Thread: Casting

Ok, so someone please explain the following to me.

Code:

#include <stdint.h>

int64_t a = 0;
uint32_t b = 1;

int32_t main(void)
{
  return (signed)a;
}

Does this code truncate down to a 32-bit variable because that is the return type? Or because when I say "signed" and do not specify a type, it assumes int?

signed defaults to int, yes. Whether that is int32 or int64, only your compiler knows (and you, if you bother to read the header or print out sizeof information, or find out in some other way). If signed does not match int32_t, then it will return the least significant 32 bits of the answer (or pad int out to 32 bits, whichever is necessary).

Ok, but is the truncation because of the return type or because of the lack of specifying what type is signed?

Could be either. If just plain int is int32, then truncation will happen at the cast. If just plain int is int64, then truncation will happen at the return.

Ok, so what happens here?

Code:

int64_t x = (unsigned)-1;

On a 32-bit system? x contains 01777777777777777777777 or 037777777777?

Here the cast would happen first, so (unsigned)-1 is 0xffffffff; this fits in 64 bits just fine, so nothing more happens.

ok so to get 01777777777777777777777 I need 2 do uint64_t x = (uint64_t)-1?

I'm not sure why you think the cast is necessary (other than to shut the compiler up). And for that matter, if your goal is that number, then just do it:

Code:

uint64_t x = 01777777777777777777777LL;