Confusion in ternary operator and typecasting

Question

I gone through this question -

why the result of : 1 ? (int *)0 : (void *)0
differs to the result of : 1 ? (int *)0 : (void *)1

Answer 1

So effeffe already answered on why the two expressions differ:

1 ? (int *) 0 : (void *) 0  // yield (int *) 0
1 ? (int *) 0 : (void *) 1  // yield (void *) 0

To answer this question now:

How to check the result ?

With gcc you can use typeof extension and __builtin_types_compatible_p builtin function:

// The two expressions differ
printf("%d\n", __builtin_types_compatible_p(typeof(1 ? (int *) 0 : (void *) 0), typeof(1 ? (int *) 0 : (void *) 1)));

// First expression yield (int *) 0
printf("%d\n", __builtin_types_compatible_p(typeof((int *) 0), typeof( 0 ? (int *)0 : (void *)0  )));

// Second expression yield (void *) 0
printf("%d\n", __builtin_types_compatible_p(typeof((void *) 0), typeof( 1 ? (int *)0 : (void *)1  )));