroup\$

Binary Lambda Calculus 112

The program shown in the hex dump below

00000000  44 51 a1 01 84 55 d5 02  b7 70 30 22 ff 32 f0 00  |DQ...U...p0".2..|
00000010  bf f9 85 7f 5e e1 6f 95  7f 7d ee c0 e5 54 68 00  |....^.o..}...Th.|
00000020  58 55 fd fb e0 45 57 fd  eb fb f0 b6 f0 2f d6 07  |XU...EW....../..|
00000030  e1 6f 73 d7 f1 14 bc c0  0b ff 2e 1f a1 6f 66 17  |.os..........of.|
00000040  e8 5b ef 2f cf ff 13 ff  e1 ca 34 20 0a c8 d0 0b  |.[./......4 ....|
00000050  99 ee 1f e5 ff 7f 5a 6a  1f ff 0f ff 87 9d 04 d0  |......Zj........|
00000060  ab 00 05 db 23 40 b7 3b  28 cc c0 b0 6c 0e 74 10  |....#@.;(...l.t.|
00000070

expects its input to consist of a Brainfuck program (looking only at bits 0,1,4 to distinguish among ,-.+<>][ ) followed by a ], followed by the input for the Brainfuck program.

Save the above hex dump with xxd -r > bf.Blc

Grab a blc interpreter from https://tromp.github.io/cl/cl.html

cc -O2 -DM=0x100000 -m32 -std=c99 uni.c -o uni
echo -n "++++++++++[>+++++++>++++++++++>+++>+<<<<-]>++.>+.+++++++..+++.>++.<<+++++++++++++++.>.+++.------.--------.>+.>.]" > hw.bf
cat bf.Blc hw.bf | ./uni

Hello World!

\$\endgroup\$
  • 1
    \$\begingroup\$ Why does this even exist? Apparently, it even exists in the realm of research. O.o \$\endgroup\$ – Isiah Meadows Dec 4 '14 at 4:11
  • \$\begingroup\$ So this wouldn't work with commented brainfuck programs? \$\endgroup\$ – kamoroso94 Feb 15 '18 at 20:05
  • \$\begingroup\$ No, not without stripping out the comments first. \$\endgroup\$ – John Tromp Feb 16 '18 at 22:27
18
\$\begingroup\$

Retina 0.8.2, 386 391 386 bytes

Code contains unprintable NUL (0x00) characters. It's also not super golfed yet, because it's already really slow, and if I golf it more, I don't know how long it'd take to finish. Appears to time out on the prime-finding sample.

There may be bugs in the online interpreter or in my program (leading new lines don't show in the output?).

Takes input like <code>│<input>. No, that is not a pipe (|). It's the Unicode character U+2502. The code also uses the Unicode characters ÿ▶◀├║. Unicode characters are used in order to support input of all ASCII characters. Therefore, these characters need to be separated from the code by a non-ASCII character.

Try it online

s`^.*
▶$0├║▶
s{`(▶>.*║.*)▶(.)(.?)
$1$2▶$3
▶$
▶
║▶
║▶
(▶<.*║.*)(.)▶
$1▶$2
T`ÿ-`o`(?<=▶\+.*║.*▶).
^\+

T`-ÿ`ÿo`(?<=▶-.*║.*▶).
^-

(▶\..*├.*)(║.*▶)(.)
$1$3$2$3
(▶,.*│)(.?)(.*├.*▶).
$1$3$2
▶\[(.*║.*▶)
[▶▶${1}
{`(▶▶+)([^[\]]*)\[
$2[$1▶
}`▶(▶+)([^[\]]*)\]
$2]$1
r`([[\]]*)▶\](.*║.*▶[^])
$1◀◀]$2
r{`\[([^[\]]*)(◀+)◀
$2[$1
}`\]([^[\]]*)(◀◀+)
$2◀]$1
◀
▶
}`▶([^│])(.*║)
$1▶$2
s\`.*├|║.*

Note there is a trailing newline there.

Brief Explanation:

Zeros 0x00 are used for the tape, which is infinite. The first replacement sets up the interpreter in the form ▶<code>│<input>├<output>║▶<tape>, where the first is the pointer for the code, and the second one is the pointer for the tape.

ÿ is 0xFF (255), which is used for Transliteration (used to implement + and -) to wrap the cells back around to zero.

is only used for readability (in case the program is stopped in the middle or you want to see the program mid-execution). Otherwise, you couldn't tell which way the pointer was moving.

Commented Code:

s`^.*                       # Initialize
▶$0├║▶
s{`(▶>.*║.*)▶(.)(.?)        # >
$1$2▶$3
▶$
▶
║▶                          # <
║▶
(▶<.*║.*)(.)▶
$1▶$2
T`ÿ-`o`(?<=▶\+.*║.*▶).      # +
^\+

T`-ÿ`ÿo`(?<=▶-.*║.*▶).      # -
^-

(▶\..*├.*)(║.*▶)(.)         # .
$1$3$2$3
(▶,.*│)(.?)(.*├.*▶).        # ,
$1$3$2
▶\[(.*║.*▶)                 # [
[▶▶${1}
{`(▶▶+)([^[\]]*)\[
$2[$1▶
}`▶(▶+)([^[\]]*)\]
$2]$1
r`([[\]]*)▶\](.*║.*▶[^])    # ]
$1◀◀]$2
r{`\[([^[\]]*)(◀+)◀
$2[$1
}`\]([^[\]]*)(◀◀+)
$2◀]$1
◀
▶
}`▶([^│])(.*║)              # next instruction
$1▶$2
s\`.*├|║.*                  # print output

Click here for the code with zeros in place of null bytes. Any occurrences of $0 should not be replaced with nulls.

Edit: Now supports empty input and suppresses trailing newline.

Infinite output is now supported. (403 bytes)

\$\endgroup\$
  • \$\begingroup\$ I kind of wish that I'd placed the <code> and the <tape> next to each other (though it'd be more characters) so that transitioning to an SMBF interpreter would be easier, if I ever decide to do that. \$\endgroup\$ – mbomb007 Jul 20 '16 at 16:43
14
\$\begingroup\$

TI-BASIC, 264 bytes

Because of limitations in TI-BASIC, this actually doesn't qualify for this challenge since it breaks rule 2; the calculators' RAM is very limited, and doing something like 30000->dim(L1 (I use L1 for the stack/array) will force it to throw an ERR:MEMORY. As such, the stack/array starts at a size of 1 and grows if the pointer is pointing to an element past the end of it. It also breaks rule 3, because it's already breaking rule 2 so I may as well not bother with a cell-size limit.

