40
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Task

Given a non-negative integer n, output 1 if n is 0, and output the value of n otherwise.

Input

A non-negative integer.

  • If you would like to accept the string as input, the string would match the following regex: /^(0|[1-9][0-9]*)$/, i.e. it must not have any leading zeroes, except when it is 0.
  • If you accept a real integer as input, you may assume that the integer is within the handling capability of the language.

Output

A positive integer, specified above. Leading zeroes are not allowed. Your output should match the regex /^[1-9][0-9]*$/.

Testcases

input output
    0      1
    1      1
    2      2
    3      3
    4      4
    5      5
    6      6
    7      7

Scoring

This is , so shortest answer in bytes wins.

Standard loopholes apply.

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  • 1
    \$\begingroup\$ You should probably put a link to the TNB CMC, since that's where this challenge came from. \$\endgroup\$ – mbomb007 May 3 '17 at 20:53
  • \$\begingroup\$ Does the answer need to be a full function, or can it be the body? \$\endgroup\$ – Caleb Kleveter May 3 '17 at 20:58
  • 1
    \$\begingroup\$ @CalebKleveter The default rule in PPCG is that the answer is either a function or a full program, but not snippets. \$\endgroup\$ – Leaky Nun May 4 '17 at 1:56
  • \$\begingroup\$ Can we print the output with a leading zero? \$\endgroup\$ – MD XF Dec 26 '17 at 21:54
  • \$\begingroup\$ @MDXF yes, you can. \$\endgroup\$ – Leaky Nun Dec 26 '17 at 22:30

118 Answers 118

18
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C (gcc), 14 13 bytes

f(n){n=n?:1;}

Thanks to @betseg for reminding me of the n?:1 trick in the comments of the other C answer!

Try it online!

C, 17 bytes

f(n){return!n+n;}

Try it online!

C, 16 bytes

#define f(n)!n+n

Try it online!

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  • \$\begingroup\$ tio.run/nexus/… hmmm... \$\endgroup\$ – betseg May 2 '17 at 16:54
  • 1
    \$\begingroup\$ @betseg That's because it's a macro. The compiler sees it as 3*!n+n which equals 3*0+5. \$\endgroup\$ – Steadybox May 2 '17 at 16:56
  • 1
    \$\begingroup\$ I know, but I think you should be able to apply arithmetic operators to the "return" values directly, that's why it's common practice to put parentheses around macros. I just don't think that the macro is valid. \$\endgroup\$ – betseg May 2 '17 at 16:58
  • 4
    \$\begingroup\$ @betseg I don't think that's a requirement in code golf. I've never seen a code golf answer with C macros do that. \$\endgroup\$ – Steadybox May 2 '17 at 17:00
  • 1
    \$\begingroup\$ @hucancode See the TIO links. You need to add a main from which the function/macro f is called. A solution doesn't need to be a full program by default. The gcc-specific version may or may not compile on another compiler, and it may or may not run correctly when compiled on another compiler. \$\endgroup\$ – Steadybox May 3 '17 at 10:13
17
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Japt, 2 bytes

ª1

Try it online!

Explanation

ª is a shortcut for JS's || operator. Japt has implicit input, so this program calculates input||1, and the result is implicitly sent to STDOUT.

w1 would work as well, taking the maximum of the input and 1.

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16
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Alice, 7 bytes

1/s
o@i

Try it online!

Explanation

1   Push 1. Irrelevant.
/   Reflect to SE. Switch to Ordinal.
i   Read all input as a string.
    Reflect off bottom right corner. Move back NW.
/   Reflect to W. Switch to Cardinal.
1   Push 1.
    IP wraps around to last column.
s   Sort swap: implicitly convert the input to an integer. Then, if the top stack 
    element is less than the one below, the two are swapped. It basically computes
    min and max of two values at the same time, with max on top.
/   Reflect to NW. Switch to Ordinal.
    Immediately reflect off the top boundary. Move SW.
o   Implicitly convert the result to a string and print it.
    Reflect off bottom left corner. Move back NE.
/   Reflect to S. Switch to Cardinal.
@   Terminate the program.
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15
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JavaScript (ES6), 7 bytes

n=>n||1
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  • 5
    \$\begingroup\$ Alternative: n=>n+!n (At least I think) \$\endgroup\$ – lol May 2 '17 at 16:45
  • \$\begingroup\$ @SIGSEGV Yes, that would work indeed. (That could also be n|!n, although this one is limited to a 31-bit quantity.) \$\endgroup\$ – Arnauld May 2 '17 at 16:48
  • \$\begingroup\$ this can be simplified to n||1. Only thing that evaluates to false is 0. \$\endgroup\$ – ansiart May 3 '17 at 22:42
  • 1
    \$\begingroup\$ @ansiart If your point is that n=>n||1 could be simplified to n||1, then no. Acceptable answers are either full programs or functions. n=>do_something_with(n) is an arrow function in ES6 syntax. \$\endgroup\$ – Arnauld May 4 '17 at 1:40
  • 1
    \$\begingroup\$ @StanStrum We're required to return the original value of n if it's not zero. A bitwise OR would modify n whenever the least significant bit is not set (e.g. (4|1) === 5). \$\endgroup\$ – Arnauld Dec 17 '17 at 23:26
14
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Pyth, 2 bytes

+!

