Behavior of "comma" operator in sizeof() operator In C [duplicate]

Question

I wrote the code about sizeof operator. If I write something like:

#include <stdio.h>

int main() {
    char a[20];
    printf("%zu\n", sizeof(a));
    return 0;
}

Output:

20 // Ok, it's fine

But, If I use the comma operator like this:

#include <stdio.h>

int main() {
    char a[20];
    char b;
    printf("%zu\n", sizeof(b, a));
    return 0;
}

Output:

8 // Why the output 8?

So, I have a questions:

• Why does compiler give an output 8 in second example?
• What is the behavior of comma operator into sizeof() operator?

Answer 1

The comma operator has no special meaning to sizeof.

sizeof(b, a) examines the complete expression (b, a), works out the resultant type, and computes the size of that type without actually evaluating (b , a). As noted by chqrlie in comments, the () are part of the expression for which the size (of the result) is evaluated.

In this case, b is a char and a is an array. If the expression b, a was to be evaluated, b would be evaluated first, the result discarded. Then a would converted to a pointer (char *) with value equal to &a[0] which would be the result of the expression (b, a).

Since the result of b, a is of type char *, sizeof(b,a) is equal to sizeof (char *). That is an implementation defined value but, for your compiler, has a value of 8 (which probably means the code is being built as a 64-bit application).

Answer 2

In most cases, arrays decay into pointers. So the type of b,a with a comma operator is a char* (not a char[20] anymore). And pointers are 8 bytes on your machine.

BTW, I think that using sizeof on some comma operator is really confusing to the reader. I recommend using sizeof on simple expressions or on types.

^{(and I just discovered this is one of the tricky differences between C and C++; see this explanation)}