C ++中8位整数和32位整数之间的按位AND

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.

Answer 2

无论语言指定什么，问题的答案是，如果高位24位在按位之前被丢弃并且被执行则无关紧要，因为它们在一个操作数中是全零位，因此在结果中。

Answer 3

8 bit integer is typecast to a 32 bit integer. Bitwise operators trigger the usual arithmetic conversions (which means the smaller type is promoted to match the larger). That's defined in chapter 5.1 of the spec.

Answer 4

Integer types smaller than int are promoted to int before any operation is performed on them. You might want to look at what the CERT Secure Coding Standard says about "understanding integer conversion rules".