C + + Union memory

See an example:

#include <iostream>#include<stdio.h>using namespacestd;union A {intA; struct {         Shortb;  ShortC; };};intMain () {A s; S.A.=0x12345678; printf ("%x%x", s.b, S.C); return 0;}

Output Result:

5678 1234

Why is that so?

Because a is union, the format that is stored in memory is:

High address------------> Low address

12 34 56 78

00010010 00110100 01010110 01111000

S.B occupies a low address of two bytes

S.C occupies a high address of two bytes

So:

s.b = 5678

S.C = 1234

To prove it, and to see more clearly, look at the program below.

#include <iostream>#include<stdio.h>using namespacestd;union A {intA; struct {         Shortb;  ShortC; };};intMain () {A s; S.A.=0x12345678; printf ("B =%x; c =%x\n", s.b, S.C); printf ("&a =%x; &b =%x; &c =%x\n", &AMP;S.A, &s.b, &S.C); return 0;}

Results:

5678 1234&a = 28ff1c; &b = 28ff1c; &c = 28ff1e

Isn't it obvious?

C + + Union memory