如何用位域计算结构的大小？

Question

#include <stdio.h> 
struct test { 
unsigned int x; 
long int y : 33; 
unsigned int z; 
};  
int main() 
{ 
struct test t; 
printf("%d", sizeof(t)); 
return 0; 

}

I am getting the output as 24. How does it equate to that?

Answer 1

As your implementation accepts long int y: 33; a long int hase more than 32 bits on your system, so I shall assume 64.

If plain int are also 64 bits, the result of 24 is normal.

If they are only 32 bits, you have encountered padding and alignment . For performance reasons, 64 bits types on 64 bits systems are aligned on a 64 bits boundary. So you have:

• 4 bytes for the first int
• 4 padding bytes to have a 8 bytes boundary
• 8 bytes for the container of the bit field
• 4 bytes for the second int
• 4 padding bytes to allow proper alignment of arrays

Total: 24 bytes