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You have one string of input bytes, output only the last byte in it.
Your submission may be a program or function outputting the last byte in the input which
I was trying to solve this with brainfuck, however all languages are allowed to participate. This is code-golf.
"?" -> "?"
"29845812674" -> "4"
Last
Try it online! (If the input could be a list of characters, &/S could work.)
5 bytes: `@&-1
8 bytes: &/S@List
10 bytes: `@«_,-1»
10 bytes: Fold!Right
10 bytes: `@<~_,-1~>
10 bytes: `^^&:Right
10 bytes: {Right^^_}
11 bytes: Get«_,-1»
11 bytes: Get<~_,-1~>
12 bytes: `@«_,#_-1»
12 bytes: `@<~_,#_-1~>
13 bytes: Get«_,#_-1»
13 bytes: Get<~_,#_-1~>
Last inputted byte. The programs contents fit with the challenge
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Mar 19, 2019 at 23:53
As @CodyGray correctly points out, taking input as a string and output to a register removes the bulk of the standalone program version.
Input string is in SI, length in CX and output character is in AL:
F3 AC REPZ LODSB ; start at memory location pointer in SI, put next value in AL,
; loop CX number of times. The last char will be in AL when done.
Or 4 bytes as a "Pascal string" (length is prepended to beginning of string):
AC LODSB ; first byte is string length
91 XCHG AX, CX ; move length to CX for loop
F3 AC REPZ LODSB ; start at memory location pointer in SI, put next value in AL,
; loop CX number of times. The last char will be in AL when done.
Or 5 bytes as a "C string" (zero/null terminated), input in DI:
F2 AE REPNZ SCASB ; scan for value in AL (0), end when found and advance DI
8A 45 FE MOV AL, [DI-2] ; DI is now two bytes ahead of last, put value of DI-2 into AL
Or as complete program as IBM PC DOS executable. Input is from command line, output is to console.
B3 80 MOV BL, 80H ; BX to DOS PSP at 80H
8A 07 MOV AL, BYTE PTR[BX] ; get command line tail length
D7 XLAT ; AL = [BX+AL]
B4 0E MOV AH, 0EH ; PC BIOS write to screen function
CD 10 INT 10H ; display
C3 RET ; exit to DOS
Output:
SI, length in CX output char is in AL" and then I think the only code that would be necessary is REPZ LODSB (2 bytes) and we'd be done. Of course this approach wouldn't be how you do it if you were coding for efficiency, not size. Your point is very well taken though, I'll post it also as a function that does the meat of the work.
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,[>,]<.
-1 as the EOF. +[>,+]<-. should work
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0)
MATL uses 1-based modular indexing so this solution grabs the element in the 0-th position of the input which is the same as the last since the 0 wraps around to the end.
Try it out at MATL Online
Explanation
% Implicitly grab the input
0 % Push the literal 0 to the stack
) % Use this zero to grab the character at the end of the string
% Implicitly display the result
DO,1<-#1PLEASECOMEFROM(2)DOWRITEIN,1DO.1<-,1SUB#1DO.5<-#1$.1~#256DO.2<-.3DO(1)NEXTPLEASE.2<-'"!3~#1'$.3~#16"$!3~#4'$.3~#64'$"!3~#2'$.3~#32"$!3~#8'$.3~#128DO(1010)NEXTPLEASE,1SUB#1<-.3DOREADOUT,1(1)DO(1002)NEXTDO(1009)NEXTDO.3<-.3~#255(2)DOFORGET#1
-8 thanks to 鳴神裁四点一号 removing DOGIVEUP to terminate by NEXT stack explosion, opening up another -4 changing a PLEASE to DO.
-11 removing some grouping from the bit-reversing expression from hell. It seems that binary operators generally act right-associative in C-INTERCAL, though having no precedence among themselves.
Writing this was... interesting. I was thinking I might want to use INTERCAL to INTERCALate, but I'm a bit less sure now.
DO ,1<-#1 PLEASE NOTE We want the input array to only have space for one element, so it will only take one at a time
DO COME FROM (2)
DO WRITE IN ,1 PLEASE NOTE If this is the first byte of the input, it'll write its value... but if not, it'll write the
previous value minus its value mod 256.