Could probably still be golfed, by the way... I've made one or two edits to that end since first submitting, but if the version below doesn't work then head back to the edit from May 6 '15 and use that code instead. Also, as there's really no ASCII in TI-BASIC, this takes numbers of any size (and anything that returns a number, like a variable or expression) as input, and outputs numbers in turn.

Use SourceCoder to build it into a .8xp file then send it to your calculator with TI-Connect or TILP or something, and run it by including your brainfuck program in double quotes followed by a colon and whatever you named the TI-BASIC program. For instance, if you named it BRAINF, you'd run a program like this: "brainfuck goes here":prgmBRAINF. If you have a shell on your calc that intercepts other commands when it detects the prgm token, though, do this: "brainfuck goes here" -> press ENTER -> prgmBRAINF.

seq(inString("<>-+.,[]",sub(Ans,S,1)),S,1,length(Ans->L2
cumSum((Ans=7)-(Ans=8->L3
seq(Ans(X),X,dim(Ans),1,~1->L4
1->P:DelVar L11->dim(L1 //this is the same as DelVar L1:1->dim(L1 as DelVar does not require a colon or newline after its argument
For(S,1,dim(L2
L2(S->T
P-(T=1)+(T=2->P
dim(L1
Ans+(P-Ans)(P>Ans->dim(L1
L1(P)-(T=3)+(T=4->L1(P
If T=5
Disp Ans
If T=6:Then
Input V
V->L1(P
End
If T=7 and not(L1(P
S+2+sum(not(cumSum(L3(S)-1=seq(L3(X),X,S+1,dim(L3->S
1-S+dim(L3
If T=8 and L1(P
S-sum(not(cumSum(L4(Ans)=seq(L4(X),X,Ans+1,dim(L4->S
End

If you don't have a way of connecting your calculator to your computer and want to type this out on-calc instead (I can't imagine why you'd want to, but I digress) note that -> is the STO> button above the ON key, ~ is the negative symbol next to ENTER, and to replace all instances of L<number> with the corresponding list token found on 2ND -> <number on keypad>

Thanks to thomas-kwa (at least, I think that's his Stack username) for helping me optimize this, especially with the [ and ] instructions.

\$\endgroup\$
  • 1
    \$\begingroup\$ Do you need the parens around Ans+S? \$\endgroup\$ – Zacharý Jul 7 '17 at 13:16
  • \$\begingroup\$ @Zacharý Good catch, no. I must have been unsure about how PEMDAS works or something... I'll refrain from editing, however, because it's been so long that it definitely isn't worth it to bump this post up to the front and because a two-byte reduction isn't going to give the answer any sort of advantage over the others lol. \$\endgroup\$ – M. I. Wright Aug 6 '17 at 23:23
  • 1
    \$\begingroup\$ I remember like 2-3 years ago when I used this program to interpret Brainf*** on my calculator. And, it's an interpret brainf*** question, I think it should be at the top to be honest. \$\endgroup\$ – Zacharý Aug 6 '17 at 23:42
  • 1
    \$\begingroup\$ Actually, I think that whole line could be S-sum(not(cumSum(L4(Ans)=seq(L4(X),X,Ans+1,dim(L4->S. (a-a=0). And hey, don't worry about forgetting ONE order of operation thing here, I've seen a whole host of people forget order of operations for % (mod) on a challenge. \$\endgroup\$ – Zacharý Aug 6 '17 at 23:46
  • 1
    \$\begingroup\$ Oh dang, yeah. Okay, that gives at least 10 bytes off since the if can be made a one-liner as well, plus some other things... may as well edit, then. You've made me whip my calculator out for the first time in like a year to check this stuff, haha \$\endgroup\$ – M. I. Wright Aug 7 '17 at 0:12
13
\$\begingroup\$

Python 275 248 255

I decided to give it a try.

import sys
i=0
b=[0]*30000
t=''
for e in open(sys.argv[1]).read():
 t+=' '*i+['i+=1','i-=1','b[i]+=1','b[i]-=1','sys.stdout.write(chr(b[i]))','b[i]=ord(sys.stdin.read(1))','while b[i]:','pass','']['><+-.,['.find(e)]+'\n'
 i+=(92-ord(e))*(e in'][')
exec t 
\$\endgroup\$
  • 12
    \$\begingroup\$ Neat, you are generating python source code using brainfuck. \$\endgroup\$ – user11 Jan 30 '11 at 2:00
  • 1
    \$\begingroup\$ You may strip 1 char, "import sys as s" and replace "sys" to "s" in the rest \$\endgroup\$ – YOU Feb 6 '11 at 6:52
  • \$\begingroup\$ Note that this is actually 247 chars. (See the nasty space after exec t?). If you use S.Mark's tip and also make the whole for cycle into one line, you can shrink this to 243 chars. \$\endgroup\$ – Oleh Prypin Apr 4 '11 at 15:18
  • \$\begingroup\$ This fails on any input containing [], a valid though trivial bf program. I've suggested an edit which fixes this, but increases the character count. To further reduce the character count, you can from sys import *, and use 'i+=1,...'.split(',') instead of ['i+=1',...]. \$\endgroup\$ – boothby Jul 5 '11 at 20:19
  • 7
    \$\begingroup\$ I'd +1, but many improvements have been suggested and not implemented. \$\endgroup\$ – mbomb007 Nov 15 '16 at 17:01
12
\$\begingroup\$

Haskell, 457 413 characters

import IO
import System
z=return
'>'#(c,(l,d:r))=z(d,(c:l,r))
'<'#(d,(c:l,r))=z(c,(l,d:r))
'+'#(c,m)=z(succ c,m)
'-'#(c,m)=z(pred c,m)
'.'#t@(c,_)=putChar c>>hFlush stdout>>z t
','#(_,m)=getChar>>=(\c->z(c,m))
_#t=z t
_%t@('\0',_)=z t
i%t=i t>>=(i%)
b('[':r)=k$b r
b(']':r)=(z,r)
b(c:r)=f(c#)$b r
b[]=(z,[])
f j(i,r)=(\t->j t>>=i,r)
k(i,r)=f(i%)$b r
main=getArgs>>=readFile.head>>=($('\0',("",repeat '\0'))).fst.b

This code "compiles" the BF program into an IO action of the form State -> IO State the state is a zipper on an infinite string.

Sad that I had to expend 29 characters to turn buffering off. Without those, it works, but you don't see the prompts before you have to type input. The compiler itself (b, f, and k) is just 99 characters, the runtime (# and %) is 216. The driver w/initial state another 32.

>ghc -O3 --make BF.hs 
[1 of 1] Compiling Main             ( BF.hs, BF.o )
Linking BF ...

