C Pre-Processor Macro code with () and {}

Question

#include <stdio.h>
#define a (1,2,3)
#define b {1,2,3}

int main()
{
    unsigned int c = a;
    unsigned int d = b;
    printf("%d\n",c);
    printf("%d\n",d);
    return 0;
}

Above C code will print output as 3 and 1.

But how are #define a (1,2,3) and #define b {1,2,3} taking a=3 and b=1 without build warning, and also how () and {} are giving different values?

Answer 1

Remember, pre-processor just replaces macros. So in your case you code will be converted to this:

#include <stdio.h>

int main()
{
    unsigned int c = (1,2,3);
    unsigned int d = {1,2,3};
    printf("%d\n",c);
    printf("%d\n",d);
    return 0;
}

In first case, you get result from , operator, so c will be equal to 3. But in 2nd case you get first member of initializer list for d, so you will get 1 as result.

2nd lines creates error if you compile code as c++. But it seems that you can compile this code in c.

Answer 2

In addition to other answers,

unsigned int d = {1,2,3};

(after macro substitution)

is not valid in C. It violates 6.7.9 Initialization:

No initializer shall attempt to provide a value for an object not contained within the entity being initialized.

With stricter compilation options (gcc -std=c17 -Wall -Wextra -pedantic test.c), gcc produces:

warning: excess elements in scalar initializer
     unsigned int d = {1,2,3};
                         ^

However, note that

unsigned int d = {1};

is valid because initializing scalar with braces is allowed. Just the extra initializer values that's the problem with the former snippet.