questions about operator “&”

Question

Probably have wrong title,but I don't know how to describe it.

int x=0x4a;
h=!(0x80000000000&(0x39+(~x+1)));

the result is h=0. The question is x has 32 bits, while 0x80000000000 has more than 32 bit. If I set variable i to 0x80000000000 and print it as hex format,it will show 0. So in that case, why not the result of h is 1 because 0x80000000000 turns to be 0?

The language I used is C

Answer 1

0x8000000000 is a long long int constant. Therefore the C compiler promotes the types in an arithmetic expression to the highest precision that occurs within an expression.

If instead you wrote:

  int x=0x4a;
  int h1=(0x80000000000&(0x39+(~x+1)));
  int h = !h1;

Then h would become 1. And perhaps a warning on missing precision would be issued.