How to set first three bytes of integer? in C++

问题

I want to set first three bytes of an integer to 0 in C++. I tried this code, but my integer variable a is not changing, output is always -63. What am I doing wrong?

#include <iostream>
#include <string>

int main()
{
  int a = 4294967233;
  std::cout << a << std::endl;
  for(int i = 0; i< 24; i++)
    {
        a |= (0 << (i+8));
        std::cout <<  a << std::endl;
    }

}

回答1:

Just use bitwise and (&) with a mask, there is no reason for loop:

a &= 0xFF000000; // Drops all but the third lowest byte
a &= 0x000000FF; // Drops all but the lowest byte

(Thanks to @JSF for the corrections)

As noted by @black, you may use Digit separators since C++14 in order to make your code more readable:

a &= 0xFF'00'00'00; // Drops all but the third lowest byte
a &= 0x00'00'00'FF; // Drops all but the lowest byte

回答2:

You need &= instead of |=.

回答3:

union also could be a choice

union my_type{
    int i;
    unsigned int u;
    ...
    unsigned char bytes[4];
};
...
my_type t;
t.u = 0xabcdefab;
t.bytes[0] = 5; // t.u = 0x05cdefab

回答4:

If you really mean "first" not lowest, then take advantage of the fact that alias rules let char alias anything:

  int a = 4294967233;
  char* p=&a;
  ...
  p[0] = whatever you wanted there
  p[1] = whatever you wanted there
  p[2] = whatever you wanted there

来源：https://stackoverflow.com/questions/34357968/how-to-set-first-three-bytes-of-integer-in-c