Question

I was trying a program to understand the behavior of structure variable,

Sample code:

struct temp
{
   int a;
   int b;
}obj;

int main()
{
   obj.a = 10;
   obj.b = 7;
   /* to see whether obj and &obj both are same 
    * I was verifying whether structure variables behave as arrays
    */
   printf("%d -- %p",obj,&obj); 
   return 0;
}

I was expecting output to be 10 and address of obj
But to my surprise,
Actual output is
10 and 00000007

This is bugging me a lot now!!!

Could anyone please help me understand why printf is taking second member and printing its value.

Answer 1

This happens because the first parameter is expected to have 4 bytes (an int) but it has 8 bytes (struct obj).

When you pass obj to printf function it will place on the stack the whole obj structure (including b member). So when printing it will take the first 4 bytes on the stack for 1st parameter (obj.a) and the next 4 bytes on the stack (obj.b) for 2nd parameter.

Check this out:

printf("%d %p %p\n", obj, &obj, &obj);