Pre-processor query

[Aftabpasha]

I've seen the following code snippet.

#include <stdio.h>
 
#define f(a,b) a##b
#define g(a) #a
#define h(a) g(a)
 
int main()
{
printf("%s\n",h(f(1,2)));
printf("%s\n",g(f(1,2)));
return 0;
}
 

The output is

12
f(1,2)

I GUESS that in the first printf(), firstly, it is replacing h(f(1,2)) by g(f(1,2)), and then immediately in the same pass, it calculated its argument i.e. f(1,2), resulting in "12". Thus getting g("12") in place of h(f(1,2)). This g("12") results in a string "12".

In short, in first printf()
h(f(1,2)) >> g("12") >> "12"

In the second printf(),
g(f(1,2)) >> "f(1,2)"

I'm not sure about this guesswork. Still some questions remain. Please put some light on it. How many passes does a C pre-processor performs? And exactly how it works? Please explain it.

Thank you.

[JosAH]

You guessed right: the # operator stringizes its operand, i.e. it puts double quotes around it. The ## operator 'glues' its operands together.

The preprocessor expands macro definitions in the text. It keeps expanding macros as long as it finds any; to block recursion it won't expand a macro that is being expanded already.

kind regards,

Jos

[Aftabpasha]

Thanks for replying.
Still I've a confusion.

Does a C Preprocessor work in multiple passes or in a single pass?

Thanks.

[JosAH]

@Aftabpasha
It uses a single pass and it rescans the text that results from macro expansion again and again as long as more macros to be expanded are found in that text.

kind regards,

Jos

[Aftabpasha]

Thanks for fruitful reply.