Question

If I perform a Bitwise AND between a 8 bit integer (int8_t) and 32 bit integer (int) will the result be a 8 bit integer or a 32 bit integer?

I am using GNU/Linux and GCC compiler

To put the question slightly differently, before performing the bitwise AND, are the first 24 bits of the 32 bit integer discarded, or is the 8 bit integer first typecast to a 32 bit integer ?

EDIT: In this little code

#include <iostream>
#include <stdint.h>
int main()
{
  int    i=34;
  int8_t j=2;

  std::cout<<sizeof((i&j))<<std::endl;//Bitwise and between a 32 bit integer and 8 bit integer
  return 0;
}

I get the output as 4. I would assume that means that the result is a 32 bit integer then. But I don’t know if the result depends on the machine, compiler or OS.

Answer 1

For the & operator (and most other operators), any operands smaller than int will be promoted to int before the operation is evaluated.

From the C99 standard (6.5.10 – describing the bitwise AND operator):

The usual arithmetic conversions are performed on the operands.

(6.3.1.8 – describing the usual arithmetic conversions):

the integer promotions are performed on both operands

(6.3.1.1 – describing the integer promotions):

If an int can represent all values of the original type, the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer
promotions.