Question

In the following code

#include <iostream>

using namespace std;
struct field
{
   unsigned first : 5;
   unsigned second : 9;
};
int main()
{
   union
   {
      field word;
      int i;
   };
   i = 0;
   cout<<"First is : "<<word.first<<" Second is : "<<word.second<<" I is "<<i<<"\n";
   word.first = 2;
   cout<<"First is : "<<word.first<<" Second is : "<<word.second<<" I is "<<i<<"\n";
   return 0;
}

when I initialize word.first = 2, as expected it updates 5-bits of the word, and gives the desired output. It is the output of ‘i’ that is a bit confusing. With word.first = 2, i gives output as 2, and when I do word.second = 2, output for i is 64. Since, they share the same memory block, shouldnt the output (for i) in the latter case be 2?

Answer 1

This particular result is platform-specific; you should read up on endianness.

But to answer your question, no, word.first and word.second don’t share memory; they occupy separate bits. Evidently, the underlying representation on your platform is thus:

bit   15 14 13 12 11 10  9  8  7  6  5  4  3  2  1  0
     +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
     |     |         second           |    first     |
     +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
     |<------------------- i ----------------------->|

So setting word.second = 2 sets bit #6 of i, and 2⁶ = 64.