DO .1<-,1SUB#1
DO .5<-#1$!1~#256' PLEASE NOTE .5 is 3 if the input is 256, 2 otherwise
DO .2<-.3
DO (1) NEXT
PLEASE NOTE If we're here, we've found the end of the input. Now, we need to print it back out... C-INTERCAL's
array I/O, in order to determine what it will actually print, subtracts the value it's going to print from the
previous one (still mod 256, and with the previous value defaulting to 0), and then reads the bits of the byte
backwards. So in order to go from the value we want to display to the value we need to feed into READ OUT, we
reverse the bits and then subtract from 256. The nightmarish expression on the following line reverses the
bits the best way I could think to: individually select each one out and then mingle them all back
together. It may be possible to emulate the method used in cesspool.c, by using mingle and unary AND as a
substitute for binary AND where we can't afford for select to rearrange it, but it might end up longer...
DO .2 <- '"'.3~#1'$'.3~#16'"$"'.3~#4'$'.3~#64'"'$'"'.3~#2'$'.3~#32'"$"'.3~#8'$'.3~#128'"'
DO (1010) NEXT PLEASE NOTE .1 already has 256 in it, which is very convenient for when you need to subtract .2 from 256.
DO ,1SUB#1 <- .3 PLEASE NOTE If we just read .3 out, we'd get a Roman numeral instead of the correct output.
DO READ OUT ,1
DON'T GIVE UP PLEASE NOTE Logical end of program. However, if we don't gracefully terminate here, nothing more is output,
and the FORGET can't keep up with the NEXTs.
(1) DO (1002) NEXT PLEASE NOTE that that line in syslib does 1001 next, which pops .5 entries off the next-stack and returns
control flow to the last one, such that if .5 is 2 flow will come back here, but if it's 3 then it'll go back
to the line that nexted to this one.
Here we add .1 and .2 into .3, then truncate it to a byte before looping back (while managing the next-stack
responsibly so the program doesn't disappear into the black lagoon for any input over 79 (?) bytes)
DO (1009) NEXT
DO .3<-.3~#255
(2) DO FORGET #1
<?=$argn[-1];
Run with php -nF input is STDIN. Example:
$ echo 29845812674|php -nF lost.php
e in the header). For your second question, the header is e=\ , which basically means e=lambda x:x[-1]
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Mar 22, 2019 at 20:14
e=\ but Markdown escapes the code character so I have to add a trailing space
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Mar 22, 2019 at 20:16
a=>a.slice(-1)
[0]. Or maybe if there is a short way to get array length. Different approach: a=>[...a].pop() (15bytes)
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tail -c1
Input is from stdin, output is to stdout.
1 0 1 1 0 0 0
0 0 0 1 1 0 0
1 1 1 1 0 0 0
0 1 0 1 2 0 0
1 2 1 1 0 0 0
0 2 0 1 3 0 0
1 3 1 1 0 0 0
0 3 0 1 4 0 0
1 4 1 1 0 0 0
0 4 0 1 5 0 0
1 5 1 1 0 0 0
0 5 0 1 6 0 0
1 6 1 1 0 0 0
0 6 0 1 7 0 0
1 7 1 1 0 0 0
0 7 0 1 8 0 0
1 8 1 1 0 0 0
0 8 0 0 9 0 0
0 9 0 0 a 0 0
0 a 0 0 b 0 0
0 b 0 0 c 0 0
0 c 0 0 d 0 0
0 d 0 0 e 0 0
0 e 0 0 f 0 0
0 f 0 0 h 0 0
0 h 0 0 g 0 0
0 g 0 0 0 1 1
1 g 1 0 0 1 1
EXPLANATION
Detect eight zero bits (which will occur at the end of the input, since TMBWW uses an infinite tape of bits.)
1 1 1 1 0 0 0
0 1 0 1 2 0 0
1 2 1 1 0 0 0
0 2 0 1 3 0 0
1 3 1 1 0 0 0
0 3 0 1 4 0 0
1 4 1 1 0 0 0
0 4 0 1 5 0 0
1 5 1 1 0 0 0
0 5 0 1 6 0 0
1 6 1 1 0 0 0
0 6 0 1 7 0 0
1 7 1 1 0 0 0
0 7 0 1 8 0 0
1 8 1 1 0 0 0
0 8 0 0 9 0 0
-------------
When eight 0 bits are detected, move back to the final byte of the input and print it out while halting the program.
