CODEWITHSUNDEEP
cmacrosconcatenationpreprocessor-directive
123
123

Reputation: 31

Issue with concatenation using `##` operator

I am concatenating two numbers using the ## operator. The code gives error when I use variables but works fine when I give direct input.

Gives an error.

[Error] 'xy' undeclared (first use in this function) 

#include<stdio.h>
#define join(a, b)  a##b ; 

int main()
{
int x = 10;
int y = 5;
int res ;
res = join(x, y);
printf("%d",res) ;
return 0 ;
}

Works Fine:

#include<stdio.h>
#define join(a, b)  a##b ; 

int main()
{
int res ;
res = join(99, 10);
printf("%d",res) ;
return 0 ;
}

Upvotes: 2

Views: 99

Answers (2)

Sadique
Sadique

Reputation: 22821

The ## preprocessing operator performs token pasting. And tokens are checked at Compile time.

What do you think this res = join(x, y); is supposed to mean?

res = xy;

The C compiler should give a compilation error.

This res = join(99, 10); means

res = 9910;

Which is valid in C syntax. Also remember that Macros are not type-safe so use casts like this:

res = (int) join(99, 10);

Your code would be valid if you had xy declared before something like this:

#include<stdio.h>
#define join(a, b)  a##b ; 

int main()
{
    int xy = 100;
    int res ;
    res = (int) join(x, y);
    printf("%d",res) ;
    return 0 ;
}

Upvotes: 2

Sourav Ghosh
Sourav Ghosh

Reputation: 134326

Well, preprocessor MACROS are textual substitution, so in case when you're using variables, after preprocessing your code looks like

res = xy;

now this xy is an undefined identifier in your code.

OTOH, when you're passing integer constants, you code looks like

res = 9910;

which is perfectly valid.

To elaborate on the working of the ## operator, quoting C11, chapter §6.10.3.3 (emphasis mine)

in the replacement list of a function-like macro, a parameter is immediately preceded or followed by a ## preprocessing token, the parameter is replaced by the corresponding argument’s preprocessing token sequence; [...]

Upvotes: 4

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