Is apparent NULL pointer dereference in C actually pointer arithmetic?

Question

I've got this piece of code. It appears to dereference a null pointer here, but then bitwise-ANDs the result with unsigned int. I really don't understand the whole part. What is it intended to do? Is this a form of pointer arithmetic?

struct hi  
{
   long a;  
   int b;  
   long c;  
};  

int main()  
{  
    struct hi ob={3,4,5};  
    struct hi *ptr=&ob;  
    int num= (unsigned int) & (((struct hi *)0)->b);  

   printf("%d",num);  
   printf("%d",*(int *)((char *)ptr + (unsigned int) & (((struct hi *)0)->b)));  
}  

The output I get is 44. But how does it work?

Answer 1

This is not an "and", this is taking the address of the right hand side argument.
This is a standard hack to get the offset of a struct member at run time. You are casting 0 to a pointer to struct hi, then referencing the 'b' member and getting its address. Then you add this offset to the pointer "ptr" and getting real address of the 'b' field of the struct pointed to by ptr, which is ob. Then you cast that pointer back to int pointer (because b is int) and output it. This is the 2nd print. The first print outputs num, which is 4 not because b's value is 4, but because 4 is the offset of the b field in hi struct. Which is sizeof(int), because b follows a, and a is int... Hope this makes sense :)