How much memory does a union need

Question

I have the following union

typedef union rem{
    int addr;
    char addrbuf[32];
} foo;

A sizeof on the union provides the following output

foo addr;
printf("size is: %d\n",sizeof addr);

size is: 32

Does it mean that a union allocates or needs the memory equal to biggest element in the union at definition ?

Are you truly defining a union instance or are you mapping a pointer to a struct that contains this union, where the first struct member (m_eAddrType) tells you how to interpret the next member of the struct that is this union (i.e. look at .addr vs. looking at .addrBuff? I have implemented this "variant record" data-structure using struct that contain a union. — franji1, Jan 21 '15 at 18:48

haccks · Accepted Answer · 2015-01-21T19:03:37.423

8

Does it mean that a union allocates or needs the memory equal to biggest element in the union at definition ?

Yes. Memory for the largest member is allocated. That's why the members of a union are stored at the same location.

Assuming int requires 4 bytes, for union

union{
    char c;
    int i;
} U;

memory will be allocated as:

enter image description here

Note that how c and i overlap. In fact c is first byte of i.

edited Jan 21 '15 at 19:03

answered Jan 21 '15 at 18:28

haccks

104,019
25
176
264

4

There may also be space allocated for padding. – Roland Rabien Jan 21 '15 at 18:31

score 1 · Answer 2 · answered Jan 21 '15 at 18:34

Memory occupied by union unlike structure will be memory for the largest member of the union. But depending on the implementation there might be extra padding at the end so there is no guarantee that the size of a union = size of the largest member. So you can say size of union will be atleast the size of the largest member in the union.

Side note: Since the memory is shared between all the members of the union only one member is active at a time.

How much memory does a union need

2 Answers2