>./BF HELLO.BF 
Hello World!

>./BF PRIME.BF 
Primes up to: 100
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 

update 2011-02-15: Incorporated J B's suggestions, did a little renaming, and tightened up main

\$\endgroup\$
  • 1
    \$\begingroup\$ You should be able to get the buffering from just IO, and the arguments from just System (-19). The buffering issue bothers me as well, as the spec doesn't really mention it and the top-voted answer doesn't even do I/O. If you must keep it, it's probably shorter to hFlush after each write than change the global buffering mode (-34+15). \$\endgroup\$ – J B Feb 15 '11 at 10:49
11
\$\begingroup\$

Conveyor, 953

This might be the most beautiful code you will ever see:

0

:I\1\@p
>#====)
^#====<
PP0
P<=======================<
00t:)01t1  a:P:P:P:P:P:P:^
>===========">">2>">2>">"^
^           +^-^5^ ^5^]^.^
^           "^"^*^"^*^"^"^
^           -^-^6^-^6^-^-^
^           #^#^*^#^*^#^#^
^           P P -^P )^P P
^           P P #^P )^P P
^t1\)t0:))t01   P   -^  1
^===========<   P   #^  0
^  t1\(t0:))t01     P   t
^=============<     P   )
^         t11(t01   0 0 )
^===============<. t P 10
^                 FT#T#=<
^=================< P 
^             t11)t01 
^===================< 10t))0tP00t:(01t(1a:P:
^                     >=====#=>==========">"
^                             ^          ]^[
^                           P ^          "^"
^===========================<=^#=====<   -^-
                            ^==<     ^ PP#^#=
                                     ^===PTPT<
                                     ^  )P P
                                     ^=<=< (
                                       ^===<
\$\endgroup\$
  • 8
    \$\begingroup\$ Could you add an explanation and a link to an implementation? I want to understand the beauty. ;) \$\endgroup\$ – DLosc May 12 '15 at 22:47
  • 1
    \$\begingroup\$ Well, I'm currently developing it, there is a compiler and a very bad explanation at github.com/loovjo/Conveyor. The source is pretty readable if you want to understand it. \$\endgroup\$ – Loovjo May 13 '15 at 8:30
9
\$\begingroup\$

C 284 362 (From a file)

#include <stdio.h>
char b[30000],z[9999],*p=b,c,*a,i;f(char*r,int s){while(c=*a++){if(!s){(c-62)?(c-60)?(c-43)?(c-45)?(c-46)?(c-44)?0:(*p=getchar()):putchar(*p):--*p:++*p:--p:++p;if(c==91)f(a,!*p);else if(c==93){if(!*p)return;else a=r;}}else{if(c==93){--s;if(!*p&&!s)return;}else if(c==91){s++;}}}}main(int c,char**v){fread(z,1,9999,fopen(*++v,"r"));a=z;f(0,0);}

Primes:

Primes up to: 100
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
Press any key to continue . . .

Compiled and ran successfully VS2008

Original solution failed to recognize loops that were initially set to zero. Still some room to golf. But finally solves the Prime Number program.

Ungolfed:

#include <stdio.h>
char b[30000],z[9999],*p=b,c,*a,i;
f(char*r,int s)
{
    while(c=*a++)
    {   
        if(!s)
        {
            (c-62)?(c-60)?(c-43)?(c-45)?(c-46)?(c-44)?0:(*p=getchar()):putchar(*p):--*p:++*p:--p:++p;
            if(c==91)f(a,!*p);
            else if(c==93){if(!*p)return;else a=r;}
        }
        else
        {
            if(c==93)
            {
                --s;
                if(!*p&&!s)return;
            }
            else if(c==91)
            {
                s++;
            }
        }
    }
}

main(int c,char**v){
    fread(z,1,9999,fopen(*++v,"r"));
    a=z;
    f(0,0);
}

Tests:

Hello World

Rot13

\$\endgroup\$
  • \$\begingroup\$ Are you checking the same pointer (l) every time you loop? I think you are supposed to check the current location of the head (p). \$\endgroup\$ – Alexandru Jan 30 '11 at 19:51
  • \$\begingroup\$ I pass the pointer to the buffer and the pointer to the stream. It checks at the end of the loop to see if the pointer l in the buffer has reached zero and breaks else it resets the stream back to the original loop [. This is necessary for nested [ loops. \$\endgroup\$ – snmcdonald Jan 30 '11 at 20:00
  • 1
    \$\begingroup\$ Yeah. I thought so. You should not check the value at pointer at first enter in the loop, but the value at the current pointer. Check the test in the question. Your program hangs. \$\endgroup\$ – Alexandru Jan 30 '11 at 20:10
  • 1
    \$\begingroup\$ You can replace break;else by return;. \$\endgroup\$ – Alexandru Jan 30 '11 at 20:56
  • 3
    \$\begingroup\$ I think you can replace (c==62)?a:b with (c-62)?b:a. \$\endgroup\$ – Alexandru Jan 30 '11 at 22:01
9
\$\begingroup\$

PHP 5.4, 296 294 273 263 261 209 191 183 178 166 characters:

I gave it a shot without using eval, but I eventually had to use it

<?$b=0;eval(strtr(`cat $argv[1]`,["]"=>'}',"["=>'while($$b){',"."=>'echo chr($$b);',","=>'$$b=fgetc(STDIN);',"+"=>'$$b++;',"-"=>'$$b--;',">"=>'$b++;',"<"=>'$b--;']));

All commands are working. This heavily abuses variable variables, and spews warnings. However, if one changes their php.ini to squelch warnings (or pipes stderr to /dev/null), this works great.

Verification (It's the "Hello World!" example from Wikipedia): http://codepad.viper-7.com/O9lYjl

Ungolfed, 367 365 335 296 267 characters:

<?php
$a[] = $b = 0;
$p = implode("",file($argv[1])); // Shorter than file_get_contents by one char
$m = array("]" => '}', "[" => 'while($a[$b]){',"." => 'echo chr($a[$b]);', "," => '$a[$b]=fgetc(STDIN);', "+" => '$a[$b]++;', "-" => '$a[$b]--;', ">" => '$b++;', "<" => '$b--;');
$p = strtr($p,$m);
@eval($p);

This should be run via the command line: php bf.php hello.bf

\$\endgroup\$
8
\$\begingroup\$

Windows PowerShell, 204

'$c=,0*3e4;'+@{62='$i++
';60='$i--
';43='$c[$i]++
';45='$c[$i]--
';44='$c[$i]=+[console]::ReadKey().keychar
';46='write-host -n([char]$c[$i])
';91='for(;$c[$i]){';93='}'}[[int[]][char[]]"$(gc $args)"]|iex

Fairly straightforward conversion of the instructions and then Invoke-Expression.

