CODEWITHSUNDEEP
cstructmemory-alignment
Rajesh Kumar G
Rajesh Kumar G

Reputation: 1534

Memory allocated for structures

I have the structure

typedef struct EData
{
    int a;
    char c;
}
Edata obj;

a is the integer variable so it takes 4 bytes and the c is the char variable so it takes 1 byte, totalling 5 bytes

But when I print sizeof(obj) it shows 8 bytes.

What is the reason?

Upvotes: 1

Views: 2234

Answers (4)

Cratylus
Cratylus

Reputation: 54074

The increase in size you notice is due to compiler's padding.
Compiler adds extra bytes to enforce correct byte boundaries.
So compiler adds the extra bytes to enforce the proper location for each member according to its type.
There is an option to stop compiler to do this (packed directive) but it is best to avoid it (except in corner-cases)

Upvotes: 1

sjngm
sjngm

Reputation: 12861

If it's a problem for you, you can either use #pragma or a compiler switch (a variety of compilers has such a switch).

Upvotes: 0

ismail
ismail

Reputation: 47592

Because on 32bit systems memory is aligned at 4byte (32bit) boundaries so it has to multiple of 4bytes, see Data structure alignment

Upvotes: 4

moinudin
moinudin

Reputation: 138357

int is 4 bytes, char is 1 byte. However, your compiler aligns each struct to a word (a word is 4 bytes on 32 bit architecture) as it improves performance. Therefore each instance of EData will be rounded up to 2 words (or 8 bytes).

What you end up with is something like this:

typedef struct EData {
    int a;
    char c;
    char padding[3];
}

Upvotes: 3

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