Stay away from zero

Question

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 code-golf, so shortest answer in bytes wins.

Standard loopholes apply.

Answer 1

R, 13 bytes

n=scan();n+!n

Here, scan is used to read the input value n. The negation of n (i.e., !n, 0 or 1) is added to n.

Answer 2

Braingolf, 8 bytes

!?_:1_;

Explanation:

 ?       If last element on stack > 0
!        Prevent if check from consuming last element on stack
  _      Pop last element on stack and print
   :     Else
    1    Add int literal 1 to end of stack
     _   Pop last element on stack and print
      ;  Prevent automatic pop of last element on stack

Answer 3

Ruby, 11 bytes

->n{n|1>>n}

Answer 4

GCC command line, 9 bytes

-Df(x)x?:

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-D in GCC checks if the following content has an =; if not, it add the code

#define <followed_content> 1

which in this case turns into

#define f(x)x?: 1

Answer 5

Sinclair ZX81/Timex TS1000/1500 BASIC ~19 tokenized BASIC bytes

 1 INPUT A
 2 PRINT A+NOT A

This is a more efficient solution than the one below; essentially it takes the NOT value of A (which would be zero in all cases but zero) and adds that to the value of A.

Old answer:

Sinclair ZX81/Timex TS1000/1500 BASIC, ~31 tokenized BASIC bytes

 1 INPUT A
 2 IF NOT A THEN LET A=NOT A
 3 PRINT A

This takes a numeric input from the user and will display 1 if zero is entered by making A NOT A; otherwise it does nothing and displays the numeric value.

Answer 6

Binary Lambda Calculus, 2.125 Bytes

00010110 00110001 0

Translation into lambda calculus:

\n. n (\x. n) (\x. x)

Answer 7

Forte, 25 bytes

9LET0=1:INPUT2:PRINT2:END

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Explanation

Clearly, the best way to approach this challenge is to define 0 to be 1. Problem solved.

Seriously. That's exactly what this program does.

9          Line number (doesn't matter as long as it's not 0 or 2)
 LET 0=1   Redefine 0 to be 1; all future occurrences of 0 will evaluate as 1
 INPUT 2   Redefine 2 to be a number read from input; all future occurrences of 2 will
           evaluate as the input number, though of course if that number was 0 it
           actually evaluates as 1
 PRINT 2   Print that number
 END       End the program

Answer 8

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.

Answer 9

x86-64 Assembly, 10 9 bytes

Following the standard System V AMD64 calling convention, this function accepts a 32-bit unsigned integer parameter via the EDI register, and returns the result via the EAX register:

89 F8  |     mov   eax, edi    ; move parameter from EDI to EAX
85 C0  |     test  eax, eax    ; test input and set flags
75 01  |     jnz   Finished    ; if input is non-zero, then jump to end
FF 40  |     inc   eax         ; otherwise, input is 0, so increment it to 1
       |  Finished:
C3     |     ret               ; return, leaving result in EAX

---

x86-32 Assembly, 8 bytes

We could also write this as 32-bit code, since we're only dealing with 32-bit values.

The nice thing about this is that the INC instruction becomes only 1 byte in length. (In 32-bit mode, inc eax can be encoded simply as 40. In 64-bit mode, 40 is interpreted as a REX prefix, so the leading FF is needed. See Intel's documentation for this instruction.)

The caveat is that most 32-bit calling conventions pass parameters on the stack, rather than in registers, and loading a value from the stack (memory) takes many more bytes. If we can be allowed a __fastcall-style calling convention that passes parameters in registers (supported by virtually all C compilers, so not really cheating, just a bit less standard), then the integer parameter is passed in ECX and we get the following code, for a total of 8 bytes:

89 C8  |     mov  eax, ecx
85 C0  |     test eax, eax
75 01  |     jnz  Finished
40     |     inc  eax
       |  Finished:
C3     |     ret

Answer 10

x86 opcode, 6 5 bytes

       41                         INC     ECX
entry: E3 FD                      JECXZ   SHORT $-1
       91                         XCHG    EAX,ECX
       C3                         RETN

ECX -> EAX

Answer 11

Java 8, 9 bytes

n->n|1>>n

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C# .NET, 9 bytes

n=>n|1>>n

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Different approach (and 1 byte shorter) than the existing Java/C# .NET answers. (Also works with negative inputs.)

Explanation:

i->       // Method with integer as both parameter and return-type
  i|      //  Return the input bit-wise OR-ed with:
    1>>i  //  1 bit-wise right-shifted by the input

1>>i will be 1 when the input is 0, and 0 for every other input.
0|1 for input 0 will therefore result in 1, and n|0 for every other input will therefore result in n.