History:

  • 2011-02-13 22:24 (220) First attempt.
  • 2011-02-13 22:25 (218) 3e4 is shorter than 30000.
  • 2011-02-13 22:28 (216) Unnecessary line breaks. Matching on integers instead of characters is shorter.
  • 2011-02-13 22:34 (207) Used indexes into a hash table instead of the switch.
  • 2011-02-13 22:40 (205) Better cast to string removes two parentheses.
  • 2011-02-13 22:42 (204) No need for a space after the argument to Write-Host.
\$\endgroup\$
8
\$\begingroup\$

C, 333 characters

This is my first BF interpreter and the first golf I actually had to debug.

This runs the prime number generator on Mac OS X/GCC, but an additional #include<string.h> may be necessary at a cost of 19 more characters if the implicit definition of strchr doesn't happen to work on another platform. Also, it assumes O_RDONLY == 0. Aside from that, leaving int out of the declaration of M saves 3 characters but that doesn't seem to be C99 compliant. Same with the third * in b().

This depends on the particulars of ASCII encoding. The Brainfuck operators are all complementary pairs separated by a distance of 2 in the ASCII code space. Each function in this program implements a pair of operators.

#include<unistd.h>
char C[30000],*c=C,o,P[9000],*p=P,*S[9999],**s=S,*O="=,-\\",*t;
m(){c+=o;}
i(){*c-=o;}
w(){o<0?*c=getchar():putchar(*c);}
b(){if(o>0)*c?p=*s:*--s;else if(*c)*++s=p;else while(*p++!=93)*p==91&&b();}
int(*M[])()={m,i,w,b};
main(int N,char**V){
read(open(V[1],0),P,9e3);
while(o=*p++)
if(t=strchr(O,++o&~2))
o-=*t+1,
M[t-O]();
}
\$\endgroup\$
  • \$\begingroup\$ I think you can shrink it more by using the 'e' notation for all the big numbers. \$\endgroup\$ – luser droog Aug 16 '11 at 4:34
  • \$\begingroup\$ @luser: I was initially surprised too, but the language and compiler won't allow that. I did manage to shrink another 4 chars with tweaks, and using a #define instead of the function table would also probably be terser. I just like the number 333 and the table :v) . \$\endgroup\$ – Potatoswatter Aug 16 '11 at 5:56
  • \$\begingroup\$ Oh, right. I really should've known that. E-notation is in the production for a floating-point constant, whereas a declaration requires an integer. BTW, this may be cheating, but check out nieko.net/projects/brainfuck for Urban Müller's version. The biggest gain appears to be heavy use of ||. \$\endgroup\$ – luser droog Aug 16 '11 at 6:10
8
\$\begingroup\$

CJam, 75 bytes

lq3e4Vc*@{"-<[],.+>"#"T1$T=(t T(:T; { _T=}g \0+(@T@t _T=o "_'(')er+S/=}%s~@

Try it online: string reverser, Hello World.

Explanation

Takes code on the first line of STDIN and input on all lines below it.

l            Read a line from STDIN (the program) and push it.
 q           Read the rest of STDIN (the input) and push it.
  3e4Vc*     Push a list of 30000 '\0' characters.
        @    Rotate the stack so the program is on top.

{               }%   Apply this to each character in prog:
 "-<[],.+>"#         Map '-' to 0, '<' to 1, ... and everything else to -1.
            ...=     Push a magical list and index from it.

s~       Concatenate the results and evaluate the resulting string as CJam code.
  @      Rotate the top three elements again -- but there are only two, so the
         program terminates.

What about that magical list?

"T1$T=(t T(:T; { _T=}g \0+(@T@t _T=o "  Space-separated CJam snippets.
                                        (Note the final space! We want an empty
                                        string at the end of the list.)
_'(')er+                                Duplicate, change (s to )s, append.
        S/                              Split over spaces.

The resulting list is as follows:

T1$T=(t    (-)
T(:T;      (<)
{          ([)
_T=}g      (])
\0+(@T@t   (,)
_T=o       (.)
T1$T=)t    (+)
T):T;      (>)
{          (unused)
_T=}g      (unused)
\0+(@T@t   (unused)
_T=o       (unused)
           (all other characters)

We generate the snippets for + and > from those for - and <, simply by changing left parens (CJam’s “decrement”) into right parens (CJam’s “increment”).

\$\endgroup\$
  • \$\begingroup\$ Shortest answer & biggest winner \$\endgroup\$ – Jack Giffin May 15 '18 at 0:32
7
\$\begingroup\$

F#: 489 chars

The following program doesn't jump at '[' / ']' instructions, but scans the source code for the next matching token. This of course makes it kind of slow, but it can still find the primes under 100. F# integer types don't overflow but wrap.

Here's the short version:

[<EntryPoint>]
let M a=
 let A,B,i,p,w=Array.create 30000 0uy,[|yield!System.IO.File.ReadAllText a.[0]|],ref 0,ref 0,char>>printf"%c"
 let rec g n c f a b=if c then f i;if B.[!i]=a then g(n+1)c f a b elif B.[!i]=b then(if n>0 then g(n-1)c f a b)else g n c f a b
 while !i<B.Length do(let x=A.[!p]in match B.[!i]with|'>'->incr p|'<'->decr p|'+'->A.[!p]<-x+1uy|'-'->A.[!p]<-x-1uy|'.'->w x|','->A.[!p]<-byte<|stdin.Read()|'['->g 0(x=0uy)incr '['']'|']'->g 0(x>0uy)decr ']''['|_->());incr i
 0

A nasty gotcha was that the primes.bf program chokes on windows newlines. In order to run it I had to save the input number to a UNIX formatted text document and feed it to the program with a pipe:

interpret.exe prime.bf < number.txt

Edit: entering Alt+010 followed by Enter also works in Windows cmd.exe

Here's the longer version:

