Size of struct NOT equal to the size of its content

Possible Duplicate:
Why isn't sizeof for a struct equal to the sum of sizeof of each member?

I have the next code:

http://ideone.com/brmRy

#include <stdio.h>
#include <stdlib.h>

typedef struct Test
{
        int a;
        int b;
        char c;
} Test;

int main(void)
{
        Test *obj = (Test*)malloc(sizeof(Test));

        printf("Size results:rnrnstruct: %irnint #1: %irnint #2: %irnchar #1: %irn", 
                sizeof(Test), sizeof(obj->a), sizeof(obj->b), sizeof(obj->c));

        return 0;
}

The result is:

Size results:

struct: 12

int #1: 4

int #2: 4

char #1: 1

Why DOES struct size 12 bytes??? int - 4 bytes char - 1 byte

2 int + 1 char = 2 * 4 bytes + 1 byte = 9 bytes.

Why does 12 ???

---

Memory is commonly aligned on 4-byte boundaries, so even though the char only takes up 1 byte, it's padded by 3 bytes to meet this segmentation requirement. Notably, individual elements of a struct don't have to be aligned, so if you made one of the ints into a short, you could reduce the struct size from 12 to 8 bytes. I believe you'd have to put the short next to the char in the struct declaration to receive this bonus, though.

---

If you're using gcc you can force "packing". This means that no nalignment is made and the struct entries stand next to each other. Try

typedef struct Test
{
    int a;
    int b;
    char c;
} __attribute__((packed)) Test;

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