Answer 12

Zsh, 13 bytes

<<<$[$1?$1:1]

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Bash, 15 bytes

echo $[$1?$1:1]

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Dash (or other POSIX-compliant shell), 17 bytes

echo $(($1?$1:1))

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Answer 13

Labyrinth, 6 bytes

?"1
@!

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?   Take number input and push it
"   No-op, but a 3-way junction:
    If top is 0, go straight
1     Convert the 0 to 1 and bounce off the dead end
"     Enter the junction again, turn right (reflected to south)
!@    Pop and print as number, turn around the corner and halt

    Otherwise, top is positive; turn right
!@    Pop and print as number and halt

Answer 14

Vyxal, 2 bytes

¬+

Straightforward answer:

¬  # nots the input (x == 0 ? 1 : 0)
+  # adds it to the input

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Answer 15

Lolwho.Cares, 62 bytes

*2*+21*1*+21010,**>*2+00210`v
0 ^02210012+*1*<  ^<0210002<<
321

Old version:

*2*+210*1*+210101,**>*2+00210`v
0 ^ 102210012+*1*<  ^<0210002<<
253

The last line isn't code; it's the input number.

Online interpreter: https://valthe.scheffers.net/testing/lolwhocares

Answer 16

RegEx, 3 Bytes

Match pattern (2 Bytes):

^0

Substitution pattern (1 Byte):

1

Takes input as string without any leading zeros and substitutes it for 1 if it is 0.

Try on Regex101

Answer 17

JavaScript (ES6), 8 bytes

Beaten to the shortest solution (as usual!), so here's an alternative.

n=>n?n:1

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f=
n=>n?n:1
console.log(f(0))
console.log(f(1))
console.log(f(8888))

Answer 18

PHP, 11 Bytes

<?=$argn?:1;

Online Version

Answer 19

Jelly, 2 bytes

o1

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An alternative to the other Jelly solution. o provides a default value for a zero/empty argument; in this case, we default it to 1.

Answer 20

Binary-Encoded Golfical, 12 bytes

This binary encoding can be converted back to the standard graphical representation using the encoder provided in the Golfical github repo, or run directly using the interpreter by adding the -x flag.

Hexdump of binary encoding:

00 40 02 15 17 14 00 01 23 1D 17 14

Original image:

Magnified 120x, with color labels:

Answer 21

Whirl, 38 bytes

01100011100011110011111100001000111100

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Explanation:

01100     op.ccw, op.intio    Mem1 = STDIN
011100    math.ccw, math.=    If (Mem1 = 0) Then (Math.Val = 1) Else (Math.Val = 0)
0111100   op.cw, op.one       Op.Val=1
11111100  math.add            Math.Val = Math.Val + Mem1
00        op.one              Op.Val=1 (Cheapest way to loop back to the Math wheel)
100       math.store          Mem1 = Math.Val
0111100   op.ccw, op.intio    STDOUT = Mem1

Answer 22

k, 2 bytes

1|

Finds the maximum of 1 and whatever number is given.

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Answer 23

Perl 5, 10 +1 byte for -p flag = 11 bytes

$_=$_*1||1

Run with -p flag.

Answer 24

GolfScript, 4 3* bytes

~1|

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*thanks to user a'_'

Answer 25

Keg, -hr 4 bytes

:0=+

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Explanation

:     # Take 2 inputs
  0=  # Check whether the input is 0
      # (Yields 1 if the input is 0)
      # (Yields 0 otherwise)
    + # Add the input with this result and print it rawly
      # i.e. 0->1, 5->5

Answer 26

convey, 6 bytes

{+1
}.

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Outputs the max of the input and 1. I think this is the optimal answer, since to use any binary function, there must be a second line with at least two characters.

You can also do the more traditional x+!x for 8 bytes.

Answer 27

Sed, 7 bytes

Input in any number base, without any leading zeros. (It doesn't even need to be a consistent base!)

s/^0/1/

0 is the only number that begins with 0; just change that to 1.

Answer 28

Python 3, 16 bytes

lambda x:x+(x<1)

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Thanks caird coinheringaahing for saving an extra byte!

Answer 29

AWK, 8 bytes

$0||$0=1

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Slightly different approach to this answer, reducing 2 bytes.

$0||$0=1      Basically, an OR logic gate. Right part is not considered if the left one is true.
              $0 is the input, parsed as true if different from zero, printing the input.
              If input is zero, $0 is set to 1 (right part).
              As $0 becomes 1, it is parsed true, and then the input is printed.

Answer 30

C (gcc), 11 bytes

f(n){n?:1;}

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