[<EntryPoint>]
let Main args =
    let memory = Array.create 30000 0uy
    let source = [| yield! System.IO.File.ReadAllText args.[0] |]
    let memoryPointer = ref 0
    let sourcePointer = ref 0
    let outputByte b = printf "%c" (char b)
    let rec scan numBraces mustScan adjustFunc pushToken popToken =
        if mustScan then
            adjustFunc sourcePointer
            if source.[!sourcePointer] = pushToken then
                scan (numBraces + 1) mustScan adjustFunc pushToken popToken
            elif source.[!sourcePointer] = popToken then
                if numBraces > 0 then scan (numBraces - 1) mustScan adjustFunc pushToken popToken
            else
                scan numBraces mustScan adjustFunc pushToken popToken 

    while !sourcePointer < source.Length do
        let currentValue = memory.[!memoryPointer]
        match source.[!sourcePointer] with
            | '>' -> incr memoryPointer
            | '<' -> decr memoryPointer
            | '+' -> memory.[!memoryPointer] <- currentValue + 1uy
            | '-' -> memory.[!memoryPointer] <- currentValue - 1uy
            | '.' -> outputByte currentValue
            | ',' -> memory.[!memoryPointer] <- byte <| stdin.Read()
            | '[' -> scan 0 (currentValue = 0uy) incr '[' ']'
            | ']' -> scan 0 (currentValue > 0uy) decr ']' '['
            |  _  -> ()
        incr sourcePointer
    0 
\$\endgroup\$
  • \$\begingroup\$ I solved the Enter issue by not pressing it but Ctrl+J :-) \$\endgroup\$ – Joey Feb 14 '11 at 8:11
  • \$\begingroup\$ Ctrl+J didn't work for me, but entering Alt+010 followed by Enter did. \$\endgroup\$ – cfern Feb 14 '11 at 9:28
7
\$\begingroup\$

Delphi, 397 382 378 371 366 364 328 characters

Eat this Delphi!

328 var p,d:PByte;f:File;z:Word=30000;x:Int8;begin p:=AllocMem(z+z);d:=p+z;Assign(F,ParamStr(1));Reset(F,1);BlockRead(F,p^,z);repeat z:=1;x:=p^;case x-43of 1:Read(PChar(d)^);3:Write(Char(d^));0,2:d^:=d^+44-x;17,19:d:=d+x-61;48,50:if(d^=0)=(x=91)then repeat p:=p+92-x;z:=z+Ord(p^=x)-Ord(p^=x xor 6);until z=0;end;Inc(p)until x=0;end.

Here the same code, indented and commented :

var
  d,p:PByte;
  x:Int8;
  f:File;
  z:Word=30000;
begin
  // Allocate 30000 bytes for the program and the same amount for the data :
  p:=AllocMem(z+z);
  d:=p+z;
  // Read the file (which path must be specified on the command line) :
  Assign(F,ParamStr(1));
  Reset(F,1);
  BlockRead(F,p^,z);
  // Handle all input, terminating at #0 (better than the spec requires) :
  repeat
    // Prevent a begin+end block by preparing beforehand (values are only usable in '[' and ']' cases) :
    z:=1;                       // Start stack at 1
    x:=p^;                      // Starting at '[' or ']'
    // Choose a handler for this token (the offset saves 1 character in later use) :
    case x-43of
      1:Read(PChar(d)^);        // ','     : Read 1 character from input into data-pointer
      3:Write(Char(d^));        // '.'     : Write 1 character from data-pointer to output
      0,2:d^:=d^+44-x;          // '+','-' : Increase or decrease data
      17,19:d:=d+x-61;          // '<','>' : Increase or decrease data pointer
      48,50:                    // '[',']' : Start or end program block, the most complex part :
        if(d^=0)=(x=91)then     // When (data = 0 and forward), or when (data <> 0 and backward)
        repeat                  //
          p:=p+92-x;            // Step program 1 byte back or forward
          z:=z+Ord(p^=x)        // Increase stack counter when at another bracket
              -Ord(p^=x xor 6); // Decrease stack counter when at the mirror char
        until z=0;              // Stop when stack reaches 0
    end;
    Inc(p)
  until x=0;
end.

This one took me a few hours, as it's not the kind of code I normally write, but enjoy!

Note : The prime test works, but doesn't stop at 100, because it reads #13 (CR) before #10 (LF)... do other submissions suffer this problem too when running on CRLF OSes?

\$\endgroup\$
  • \$\begingroup\$ Wow! I never would have expected to trump C in terseness with Delphi! Not until you apply my ideas to C I guess ;-) \$\endgroup\$ – PatrickvL Mar 12 '11 at 16:30
7
\$\begingroup\$

C, 260 + 23 = 283 bytes

I created a C program which can be found here.

main(int a,char*s[]){int b[atoi(s[2])],*z=b,p;char*c=s[1],v,w;while(p=1,
*c){q('>',++z)q('<',--z)q('+',++*z)q('-',--*z)q('.',putchar(*z))q(',',*z
=getchar())if(*c=='['||*c==']'){v=*c,w=184-v;if(v<w?*z==0:*z!=0)while(p)
v<w?c++:c--,p+=*c==v?1:*c==w?-1:0;}c++;}}

Has to be compiled via gcc -D"q(a,b)"="*c-a||(b);" -o pmmbf pmmbf.c and can be called as follows: pmmbf ",[.-]" 30000 whereby the first argument (quoted) contains the bf-program to run, the second determines how large the tape should be.

\$\endgroup\$
  • 1
    \$\begingroup\$ I think that you need to add 23 characters to your count for the -D"q(a,b)"="*c-a||(b);" option, since that seems (to my limited understanding, at least) to be helping you shrink your code. \$\endgroup\$ – Gareth Aug 5 '11 at 9:23
  • \$\begingroup\$ The option is included in the posted text. The reason for it is to avoid the lengthy word define and newline, but I don't think that's really kosher. Anyway with the quotes, comment, and gcc -D I don't see the advantage at all. \$\endgroup\$ – Potatoswatter Aug 6 '11 at 3:30
5
\$\begingroup\$

C, 267

#define J break;case
char*p,a[40000],*q=a;w(n){for(;*q-93;q++){if(n)switch(*q){J'>':++p;J'<':--p;J'+':++*p;J'-':--*p;J'.':putchar(*p);J',':*p=getchar();}if(*q==91){char*r=*p&&n?q-1:0;q++;w(r);q=r?r:q;}}}main(int n,char**v){p=a+read(open(v[1],0),a,9999);*p++=93;w(1);}

Run as ./a.out primes.bf

Ungolfed Version:

#define J break;case

char*p,a[40000],*q=a; // packed so program immediately followed by data

w(n){
    for(;*q-93;q++){ // until ']'
        if(n)switch(*q){ // n = flagged whether loop evaluate or skip(0)
                J'>':++p;
                J'<':--p;
                J'+':++*p;
                J'-':--*p;
                J'.':putchar(*p);
                J',':*p=getchar();
        }
        if(*q==91){char*r=*p&&n?q-1:0;q++;w(r);q=r?r:q;} // recurse on '[', record loop start
    }
}

main(int n,char**v){
    p=a+read(open(v[1],0),a,9999);
    *p++=93; // mark EOF with extra ']' and set data pointer to next
    w(1); // begin as a loop evaluate
}
\$\endgroup\$
5
\$\begingroup\$

Python 2, 223

I admit that I recycled an old program of mine (but had to change it quite a bit, because the old version didn't have input, but error checking...).

