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"

Answer 1

VBA (Excel), 14 12 bytes

using Immediate Window and Cell A1 as input

Thanks @tsh

?[RIGHT(A1)] or ?Right([A1],1)

Answer 2

Python 3, 11 18 34 Bytes

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.

Try it online!

Answer 3

IBM/Lotus Notes Formula, 11 bytes

@Right(i;1)

Computed field formula taking its input from editable field i

Answer 4

Turing Machine Code, 72 42 bytes

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

Try it online.

Answer 5

C# 8.0, 8 bytes

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]

C# 8.0, Runs without crashing at time of writing, 22 bytes

s=>s.ToCharArray()[^1]

C# 8.0, Full Program, 78 bytes

using C=System.Console;class A{static void Main(){C.Write(C.ReadLine()[^1]);}}

Answer 6

Forth (gforth), 17 bytes

: f 1- + 1 type ;

Try it online!

Explanation

Adds string-length - 1 to the string address and then prints a string of length 1 starting at that address.

Code Explanation

: f        \ start a new word definition
  1-       \ subtract 1 from string length
  +        \ add result to string address
  1 type   \ print string of length 1 starting at the new address
;          \ end word definition

Answer 7

Retina, 10 9 bytes

(.|¶)*
$1

Try it online!

Answer 8

LiveScript, 8 bytes

(.[*-1])

Explanation:

(.[*-1])
(.[*-1]) # "BIOP": operator section à la Haskell
 .[   ]  # Index into the implicit argument
   *-1   # In [], "*" refers to the length

Answer 9

F#, 14 8 bytes

Seq.last

-6 bytes thanks to aloisdg.

Strings are treated as sequences in F#, so you can use the Seq.last function to get the last character in it.

Answer 10

C (gcc), 31 bytes

f(int*s){gets(s),printf("%s");}

Try it online!

Answer 11

Caboose, 1514 bytes

Caboose hates me, because it does. There isn't any convenient string-indexing instruction in Caboose!

var a=input();print(a.endsWith(' ')and' 'or a.endsWith('!')and'!'or a.endsWith('"')and'"'or a.endsWith('#')and'#'or a.endsWith('$')and'$'or a.endsWith('%')and'%'or a.endsWith('&')and'&'or a.endsWith("'")and"'"or a.endsWith('(')and'('or a.endsWith(')')and')'or a.endsWith('*')and'*'or a.endsWith('+')and'+'or a.endsWith(',')and','or a.endsWith('-')and'-'or a.endsWith('.')and'.'or a.endsWith('/')and'/'or a.endsWith('0')and'0'or a.endsWith('1')and'1'or a.endsWith('2')and'2'or a.endsWith('3')and'3'or a.endsWith('4')and'4'or a.endsWith('5')and'5'or a.endsWith('6')and'6'or a.endsWith('7')and'7'or a.endsWith('8')and'8'or a.endsWith('9')and'9'or a.endsWith(':')and':'or a.endsWith(';')and';'or a.endsWith('<')and'<'or a.endsWith('=')and'='or a.endsWith('>')and'>'or a.endsWith('?')and'?'or a.endsWith('@')and'@'or a.endsWith('A')and'A'or a.endsWith('B')and'B'or a.endsWith('C')and'C'or a.endsWith('D')and'D'or a.endsWith('E')and'E'or a.endsWith('F')and'F'or a.endsWith('G')and'G'or a.endsWith('H')and'H'or a.endsWith('I')and'I'or a.endsWith('J')and'J'or a.endsWith('K')and'K'or a.endsWith('L')and'L'or a.endsWith('M')and'M'or a.endsWith('N')and'N'or a.endsWith('O')and'O'or a.endsWith('P')and'P'or a.endsWith('Q')and'Q'or a.endsWith('R')and'R'or a.endsWith('S')and'S'or a.endsWith('T')and'T'or a.endsWith('U')and'U'or a.endsWith('V')and'V'or a.endsWith('W')and'W'or a.endsWith('X')and'X'or a.endsWith('Y')and'Y'or a.endsWith('Z')and'Z'or a.endsWith('[')and'['or a.endsWith('\\')and'\\'or a.endsWith(']')and']'or'~');

If I add more constants, then Caboose will say that there are too many constants in the chunk. Fortunately it passes all test cases given. Basically it (tries to) check the last character against all characters in printable ASCII.

TIO

Answer 12

Scratch 3.0, 7 blocks/68 bytes

or, as scratchblocks syntax

when gf clicked
ask()and wait
say(letter(length of(answer))of(answer

Try it on scratch

Did I mention this was done 100% on mobile? Because it was really hard making this, but I think it was worth it.

Answer 13

naz, 38 bytes

2a2x1v1x1f1r3x1v2e2x2v1f0x1x2f2v1o0x1f

Works for any input string terminated with the control character STX (U+0002).

