With a union, all members share the same memory. With a struct, they do not share memory, so a different space in memory is allocated to each member of the struct.
For example:
union foo
{
int x;
int y;
};
foo f;
f.x = 10;
printf("%d\n", f.y);
Here, we assign the value of 10 to foo::x. Then we output the value of foo::y, which is also 10 since x and y share the same memory. Note that since all members of a union share the same memory, the compiler must allocate enough memory to fit the largest member of the union. So a union containing a char and a long would need enough space to fit the long.
But if we use a struct:
struct foo
{
int x;
int y;
};
foo f;
f.x = 10;
f.y = 20;
printf("%d %d\n", f.x, f.y);
We assign 10 to x and 20 to y, and then print them both out. We see that x is 10 and y is 20, because x and y do not share the same memory.
EDIT: Also take note of Gman's comment above. The example I provided with the union is for demonstration purposes only. In practice, you shouldn't write to one data member of a union, and then access another data member. Usually this will simply cause the compiler to interpret the bit pattern as another type, but you may get unexpected results since doing this is undefined behavior.