P="";i,a=0,[0]*30000
import os,sys
for c in open(sys.argv[1]).read():x="><+-.[,]".find(c);P+=" "*i+"i+=1 i-=1 a[i]+=1 a[i]-=1 os.write(1,chr(a[i])) while+a[i]: a[i]=ord(os.read(0,1)) 0".split()[x]+"\n";i+=(x>4)*(6-x)
exec P

Runs the primes calculator fine.

I see now that Alexandru has an answer that has some similarities. I'll post mny answer anyways, because I think there are some new ideas in it.

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5
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C (gcc) Linux x86_64, 884 621 525 487 439 383 358 354 bytes

*z,*mmap();d[7500];(*p)();*j(a,g)char*a;{char*t=a,*n,c,q=0;for(;read(g,&c,!q);)t=c==91?n=j(t+9,g),z=mempcpy(t,L"\xf003e80Ƅ",5),*z=n-t-9,n:c==93?q=*t++=233,z=t,*z=a-13-t,z+1:stpcpy(t,c-62?c-60?c-43?c-45?c-46?c-44?"":"1\xc0P_\xF\5":"RXR_\xF\5":L"໾":L"۾":L"컿":L"웿");return t;}main(P,g)int**g;{p=mmap(0,1<<20,6,34,0,0);p(*j(p,open(g[1],0))=195,d,1);}

Try it online!

This is a JIT that compiles BF code into x86_64 machine language at runtime. This performs a straight translation so commonly occurring sequences such as >>>, <<<, +++ and --- aren't coalesced into faster instructions.

Less golfed version:

// size of data area
*z,c,*mmap();d[7500];(*p)();
// recursive function translates BF commands to x86_64 instructions
*j(a,g)char*a;{
  char*t=a,*n,q=0;
  for(;read(g,&c,!q);)
    t=c==91? // [
        // cmpb $0x0,(%rsi)
        // je n-t-9
        n=j(t+9,g),
        z=mempcpy(t,L"\xf003e80Ƅ",5)
        *z=n-t-9,
        n
      :
        c==93? // ]
          // jmp a-13-t
          q=*t++=233,
          z=t,
          *z=a-13-t,
          z+1
        :
          stpcpy(t,c-62? // >
                     c-60? // <
                       c-43? // +
                         c-45? // -
                           c-46? // .
                             c-44? // ,
                               ""
                             :
                               // xor %eax,%eax
                               // push %rax
                               // pop %rdi
                               // syscall
                               "1\xc0P_\xF\5"
                           :
                             // push %rdx
                             // pop %rax
                             // push %rdx
                             // pop %rdi
                             // syscall
                             "RXR_\xF\5"
                         :
                           // decb (%rsi)
                           L"໾"
                       :
                         // incb (%rsi)
                         L"۾"
                     :
                       // dec %esi
                       L"컿"
                   :
                     // inc %esi
                     L"웿");
  return t;
}
main(P,g)int**g;{
  // allocate text (executable) memory and mark as executable
  p=mmap(0,1<<20,6,34,0,0);
  // run JIT, set %rdx=1 and call code like a function
  p(*j(p,open(g[1],0))=195,d,1);
}
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4
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C, 374 368

Reads from a file. Passes PRIME.BF test.

Usage: ./a.out PRIME.BF

#include <stdio.h>
main(int c,char**v){int m[30000],s[99],p=0,i=0,n=0;char l[9999],d;FILE*f=fopen(v[1],"r");for(l[i]=0;i<9999&&l[i]!=EOF;l[i]=getc(f))i++;for(i=1;d=l[i];i++){if(!n){p+=d-62?0:1;p-=d-60?0:1;m[p]+=d-43?0:1;m[p]-=d-45?0:1;if(d==46)putchar(m[p]);if(d==44){m[p]=getchar();}if(d==93){i=s[c]-1;c--;n++;}}if(d==91){if(m[p]){c++;s[c]=i;}else{n++;}}n-=d-93?0:1;}}


Reformatted:

#include <stdio.h>
main(int c,char**v){
    int m[3000],s[99],p=0,i=0,n=0;
    char l[9999],d;
    FILE*f=fopen(v[1],"r");
    for(l[i]=0;i<9999&&l[i]!=EOF;l[i]=getc(f))i++;
    for(i=1;d=l[i];i++){
        if(!n){ // > < + - . , ] \n [ ]
            p+=d-62?0:1;
            p-=d-60?0:1;
            m[p]+=d-43?0:1;
            m[p]-=d-45?0:1;
            if(d==46)putchar(m[p]);
            if(d==44){m[p]=getchar();}
            if(d==93){i=s[c]-1;c--;n++;}
        }
        if(d==91){if(m[p]){c++;s[c]=i;}else{n++;}}
        n-=d-93?0:1;
    }
}
\$\endgroup\$
  • \$\begingroup\$ 3000 vs 30000. Your buffer is too small. The program size is too small also. \$\endgroup\$ – Alexandru Jan 31 '11 at 12:54
  • \$\begingroup\$ I made a typo, fixed. What do you mean by program size? If you mean max file size, you didn't specify a minimum it should handle. \$\endgroup\$ – jtjacques Jan 31 '11 at 15:27
4
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Lua, 285

loadstring("m,p={0},1 "..io.open(arg[1]):read"*a":gsub("[^.,<>[%]+-]",""):gsub(".",{["."]="io.write(string.char(@)) ",[","]="@=io.read(1):byte() ",["<"]="p=p-1 ",[">"]="p=p+1 @=@or 0 ",["["]="while @~=0 do ",["]"]="end ",["+"]="@=(@+1)%256 ",["-"]="@=(@-1)%256 "}):gsub("@","m[p]"))()

Somewhat readable version:

loadstring( --execute
    "m,p={0},1 ".. --initialize memory and pointer
    io.open(arg[1]) --open file
        :read"*a" --read all
            :gsub("[^.,<>[%]+-]","") --strip non-brainfuck
                :gsub(".", --for each character left
                    {["."]="io.write(string.char(@)) ", -- '@' is shortcut for 'm[p]', see below
                    [","]="@=io.read(1):byte() ",
                    ["<"]="p=p-1 ",
                    [">"]="p=p+1 @=@or 0 ", --if a before unexplored memory cell, set to 0
                    ["["]="while @~=0 do ",
                    ["]"]="end ",
                    ["+"]="@=(@+1)%256 ", --i like it overflowing
                    ["-"]="@=(@-1)%256 "
                    }
                )
                    :gsub("@","m[p]") --replace the '@' shortcut
    ) --loadstring returns a function
() --call it

