Print last inputted byte

Question

Challenge

You have one string of input bytes, output only the last byte in it.

Rules

Your submission may be a program or function outputting the last byte in the input which

• is either a string, stdin or command-line arguments, and
• is non-empty.

I was trying to solve this with brainfuck, however all languages are allowed to participate. This is code-golf.

Examples

"?" -> "?"
"29845812674" -> "4"

• ბიმო: Welcome, I changed your question to fit our format more properly (note this is what the sandbox is for, usually). However in its current state the challenge is very easy (also in bf), so not sure about that.
• MilkyWay90: I vote against closing; it may be trivial, but that doesn't make it offtopic
• Laikoni: @ბიმო We have a consensus not to edit off-topic questions to make them on-topic which I think would have applied here.
• kepe: What kind of string? Is it guaranteed to be ASCII only? Or should we handle UTF-8 (and how?) for example?
• jean: @FireCubez Yes, ASCII only

Answer 1

Attache, 4 bytes

Last

Try it online! (If the input could be a list of characters, &/S could work.)

Alternatives

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~>

• ASCII-only: :| wtf so many alternatives
• Conor O'Brien: @ASCII-only Least I could do on a simple challenge like this :p
• MilkyWay90: Print Last inputted byte. The programs contents fit with the challenge

Answer 2

x86-16 machine code, 2 bytes

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

x86-16 machine code, IBM PC DOS, (削除) 12 (削除ここまで) (削除) 11 (削除ここまで) 10 bytes

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:

enter image description here

• Cody Gray: Although this certainly gets style points, from a strict golfing point of view, it's worth noting that you are allowed to write functions that return the result in a register. So, this can get a lot shorter. You can trivially eliminate 4 bytes, and a rewrite could shrink it down even further. By the way, is that screenshot from an emulator? Which one?
• 640KB: @CodyGray, oh absolutely, the code to take input from command line and output to console is nearly all of it. Yeah, I could say "input string in 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.

Answer 3

Brainf***, 7 bytes

,[>,]<.

• jean: ideone.com/XoJLD0 It still doesn't work ;(
• SuperJedi224: @jean Try it here. (Honestly, I didn't even know ideone did BF).
• jean: Yes, in your link everything works fine. But online judge for this problem uses ideone where it doesn't work ;(
• Jo King: @jean ideone seems to use -1 as the EOF. +[>,+]<-. should work
• jean: @Jo King Sorry, but the input looks like: 29845812674[enter][EOF] how can I print the last digit?

Answer 4

INTERCAL, (削除) 270 (削除ここまで) (削除) 258 (削除ここまで) (削除) 247 (削除ここまで) (削除) 246 (削除ここまで) 245 bytes

DO,1<-#1PLEASECOMEFROM#2DOWRITEIN,1DO.1<-,1SUB#1DO.5<-#1$.1~#256DO.2<-.3DO(1)NEXTPLEASE.2<-!3~#15'$.3~#240PLEASE.2<-!2~#15'$.2~#240PLEASE.2<-!2~#15'$.2~#240DO(1010)NEXTDO,1SUB#1<-.3DOREADOUT,1(1)DO(1002)NEXTDO(1009)NEXTDO.3<-.3~#255(2)DOFORGET#1

Try it online!

-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.

-1 trying something else for bit reversal--definitely expected more savings than that! A more principled approach might be able to do it in two steps unless there's some obvious invariant I'm missing, but I'm too tired to take a stab at that just yet.

-1 thanks again to 鳴神裁四点一号, using a "computed" COME FROM where the computed label just so happens to be a constant. Looks like I wasn't reading enough tips!

Writing this was... interesting. I was thinking I might want to use INTERCAL to INTERCALate, but I'm a bit less sure now.

Ungolfed and commented:

 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. Rather than select the bits out individually and mingle them all
 in one expression, I mingle the back and front halves of the byte until it adds up to a reversal.
 DO .2 <- !3~#15' $ !3~#240'
 DO .2 <- !2~#15' $ !2~#240'
 DO .2 <- !2~#15' $ !2~#240'
 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

• IY5dVSjABEeV: DOGIVEUP can be removed to golf off 8 bytes because program halts by stack overflow and output is unchanged
• IY5dVSjABEeV: 245 bytes

Answer 5

MATL, 2 bytes

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

• Sanchises: I'd have gone for J)...

Answer 6

Python 3, 14 bytes

lambda x:x[-1]

Try it online!

• Nathan Dimmer: I feel dumb, how are you calling this without declaring it to a variable?
• Nathan Dimmer: I looked at your TIO, but it doesn’t make much sense... What are you doing in your header?
• MilkyWay90: @Bobawob For your first question, anonymous lambdas are allowed for answers (I call it using by assigning the lambda to the variable e in the header). For your second question, the header is e=\ , which basically means e=lambda x:x[-1]
• MilkyWay90: Note that in my above comment, there is not supposed to be a trailing space in e=\ but Markdown escapes the code character so I have to add a trailing space
• Nathan Dimmer: That’s really cool! Thank you!

Answer 7

PHP, 13 bytes

<?=$argn[-1];

Try it online!

Run with php -nF input is STDIN. Example:

$ echo 29845812674|php -nF lost.php

Answer 8

Javascript, 14 bytes

a=>a.slice(-1)

• Matsyir: I tried my best to find a shorter way but to my surprise I don't think this can be improved, unless there is a funky way to reverse a string in under 8 bytes, before calling [0]. Or maybe if there is a short way to get array length. Different approach: a=>[...a].pop() (15bytes)
• l4m2: @Matsyir a=>/.$/.exec(a) also 15

Answer 9

Mornington Crescent, 389 Bytes

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

Try it online!