Try it online

Explanation

+!
 !Q    1 if (implicit) input is 0, 0 otherwise.
+  Q   Add the (implicit) input.
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12
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Retina, 4 bytes

^0
1

Try it online!

If the input starts with a zero, replace that with a 1. (Works because the input is guaranteed to have no leading zeros for non-zero values.)

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12
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V, 4 bytes

é0À

Try it online!

Abuses an non-preferred but expected behavior, so I can't really call it a bug. Explanation:

In Vim, commands accept a count. For example, <C-a> will increment a number, but 7<C-a> will increment a number by 7. However, you can't use 0 as a count, because

  • 0 is already a command (go the first column), and

  • In the context of a text editor, it rarely makes sense to request that a command be run 0 times.

This is fine for a text editor, but usually obnoxious for a golfing language, so V overwrites some commands so that 0 is a valid count. For example, é, ñ, Ä, and some others. However, since <C-a> is a builtin vim command, it is not overwritten, so running this with a positive input gives:

N       " N times:
 <C-a>  "   Increment

But running with 0 as input gives:

0       " Go to column one
 <C-a>  " Increment

Full explanation:

é0          " Insert a 0
  À         " Arg1 or 1 times:
   <C-a>    " Increment
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  • 1
    \$\begingroup\$ The one time that 0 not being a count is useful. I didn't even consider it at first because I've avoided it so many times \$\endgroup\$ – nmjcman101 May 2 '17 at 16:58
12
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J, 2 bytes

^*

Try it online!

^ [argument] raised to the power of

* the sign of the argument (0 if 0 else 1)

Because 1=0^0 in J.

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12
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Haskell, 5 bytes

max 1

Usage example: (max 1) 0 -> 1.

Nothing much to explain.

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10
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dc, 7

?d0r^+p

Relies on the fact that dc evaluates 00 to 1, but 0n to 0 for all other n.

Try it online.

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9
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Cubix, 6 bytes

OI!1L@

Somehow managed to fit it on a unit cube... Test it online!

Explanation

Before being run, the code is arranged as a cube net:

  O
I ! 1 L
  @

The IP (instruction pointer) is then placed on the far-left face (I), facing to the right. The instructions run from there are:

I  Input a number from STDIN and push it to the stack.
!  If the top number is non-zero, skip the next instruction.
1  Push a 1 (only if the input was zero).
L  Turn left. The IP is now on the top face facing the !.
O  Output the top item as a number.

The IP then hits ! again, skipping the @ on the bottom face. This is not helpful, as we need to hit the @ to end the program. The IP hits the L again and goes through the middle line in reverse (L1!I) before ending up on the L one more time, which finally turns the IP onto @.

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9
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brainfuck, 8 bytes

+>,[>]<.

Try it online!

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9
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R, 13 bytes

max(1,scan())

reads n from stdin. With pmax, it can read in a list and return the appropriate value for each element in the list for +1 byte.

try it online!

I should note that there is another fine R solution in 13 bytes by Sven Hohenstein which allows for yet another 13 byte solution of

(n=scan())+!n

which makes me wonder if that's the lower limit for R.

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  • \$\begingroup\$ Another 13 bytes solution using pryr: pryr::f(n+!n). Can't find anything smaller... \$\endgroup\$ – JayCe May 17 '18 at 2:51
7
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V, 5 bytes

ç^0/<C-a>

Where <C-a> is 0x01.

Try it online!

Explanation

ç                   " On every line
 ^0/                " that begins with a zero do:
    <C-a>           " Increment the number on that line
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6
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Jelly, 2 bytes

Try it online!

Pretty much exactly my Pyth answer, but it's my first Jelly program.

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6
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Oasis, 2 bytes

Uses the following formula: a(0) = 1, a(n) = n

n1

Try it online!

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  • 1
    \$\begingroup\$ Nice. My approach was >V. \$\endgroup\$ – Leaky Nun May 2 '17 at 16:43
  • \$\begingroup\$ @LeakyNun Oh nice! \$\endgroup\$ – Adnan May 2 '17 at 16:44
6
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R 20 16 bytes

pryr::f(n+(n<1))
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  • \$\begingroup\$ Welcome to PPCG! \$\endgroup\$ – Martin Ender Jun 14 '17 at 18:48
  • \$\begingroup\$ Thanks @MartinEnder. I am already learn some tricks of the trade. \$\endgroup\$ – Shayne03 Jun 15 '17 at 14:50
5
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Brachylog, 3 bytes

∅1|

Try it online!

Explanation

If we add the implicit ? (Input) and . (Output), we have:

?∅          Input is empty (that is, [] or "" or 0 or 0.0)
  1.        Output = 1
    |       Else
     ?.     Input = Output
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5
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MarioLANG, 12 bytes

;
=[
:<+
 =:

Try it online!

How it works

Mario starts in the top left, initially walking right. He reads an int from input (;) and stores it in the current memory cell. Then he falls off the ground (=), hitting [, which makes him ignore the next command if the current cell is 0.

