Question

I’m confused by the following code:

#include <iostream> 

using namespace std;

struct bit 
{
    int a:3; 
    int b:2; 
    int c:3; 
}; 
int main(int argc, char* argv[]) 
{ 
    bit s; 
    char *c = (char*)&s; 
    *c = 0x99; 
    cout << s.a <<endl <<s.b<<endl<<s.c<<endl; 
    return 0; 
}

Since a takes 3 bits , b 2 bits , c 3 bits , when i cast this struct to a char* pointer , did i took the first 8 bytes or the last 8 bytes ?

1. On a Intel 32 bit machine , how will the compiler store an 32 bit integer ?

2. Why i get 1 , -1 , -4 as a result ?

Answer 1

0x99 is 10011001 binary is 100 11 001, which looks very much like 1,-1,-4 (in reverse order). And yes, it’s 8 least significant bits.

I believe signedness has confused you, so you may want to use unsigned int in the struct. If it’s not signedness, then please, be more specific.