Question

According to byte padding, in 32 bit OS, the CPU can address memory in multiple of 4 as given here when the computer’s word size is 4 bytes (a byte meaning 8 bits), the data to be read should be at a memory offset which is some multiple of 4. When this is not the case, e.g. the data starts at the 14th byte instead of the 16th byte, then the computer has to read two 4-byte chunks and do some calculation before the requested data has been read, or it may generate an alignment fault. Even though the previous data structure ends at the 14th byte, the next data structure should start at the 16th byte. Two padding bytes are inserted between the two data structures to align the next data structure to the 16th byte.

#include<stdio.h>
struct test {
        char a;
        char b;
        int c;
        int d;
};
void main() {
        int a,b;
        char c,d;
        printf("Address of a & b = %u & %u respectively\n",&a,&b);
        printf("Address of c & d = %u & %u respectively\n",&c,&d);      
        struct test t1;
        printf("The size of structure:::%d\n",sizeof(t1));
}

Output is:

Address of a & b = 3216087804 & 3216087808 respectively
Address of c & d = 3216087802 & 3216087803 respectively
The size of structure:::12

While when I declared structure in this way:

struct test {
        char a;
        int b;
        int c;
        char d;
};

Output in this case:

The size of structure:::16

Why alignment fault will not occur when we try to access secord char variable at odd location in memory or variable which is not exist at address which is in multiple of 4?

Answer 1

Padding is being done in both cases. Inside the code you provided, char b requires no padding and thus receives none. int c requires two bytes of padding (I’m assuming typical x86 rules) and gets it. The total sizeof the struct is 10 bytes of data + 2 bytes of padding for the indicated 12 bytes.

In the separate case, int b requires three bytes of padding and the struct as a whole requires another three to put char a on a 4-byte boundary address. 10 bytes of data + 3 + 3 bytes of padding give 16 bytes.