If the cell is not 0, he'll start walking left (<), output the current cell as an int (:), and fall to his death (end of program).

If the cell is 0, he ignores the command to turn left, and keeps walking right. He increments the current cell (+), outputs it, and falls to his death.

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5
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Brain-Flak, 22, 10 bytes

({{}}[]{})

Try it online!

Explanation:

If the input is non-zero, then {{}} will pop everything off the stack and evaluate to the input. If it is zero, nothing will be popped, and it will evaluate to zero. So running ({{}}) gives

Non-zero:

n

Zero:

0
0

At this point, we'll add the height of the stack (0 for non-zero, 1 for zero) and pop one more value off the stack. (since the stack is padded with an infinite number of 0's, this will pop either the top 0 or an extra 0)

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4
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TI-BASIC, 7 bytes

:Prompt X
:X+not(X

Alternatively,

TI-BASIC, 7 bytes

:Prompt X
:max(X,1
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4
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Hexagony, 7 6 bytes

)?<@.!

Expanded:

 ) ?
< @ .
 ! .

Try it online!

Saved 1 byte thanks to Martin!

If the number is nonzero print it, otherwise add one to it and print that instead.

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4
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APL (Dyalog), 3 bytes

1∘⌈

Try it online!

This takes the ceil of the argument and 1.

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4
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Python, 15 bytes

lambda n:n or 1
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  • \$\begingroup\$ Why not just n or 1, 6 bytes? \$\endgroup\$ – DReispt May 4 '17 at 7:57
  • 2
    \$\begingroup\$ Because that's just a snippet, while we usually answer with complete programs or functions. I'm not sure if this is stated explicitly in some rules somewhere, but at least that's the de facto standard. \$\endgroup\$ – daniero May 4 '17 at 10:47
  • \$\begingroup\$ Quoting trichoplax: The rules are not terribly clear. I think we have a consensus on meta that REPLs count, but as a separate language, which would allow snippets in many cases, but snippets are not permitted according to this meta post -> codegolf.meta.stackexchange.com/questions/2419/… \$\endgroup\$ – daniero May 4 '17 at 11:07
  • \$\begingroup\$ @trichoplax 1or n would always return 1, wouldn't it? \$\endgroup\$ – daniero May 4 '17 at 11:08
  • \$\begingroup\$ Oh of course - sorry ignore me... \$\endgroup\$ – trichoplax May 4 '17 at 11:22
4
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dc, 11 bytes

[1]sf?d0=fp

[1]sf stores a macro in register f which pushes 1 to the top of the stack, ? reads input, d0=f runs macro f if input was 0, p prints the top of the stack.

Test:

$ dc -e "[1]sf?d0=fp" <<< 0
1
$ dc -e "[1]sf?d0=fp" <<< 1
1
$ dc -e "[1]sf?d0=fp" <<< 42
42
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4
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Excel, 10 Bytes

=A1+(A1=0)

This saves 4 Bytes over the obvious 'IF' statement solution, =IF(A1=0,1,A1).

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  • 3
    \$\begingroup\$ And 1 byte less than the less obvious =A1+NOT(A1) \$\endgroup\$ – Engineer Toast May 2 '17 at 20:05
4
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Java 8, 10 bytes

i->i<1?1:i
  • Thanks to @LeakyNun for saving -1 byte
    • Didn't notice it's a non-negative integer
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  • 3
    \$\begingroup\$ i==0 can be replaced by i<1 \$\endgroup\$ – Leaky Nun May 2 '17 at 16:58
3
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Mathematica, 9 8 bytes

Per Martin Ender:

#~Max~1&

First idea:

#/. 0->1&

Pure function with replaces 0 with 1. The space is necessary or it thinks we are dividing by .0.

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3
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Perl 5, 6 + 2 bytes for the -l and -p flags

$_||=1

Takes input on separate lines from stdin. Runs with the flags -lp.

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3
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Taxi, 517 bytes

Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:n 1 l 1 l 2 r.Pickup a passenger going to Addition Alley.Pickup a passenger going to Knots Landing.Go to Knots Landing:n 2 r 2 r 1 l.Pickup a passenger going to Addition Alley.Go to Addition Alley:w 1 r 1 l 2 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:n 1 r 1 r.Pickup a passenger going to Post Office.Go to Post Office:n 1 l 1 r.

Formatted for humans:

Go to Post Office:w 1 l 1 r 1 l.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:s 1 l 1 r.
Pickup a passenger going to Cyclone.
Go to Cyclone:n 1 l 1 l 2 r.
Pickup a passenger going to Addition Alley.
Pickup a passenger going to Knots Landing.
Go to Knots Landing:n 2 r 2 r 1 l.
Pickup a passenger going to Addition Alley.
Go to Addition Alley:w 1 r 1 l 2 l.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery:n 1 r 1 r.
Pickup a passenger going to Post Office.
Go to Post Office:n 1 l 1 r.

The key location is Knots Landing which performs the NOT operation so 0 returns 1 and all other numbers return 0. Now, you can go to Addition Alley to add the result from Knots Landing to the original input.

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