What does the # operator do in macros?

Question

#include <stdio.h>

#define foo(x, y) #x #y

int main()
{
    printf("%s\n", foo(k, l));
    return 0;
}

Output:
kl

I know that ## does concatenation. From the output it seems that # also does concatenation. Am I correct?

If I am correct then what is the difference between ## operator and # operator?

@AaronDigulla, yes I know, but a duplicate of a duplicate is not necessarily an exact duplicate for the original question. The one I linked is really almost identical. — Jens Gustedt, Feb 04 '14 at 14:34

score 2 · Answer 1 · answered Feb 04 '14 at 13:56

2

# stringifies the parameter. See http://www.cs.utah.edu/dept/old/texinfo/cpp/cpp.html#SEC15

## concatenates strings. See http://www.cs.utah.edu/dept/old/texinfo/cpp/cpp.html#SEC16

answered Feb 04 '14 at 13:56

keshlam

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sepp2k · Accepted Answer · 2014-02-04T13:55:29.580

1

# turns the argument into a string. So foo(k, l) becomes "k" "l", which is the same as "kl" because in C multiple string literals that are directly next to each other are treated as a single string literals.

If # did concatenation, your printf call would become printf("%s\n", kl); which would produce an error about kl not being defined.

edited Feb 04 '14 at 13:55

answered Feb 04 '14 at 13:54

sepp2k

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score 1 · Answer 3 · edited May 23 '17 at 12:28

1

## concatenates the two arguments, # quotes ("Stringification") them. So the compiler sees:

printf("%s\n", "k" "l");

If you use GCC, use -E to see the output of the preprocessor.

This question contains details about concatenation of string literals: Implementation of string literal concatenation in C and C++

edited May 23 '17 at 12:28

Community

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answered Feb 04 '14 at 13:55

Aaron Digulla

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What does the # operator do in macros?

3 Answers3

Linked