Works perfectly

Lua, 478, w/o loadstring

local m,p,i,r,c={0},1,1,{},io.open(arg[1]):read"*a"while i<=#c do(({[43]=function()m[p]=(m[p]+1)%256 end,[45]=function()m[p]=(m[p]-1)%256 end,[62]=function()p=p+1 m[p]=m[p]or 0 end,[60]=function()p=p-1 end,[46]=function()io.write(string.char(m[p]))end,[44]=function()m[p]=io.read(1):byte()end,[91]=function()if m[p]==0 then i=select(2,c:find("%b[]",i))else r[#r+1]=i end end,[93]=function()if m[p]==0 then r[#r]=nil else i=r[#r] end end})[c:byte(i)]or function()end)()i=i+1 end

Readable version:

local m,   p, i, r,  c= --memory, pointer, brackets stack, code
      {0}, 1, 1, {}, io.open(arg[1]) --open file
              :read"*a" --read it
while i<=#c do --while there's code
    (
        (
            {
                [43]=function() -- +
                    m[p]=(m[p]+1)%256
                end,
                [45]=function() -- -
                    m[p]=(m[p]-1)%256
                end,
                [62]=function() -- >
                    p=p+1 m[p]=m[p]or 0 --if new memory cell, set it to 0
                end,
                [60]=function() -- <
                    p=p-1
                end,
                [46]=function() -- .
                    io.write(string.char(m[p]))
                end,
                [44]=function() -- ,
                    m[p]=io.read(1):byte()
                end,
                [91]=function() -- [
                    if m[p]==0 then
                        i=select(2,c:find("%b[]",i)) --find matching ]
                    else
                        r[#r+1]=i --push position to the stack
                    end
                end,
                [93]=function() -- ]
                    if m[p]==0 then
                        r[#r]=nil --pop from stack
                    else
                        i=r[#r] --go to position on the top of stack
                    end
                end
            }
        )[c:byte(i)] --transform character into code
        or function()end --do nothing on non-brainfuck
    )() --run the resulting function
    i=i+1 --go to the next opcode
end
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4
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Brainfuck, 948 bytes

Well, that took a while. I golfed a Brainfuck self-interpreter by ... not me.

->->>>-[,+>+<[->-]>[->]<+<-------------------------------------[+++++++++++++++++++++++++++++++++++++>-]>[->]<<[>++++++++[-<----->]<---[-[-[-[--------------[--[>+++++++[-<---->]<-[--[[+]->]<+[->++>]->]<+[->+>]->]<+[->+++++>]->]<+[->++++++>]->]<+[->+++++++>]->]<+[->++++>]->]<+[->++++++++>]->]<+[->+++>]->]+<+[->->]>[-<->]<]>>->>-<<<<<+++[<]>[-[-[-[-[-[-[-[-<<++++++++>>>[>]>>>>+[->>+]->,<<<+[-<<+]-<<<[<]<]>[<<<+++++++>>>[>]>>>>+[->>+]->.<<<+[-<<+]-<<<[<]]<]>[<<<++++++>>>[>]>>>>+[->>+]<<-<<+[-<<+]-<<<[<]]<]>[<<<+++++>>>[>]>>>>+[->>+]+>>-<<[-<<+]-<<<[<]]<]>[<<<++++>>>[>]>>>>+[->>+]->-<<<+[-<<+]-<<<[<]]<]>[<<<+++>>>[>]>>>>+[->>+]->+<<<+[-<<+]-<<<[<]]<]>[<++[>]>>>>+[->>+]->[<<<+[-<<+]-<<<[<]-[<<-[>->-[<+]]<+[->>[<]]<-[>-->+[<++]]<++[-->>[<]]<++>>[[-<+>]<<[->>+<<]]<[>]>]]<[<<+[-<<+]-<<<[<]>--<<++>]>]<]>[<<<+>>>[>]>>>>+[->>+]->[<<<+[-<<+]-<<<[<]]<[<<+[-<<+]-<<<[<]+[>-[<-<]<<[>>]>>-[<+<]<<[>>]>>++<[>[-<<+>>]<[->+<]]<[>]>]]>[[-<<+>>]<[->+<]>]]>]
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4
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Recall, 594 bytes

In short: Recall has no arithmetic operators in a classic sense, it only has bitwise operations. You can not just "add one" etc. Recall is also strictly stack-based.

DC505M22022M32032M606M42042M707M92092M4405022o032o06o042o07o092o044o1305022o06o042o092o52052q.q2305022o06o07o93093q.q5403206o07o14014q.q6403206o042o07o24024q.q74Yx34034z03MMMMMMMM034o3yY030401r3.4.101zyY040301r4.3.101zY01052gZ02Z040301052023s4.3.10zyY01023gZ02z030401023052s3.4.10zyY01093gZ02q20zyY01054gZ02u20zyY01014gZx20zyY01064gZ02X0zyY01024gZ03304302r33.43.20zyY01074gZ04303302r43.33.20zyyQ6205.8Y06208g6206208iZ08M808013izy062U7205.9Y07209g7207209iz09M909013izy072R53.63.82063MMMMMMMM053o63082013i53082KKKKKKKK82053063082S84.94.12.73.83t012073083TY083073012r83.73.12012084gzY012094gZt0zyy

Example 1: Print something

Input:

-[--->+<]>-----..-[----->+<]>.++++.+[->++++<]>.---[----->++<]>.---.------------.++++++++.++++++++.+[-->+++++<]>-.

Output:

PPCG rocks!

Example 2: Output square numbers up to 100

Input:

+[>++<-]>[<+++++>-]+<+[>[>+>+<<-]++>>[<<+>>-]>>>[-]++>[-]+>>>+[[-]++++++>>>]<<<[[<++++++++<++>>-]+<.<[>----<-]<]<<[>>>>>[>>>[-]+++++++++<[>-<-]+++++++++>[-[<->-]+[<<<]]<[>+<-]>]<<-]<<-]

Output:

0
1
4
9
16
25
36
49
64
81
100

This example might take a few minuted to execute and might cause a "this tab is frozen" message. Ignore that and wait.

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  • 4
    \$\begingroup\$ Your website domain has expired. Also, this answer is non-competing, because the language is newer than the challenge. \$\endgroup\$ – mbomb007 Mar 23 '16 at 20:27
3
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OCaml(lex), 497 chars

OCamllex is part of the standard distribution of OCaml.