Explanation (with 0x commands removed)

2a2x1v             # Set variable 1 equal to 2
1x1f1r3x1v2e2x2v1f # Function 1
                   # Read a byte of input
                   # Jump to function 2 if it equals variable 1
                   # Otherwise, store it in variable 2,
                   # then jump back to the start of the function
1x2f2v1o           # Function 2
                   # Load variable 2 into the register and output it
1f                 # Call function 1

Answer 14

Ruby -pl, 9 bytes

Similar to the Perl solution, but chop doesn't return the last character in Ruby. (It instead returns the rest of the string without the last character.)

$_=$_[-1]

Try it online!

Answer 15

MAWP, 2 bytes

|;

Outputs top of stack.

Try it!

Answer 16

Flurry, 14 bytes

([]{(){}{}}{})

Run example

$ echo -n "?" | ./flurry -bnb -c "([]{(){}{}}{})"
?
$ echo -n "Hello world" | ./flurry -bnb -c "([]{(){}{}}{})"
d

-bnb flag means "print stack as chars, print nothing for return value, and take stdin as chars".

If the challenge were asking for "the last value from an integer array", I would use return value output and write 2 byte solution {} (pop the stack and return it). But since character I/O is required and the only way to output a char is via the stack, I had to manually empty the stack.

How it works

(
 []     Stack height as Church numeral; Given 2 arguments `f` `x`,
          a Church numeral `n` acts as "apply `f` `n` times to `x`"
 {      Define f, with its argument pushed to the stack:
  ()      K; Given two arguments, ignore the second
  {}      Pop; the argument
  {}      Pop; pop an extra item from the stack (ignored by K)
 }      End definition of f: Return its argument unchanged,
          popping and discarding an item from the stack
 {}     Pop an item (last char)
        At this point, the stack is empty and the value is the last char
)     Push the result of the above

Answer 17

Notepad, 9 5 keystrokes

Assuming the cursor is at the end of the file.

[LeftArrow][Ctrl-Shift-LeftArrow][Delete]

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

Answer 18

vim, 6 bytes

VGJ$d^

Annotated

VGJ     # join all lines
$       # move cursor to (before) last character
d^      # delete to beginning

Try it online!

Answer 19

dotcomma, 48 bytes

[[[],],][.[[[,.][.[,]].,][[,].[,]].]][,[,].[,.]]

Try it online!

Because dotcomma is queue-based, you can't easily access the last character. My solution is to reverse the input and then output the first character of it.

Code:

                # reverse the input
[[[],],]        mark the end with 0 0
[.[             while there are characters in input
  [[,.]         put a character on the recursion stack
    [.[,]]      go to reverse section
  .,]           save the character in the reverse section
  [[,].[,]]     go to input section
.]]

[               # get first character and delete all others
  ,[,]          delete 0 and append the next character (last of input) to the end of the queue
  .[,.]         delete everything until we reach a 0
]
                implicitly output the queue

dotcomma (experimental), 17 bytes

[,][[],][.[,.]].,

Try it online!

I marked this "experimental" because the language dotcomma is still very young and I don't know if this is intentional behaviour. If you pass input as a list of strings it reverses each string, so you only need to extract the first letter.

[,]      put first character on recursion stack
[[],]    put 0 (end of queue marker) on queue
[.[,.]]  delete everything from queue
.,       write back character

Answer 20

Julia, 4 bytes

last

Try it online!

Answer 21

Wolfram Language (Mathematica), 16 bytes

#~StringTake~-1&

Try it online!

Answer 22

C# (.NET Core), 81 bytes, command-line input

class M{static void Main(string[] a){System.Console.Write(a[0][a[0].Length-1]);}}

Try it online!

If you run this from an actual command line, you will need to wrap your string in quotes if it contains spaces.

Answer 23

C# (.NET Core), 115 bytes, console input

using C=System.Console;class M{static void Main(string[] a){int c=0,d;while((d=C.Read())>-1)c=d;C.Write((char)c);}}

Try it online!

This feels kinda janky, but it does work. Interestingly, I can't save any bytes with a for loop as the code stands.

Answer 24

Perl 6, 11 bytes

*.comb[*-1]

Try it online!

Anonymous Whatever lambda that takes a string, splits it into characters, and returns the last one.

Answer 25

Batch, 24 bytes

@set s=%1
@echo %s:~-1%

Takes input as a command-line argument. Note that arguments can't include special characters or spaces, but you can fake arguments with spaces in this case by preceding it with a ", which results in a single argument that begins with ", however there is no easy solution for arguments that include special characters. Batch can't easily read "all of stdin". To read up to but not including the first newline itself would however be a byte shorter:

@set/ps=
@echo %s:~-1%

Edit: A version that handles arbitrary characters in a (quoted) argument for 92 bytes:

@set s="%~1"
@set "s=%s:~-2,1%
@if "%s%"=="" (echo ^")else for %%s in ("%s%")do @echo %%~s