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.)

Answer 10

Arn, 2 bytes

:}

Pretty simple, the suffix :} gets the last element of an array (implicit casting)

• user96495: Another alternative: _\
• user: +1 for a happy symbol

Answer 11

Java 11+

Input from STDIN, (削除) 71 (削除ここまで) 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)

• Kevin Cruijssen: You could add the code for program argument to have all three options. :) Too bad OP specifically asks for Strings, otherwise s->s[s.length-1] would have been enough with a char[] parameter-type.
• Olivier Grégoire: 61 bytes.
• Benjamin Urquhart: @OlivierGrégoire just now seeing your answer 2 years later. Thanks.

Answer 12

Bash + coreutils, 8 bytes

tail -c1

Input is from stdin, output is to stdout.

Answer 13

Turing Machine But Way Worse, 391 bytes

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

Try it online!

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

Answer 14

TI-BASIC (TI-84), 10 bytes

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.

Answer 15

Excel, 10 bytes

Pretty much equivalent to @remoel's VBA answer:

=RIGHT(A1)

Answer 16

Haskell, (削除) 9 (削除ここまで) 4 bytes

last

Try it online!

• dfeuer: Why bother with pure? Isn't last enough?
• flawr: I assumed that the output must be a string too, but you're right, OP is only talking about "bytes".

Answer 17

Seed, 11 bytes

5 370394306

Try it online!

The resulting Befunge-98 program ~2j@, was (削除) stolen (削除ここまで) borrowed from Jo King here, so credit to them for that.

Answer 18

><>, 2 bytes

Using command line args

o;

Try it online!

><>, 11 bytes

Using stdin

\~o;
/?(0:i

Try it online!

Answer 19

05AB1E, 1 byte

¤

Try it online!

θ or ` would also work.

Answer 20

Cubix, 6 bytes

pA/@po

Try it online!

 p
A / @ p
 o

Watch it run

• A Takes all the input
• / Redirect around the cube
• pp bring bottom of the stack to the top twice
• o/@ output as character, redirect and halt

Answer 21

Befunge-93, 12 (削除) 15 (削除ここまで) bytes

:1+_p1-,@>~#

Try it online!

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.

• Jo King: Actually, $$ instead of p1 should work without the warning for the same amount of bytes

Answer 22

Jelly, 1 byte

Ṫ

Try it online!

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.

• MilkyWay90: -1 Fails for any number (tio.run/##y0rNyan8///hzlX///83MTY1NjE1MQYA)
• ბიმო: @MilkyWay90: Doesn't need to be a full program, probably this will work as a Jelly function taking a string. But then again I don't know Jelly, so I might be wrong.
• MilkyWay90: Okay, I'll try seeing whether or not it will work as a link
• MilkyWay90: @ბიმო Seems to work (OP can you edit the answer so I can undo my downvote?)
• MilkyWay90: You can just define the T as a link and make it input a string, removing the need for "" in your input.

Answer 23

APL+WIN, 4 bytes

 ̄1↑⎕

Prompt for input string and select last byte.

• rak1507: 3 byte solution ⊃⌽⍞

Answer 24

Emotion, 5 bytes

😶👉😃😨👿

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.

Try it online!

• ASCII-only: in your readme, 1. i think "to emoji" should be "from emoji" 2. if it should be "from emoji" then emotinomicon and emojicode are a thing already
• Quantum64: Actually what you're looking at is output from a compiler, which can also be accessed via the online interface.
• Kyle Delaney: Isn't each emoji more than one byte? I'd think they'd be two bytes at least.
• JPeroutek: Emotion uses a custom code page. See quantum64.github.io/EmotionBuilds/1.1.0//…

Answer 25

Cascade, 9 bytes

?a|,
;.]^

Pretty happy with this, as it is only 3 bytes longer than my cat program

Expanded

 ?
 ^;.
 | |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.

Try it online!

Answer 26

Rust, (削除) 25 (削除ここまで) (削除) 16 (削除ここまで) 10 bytes

|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

• madlaina: The .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
• Aiden4: @madlaina thanks, I was testing with let bindings, and I had to be explicit there.

Answer 27

Notepad, (削除) 9 (削除ここまで) (削除) 5 (削除ここまで) 4 keystrokes

Assuming the cursor is at the end of the file.

[LeftArrow][Shift-Home][Backspace]

I know it is not a language, but I'm posting just for fun.

-1 by Steve Bennett

• user96495: The input is one line, according to the challenge specs. -6 keystrokes: [LeftArrow][Ctrl-Backspace]
• TwilightSparkle: @hi. Doesn't work on my notepad. Ctrl-Backspace just adds an unprintable character, which I suspect is \x7f.
• Jo King: @hi. I don't see anything in the specs about it only being one line of input
• Steve Bennett: I guess it depends on the platform, but wouldn't you do [LeftArrow] [Shift-Home] [Backspace]? 4 keystrokes.
• TwilightSparkle: @SteveBennett Indeed, apparently I've forgotten the Home key at that time.

Answer 28

PowerShell, 11 bytes

"$args"[-1]

Try it online!

Answer 29

SmileBASIC, 16 bytes

INPUT S$?POP(S$)

Answer 30

Twig, 37 bytes

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)

• Ven: Ok, that's the first time I see a Twig golf.
• Ismael Miguel: @Ven I've been doing it for quite a while. Here's an example: codegolf.stackexchange.com/a/174040 and here's another one: codegolf.stackexchange.com/a/166800 (I have a few more, but well)