{let a=Array.create 30000 0
let(%)f g h=f(g h)
let s v i=a.(i)<-v;i
let o d i=s(a.(i)+d)i
let p i=print_char(Char.chr a.(i));flush stdout;i
let r i=s(Char.code(input_char stdin))i
let rec w g i=if 0=a.(i)then i else w g(g i)
let n x=x}
rule t f=parse
|'>'{t(succ%f)lexbuf}
|'<'{t(pred%f)lexbuf}
|'+'{t((o 1)%f)lexbuf}
|'-'{t((o(-1))%f)lexbuf}
|'.'{t(p%f)lexbuf}
|','{t(r%f)lexbuf}
|'['{t((w(t n lexbuf))%f)lexbuf}
|']'|eof{f}
|_{t f lexbuf}
{let _=t n(Lexing.from_channel(open_in Sys.argv.(1)))0}

Save as b.mll and run with

ocamllex b.mll && ocaml b.ml prime.bf

I don't like parsing by hand, so I used the provided lexer generator. From the tokens read, we compose a function for the whole brainf*ck program.

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3
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C# (2861 char, ~84 lines)

This is not the prettiest solution to the problem, and probably not all that 'Golf-ish', since I wasn't as concerned with length as I probably should have been. (I didn't remove the comments or extra white space.) I just wanted to try something in a new language, to see if I could. If I did it again, I'd drop the use of the stack for returning from ']' and just look back. Run without command line arguments it runs the hello world program given in the problem description. It accepts one command line argument, the filename of the program to run.

using System;
using System.Collections.Generic;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            String ProgSource;
            if (args.Length > 0)
                ProgSource = System.IO.File.ReadAllText(args[0]);
            else //hello world
                ProgSource = "";

            Stack<int> stack = new Stack<int>();
            char[] bfProg = ProgSource.ToCharArray();
            char[] mem = new char[30000];
            int ptr = 0;

            for (int ip = 0; ip<bfProg.Length; ip++){
                switch (bfProg[ip])
                {
                    case ('>'): ptr++;  break;
                    case ('<'): ptr--;  break;
                    case ('+'): mem[ptr]++; break;
                    case ('-'): mem[ptr]--; break;
                    case ('.'): Console.Write(mem[ptr]); break;
                    case (','): 
                        char key = Console.ReadKey(false).KeyChar;
                        if (key == '\r')
                        {
                            key = (char)10;
                            Console.WriteLine();
                        }
                        mem[ptr] = key;
                        break;
                    case ('['):
                        if (mem[ptr] == 0)
                        {
                            int openBraces = 1;
                            //find the closing brace for this expression
                            for (int x = 1; x < (bfProg.Length - ip); x++)
                            {
                                if (bfProg[ip + x] == ']') openBraces--;
                                if (bfProg[ip + x] == '[') openBraces++;
                                if (openBraces == 0)
                                {
                                    if (stack.Peek() == ip) stack.Pop();
                                    ip += x;
                                    break;
                                }                                
                            }
                       }
                       else
                       {
                           stack.Push(ip);
                       }
                       break;
                    case (']'):
                        if (mem[ptr] == 0)
                            stack.Pop();
                        else
                        {
                            ip = stack.Peek();
                        }
                        break;
                }
            }

            Console.WriteLine("\n\n\nExecution Completed Sucessfully. Press any key to continue...");
            Console.ReadKey();

        }
    }

}

Edit: Removed unused references.

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  • 1
    \$\begingroup\$ @mbomb007 - Updated. Completely forgot I even did this one. (Didn't even realize anyone even read these old questions) \$\endgroup\$ – theB Aug 31 '15 at 21:18
  • \$\begingroup\$ Not only do people still read them, they still answer and golf them. \$\endgroup\$ – mbomb007 Nov 15 '16 at 17:08
3
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C (gcc), 273 268 bytes

main(_,a){_=fopen("w.c","w");fputs("main(){char a[30000],*p=a;",_);x:a=getchar();fputs(a-62?a-60?a-43?a-45?a-46?a-44?a-91?a-93?~a?"":"}":"}":"while(*p){":"*p=getchar();":"putchar(*p);":"--*p;":"++*p;":"--p;":"++p;",_);if(~a)goto x;fclose(_);system("cc w.c;./a.out");};

Try it online!

-5 thanks to ceilingcat

Takes input from stdin.

This relies a little bit on the environment, but is pretty consistent. This is effectively the eval solution for c. It writes an appropriate C program to the file w.c, compiles it, and runs it as the desired executable. Thus as a bonus effect this actually compiles the bf code and leaves a.out as a binary for it. Note that depending on the system you may need to modify the last string. In particular most windows c compilers call the default executable "a.exe". Luckily as far as I can tell, they all have the same length so the bytecount is the same. (though if you don't have a cc defined you may need to add a letter such as gcc to the compile command, adding 1 byte).

I am aware that this thread is a bit old, but I didn't see this style of C solution yet, so I thought I'd add it.

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2
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[EDIT]

C++11, 355, reads from file:

#include<functional>
#include<stdio.h>
main(){
char b[30000],g[9999],*f=g,*p=b,n[]="+-,.><[]",j;
std::function<void()>m[]={
[&p]{(*p)++;},
[&p]{(*p)--;},
[&p]{*p=getchar();},
[&p]{putchar(*p);},
[&p]{p++;},
[&p]{p--;},
[&p,&f]{if(!(*p))while(*f-93)f++;},
[&f,&m]{while(*f-91)f--;m[6]();}
};
fread(g,1,9999,fopen(a[1],0));
for(;*f;f++)for(j=0;n[j];j++)if(n[j]==*f)m[j]();
}

Test

http://ideone.com/b7vO4

[OLD VERSION]

C++11, 391, to see running: http://ideone.com/yZHVv

#include<functional>
#include<stdio.h>
main(int c,char **a) {
  char b[30000],g[9999],*f=g,*r=f,*p=b;
  std::function<void()>m[256];
  m['>']=[&p]{p++;};  
  m['<']=[&p]{p--;};
  m['+']=[&p]{(*p)++;};
  m['-']=[&p]{(*p)--;};
  m['.']=[p]{putchar(*p);};
  m[',']=[&p]{*p=getchar();};
  m['[']=[p,&r,&f]{*p?r=f-1:r=0;};
  m[']']=[&r,&f]{r?f=r:r=f;};
  fread(g,1,9999,fopen(a[1],"r"));
  while (c=*(f++))if(m[c]&&(r||c==']'))m[c]();
}
\$\endgroup\$

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