Explanation: The first line makes a copy of the argument in a variable and ensures that it is quoted. The second argument then takes the second last character (because the quote is now the last character). However, if that was also a quote then this results in an empty variable, so we need to special-case that and output a (quoted) quote. Otherwise, we still need to quote the character in case it is a special character as echoing "%s%" will echo the quotes and echoing %s% will actually interpret special characters, so the variable needs to be quoted to allow it to be parsed but then immediately unquoted so it can be printed. This is achieved using the for command. 86 bytes to read up to but not including the first newline from stdin while supporting special characters:

@set/ps=
@set "s=%s:~-1%
@if "%s%"=="" (echo ^")else for %%s in ("%s%")do @echo %%~s

Answer 26

Scratch 3.0, scratchblocks3 syntax

As a function, 61 bytes

define l
ask[]and wait
say(letter(length of(answer))of(answer

As a full program, 68 bytes

when gf clicked
ask[]and wait
say(letter(length of(answer))of(answer

Try both online

Answer 27

Befunge-98, 5 bytes

~2j@,

Try it online!

Explanation:

~           Take input
 2j         Skip next two instructions
~           Repeat until EOF, where it reflects
   @,       Print the last character and exit

Answer 28

C# (Visual C# Interactive Compiler), 11 bytes

x=>x.Last()

Try it online!

Answer 29

Japt, 1 byte

Ì

Try it online!

-1 byte thanks to Quintec!

Answer 30

Whitespace, 54 bytes

[N
S S N
_Create_Label_LOOP][S S S N
_Push_0][S N
S _Duplicate_0][T   N
T   S _Read_STDIN_as_character][T   T   T   _Retrieve_input][S N
S _Duplicate][S S S T   S T S N
_Push_10][T S S T   _Subtract][N
T   S S N
_If_0_Jump_to_Label_PRINT][N
S N
N
_Jump_to_Label_LOOP][N
S S S N
_Create_Label_PRINT][S N
N
_Discard_top][T N
S S _Print_as_character]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Since Whitespace can only take input as integer or character, we must add a trailing character to indicate we're done with the input-string after reading it character by character, for which I've used a newline.

Try it online (with raw spaces, tabs and new-lines only).

Example run: input = A2#

Command    Explanation                   Stack                Heap    STDIN STDOUT STDERR

NSSN       Create Label_LOOP             []
 SSSN      Push 0                        [0]
 SNS       Duplicate top (0)             [0,0]
 TNTS      Read STDIN as character       [0]                  {0:65}  A
 TTT       Retrieve at heap address (0)  [65]                 {0:65}
 SNS       Duplicate top (65)            [65,65]              {0:65}
 SSSTSTSN  Push 10                       [65,65,10]           {0:65}
 TSST      Subtract top two (65-10)      [65,55]              {0:65}
 NTSSN     If 0: Jump to Label_PRINT     [65]                 {0:65}
 NSNN      Jump to Label_LOOP            [65]                 {0:65}
 
 SSSN      Push 0                        [65,0]
 SNS       Duplicate top (0)             [65,0,0]
 TNTS      Read STDIN as character       [65,0]               {0:50}  2
 TTT       Retrieve at heap address (0)  [65,50]              {0:50}
 SNS       Duplicate top (50)            [65,50,50]           {0:50}
 SSSTSTSN  Push 10                       [65,50,50,10]        {0:50}
 TSST      Subtract top two (50-10)      [65,50,40]           {0:50}
 NTSSN     If 0: Jump to Label_PRINT     [65,50]              {0:50}
 NSNN      Jump to Label_LOOP            [65,50]              {0:50}

 SSSN      Push 0                        [65,50,0]
 SNS       Duplicate top (0)             [65,50,0,0]
 TNTS      Read STDIN as character       [65,50,0]            {0:35}  #
 TTT       Retrieve at heap address (0)  [65,50,35]           {0:35}
 SNS       Duplicate top (35)            [65,50,35,35]        {0:35}
 SSSTSTSN  Push 10                       [65,50,35,35,10]     {0:35}
 TSST      Subtract top two (35-10)      [65,50,35,25]        {0:35}
 NTSSN     If 0: Jump to Label_PRINT     [65,50,35]           {0:35}
 NSNN      Jump to Label_LOOP            [65,50,35]           {0:35}

 SSSN      Push 0                        [65,50,35,0]
 SNS       Duplicate top (0)             [65,50,35,0,0]
 TNTS      Read STDIN as character       [65,50,35,0]         {0:10}  \n
 TTT       Retrieve at heap address (0)  [65,50,35,10]        {0:10}
 SNS       Duplicate top (10)            [65,50,35,10,10]     {0:10}
 SSSTSTSN  Push 10                       [65,50,35,10,10,10]  {0:10}
 TSST      Subtract top two (10-10)      [65,50,35,10,0]      {0:10}
 NTSSN     If 0: Jump to Label_PRINT     [65,50,35,10]        {0:10}

NSSSN      Create Label_PRINT            [65,50,35,10]        {0:10}
 SNN       Discard top                   [65,50,35]           {0:10}
 TNSS      Print as character to STDOUT  [65,50]              {0:10}        #
                                                              {0:10}               error

Stops with the error: Exit not defined.