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You have one string of input bytes, output 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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– MilkyWay90
Mar 19 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.
D1 EE SHR SI, 1 ; SI to DOS PSP at 80H (SI is 0100H at runtime)
AC LODSB ; load command line length in AL
91 XCHG AX, CX ; move length to CX for loop
F3/ AC REPZ LODSB ; load next char into AL, repeat until CX = 0
B4 0E MOV AH, 0EH ; PC BIOS write to screen function
CD 10 INT 10H ; display
C3 RET ; exit to DOS
Output:
Download LAST.COM DOS executable (11 bytes)
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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– 640KB
Mar 19 at 15:04
,[>,]<.
-1 as the EOF. +[>,+]<-. should work
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– Jo King
Mar 17 at 21:21
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
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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– Matsyir
Sep 26 at 14:15
e in the header). For your second question, the header is e=\ , which basically means e=lambda x:x[-1]
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– MilkyWay90
Mar 22 at 20:14
e=\ but Markdown escapes the code character so I have to add a trailing space
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– MilkyWay90
Mar 22 at 20:16
tail -c1
Input is from stdin, output is to stdout.
<?=$argn[-1];
Run with php -nF input is STDIN. Example:
$ echo 29845812674|php -nF lost.php
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.
Input from STDIN, 71 bytes
v->{int i=0;for(;System.in.available()>0;i=System.in.read());return i;}
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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– Kevin Cruijssen
Mar 18 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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– Jo King
Mar 18 at 0:46
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
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.
Pretty much equivalent to @remoel's VBA answer:
=RIGHT(A1)
?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.
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)
😶👉😃😨👿
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.
using Immediate Window and Cell A1 as input
Thanks @tsh
?[RIGHT(A1)] or ?Right([A1],1)
import sys;print(sys.argv[-1][-1])
Usage via running the program as a python script on the command line. Input is provided as the last argument to the program.
@Right(i;1)
Computed field formula taking its input from editable field i
Assumes an input with no empty cells (spaces). Thanks to ASCII-only for saving 30 bytes.
0 * * r 1
1 * * l 2
1 _ _ l halt
2 * _ r 0
Old version in 72 bytes:
0 * * r 0
0 _ * l 1
1 * * l 2
2 * _ l 2
2 _ _ r 3
3 _ _ r 3
3 * * * halt
0 * * r 1/1 * * l 2/1 _ _ l halt/2 * _ r 0?
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– ASCII-only
Mar 18 at 0:22
Requires .NET Core 3.0, which is in beta. This currently crashes the CLR due to a bug, but once the bug is fixed, this will run as expected and fulfill the challenge requirements.
s=>s[^1]
s=>s.ToCharArray()[^1]
using C=System.Console;class A{static void Main(){C.Write(C.ReadLine()[^1]);}}
array[^n] is the same as array[array.Length - n]
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– Arcanox
Mar 18 at 20:24