0 9 0 0 a 0 0
0 a 0 0 b 0 0
0 b 0 0 c 0 0
0 c 0 0 d 0 0
0 d 0 0 e 0 0
0 e 0 0 f 0 0
0 f 0 0 h 0 0
0 h 0 0 g 0 0
0 g 0 0 0 1 1
1 g 1 0 0 1 1
sub(Ans,length(Ans),1
Gets the last character in the input string.
Input is in Ans.
Output is in Ans and is automatically printed out.
Pretty much equivalent to @remoel's VBA answer:
=RIGHT(A1)
Even a task as simple as this presents an interesting optimisation challenge when riding the London Underground.
Take Northern Line to Bank
Take Circle Line to Bank
Take Central Line to Mile End
Take Central Line to Holborn
Take Piccadilly Line to Heathrow Terminals 1, 2, 3
Take Piccadilly Line to Acton Town
Take District Line to Acton Town
Take District Line to Parsons Green
Take District Line to Mile End
Take Central Line to Bank
Take Circle Line to Bank
Take Northern Line to Mornington Crescent
Visiting Mile End station allows you to take a substring from the end of the input - but to chop just 1 character, you need to generate the integer 1 somehow. Rather than doing any arithmetic, the fastest method turns out to be to parse it from the station name "Heathrow Terminals 1, 2, 3".
To bypass that, an alternate strategy for this challenge would be to reverse the input, read the character code for the now first byte, and then turn that back into a char to output - but this approach takes 12 bytes longer. (Although there are fewer trips needed, so the tickets would be cheaper.)
_\
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Input from STDIN, 71 61 bytes
-10 bytes thanks to OlivierGrégoire
v->{var x=System.in;x.skip(x.available()-1);return x.read();}
Try it online!
Function Argument, 25 bytes
s->s.charAt(s.length()-1)
s->s[s.length-1] would have been enough with a char[] parameter-type.
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Mar 18, 2019 at 10:47
pA/@po
p
A / @ p
o
A Takes all the input/ Redirect around the cubepp bring bottom of the stack to the top twiceo/@ output as character, redirect and halt
:1+_p1-,@>~#
Thanks to @Jo King for golfing off 3 bytes.
Alternate 15 byte version that is less messy:
~:1+#v!_
@,$<
Taking strings as input in Befunge isn't the easiest. If there were a single command to take in multiple characters, it would be as simple as reading the string, popping/printing the top character, and exiting.
$$ instead of p1 should work without the warning for the same amount of bytes
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Ṫ
Not the most difficult challenge in Jelly...
Note this accepts the input as a string; if the input could be interpreted otherwise (e.g. a number, a list), then it the argument will need to be quoted (e.g. "123456" or "[123,197]"). Alternatively this can be seen as a link that takes a byte array and returns the last member of that array, in accordance with PPCG standard rules.
Thanks to @MilkyWay90 and @ბიმო for pointing this out.
😶👉😃😨👿
Explanation
😶 Push a copy of the first stack value.
👉 Push the length of the first stack value interpreted as a string.
😃 Push literal 1
😨 Push the difference of the second and first stack values.
👿 Push the character of the second stack value at the index of the top stack value.
?a|,
;.]^
Pretty happy with this, as it is only 3 bytes longer than my cat program
?
^;.
| |a
] |
a ,|
This essentially just loops through pushing input characters into the a stack until EOF is reached. Then it outputs the item at the top of the a stack using .a.
|s|s.pop()
Try it on the Rust Playground!
An anonymous function that takes in a mutable String and outputs the char at the end of the string. Minus a lot of bytes, thanks to madlaina
.unwrap() call isn't necessary as per this meta post. Also, if you use the closure in a function that requires a specific signature, the compiler can do a better job at type deduction and you can leave the :String out: example
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INPUT S$?POP(S$)
This just uses a simple "extract 1 char from the end and print it" aproach.
{%macro a(a)%}{{a[-1:1]}}{%endmacro%}
It was really easy to do, and test, but was fun!
To use it, you have to put it on a .twig file and import it:
{% import 'a.twig' as a %}
{{ a.a("string") }} {# should display "g" #}
You can test it on https://twigfiddle.com/aa19wd (testcases included)
¯1↑⎕
Prompt for input string and select last byte.
