Question

Ignoring padding/alignment issues and given the following struct, what is best way to get and set the value of member_b without using the member name.

struct mystruct {
    int member_a;
    int member_b;
}
struct mystruct *s = malloc(sizeof(struct mystruct));

Put another way; How would you express the following in terms of pointers/offsets:

s->member_b = 3;
printf("%i",s->member_b);

My guess is to

• calculate the offset by finding the sizeof the member_a (int)
• cast the struct to a single word pointer type (char?)
• create an int pointer and set the address (to *charpointer + offset?)
• use my int pointer to set the memory contents

but I get a bit confused about casting to a char type or if something like memset is more apropriate or if generally i’m aproching this totally wrong.

Cheers for any help

Answer 1

The approach you’ve outlined is roughly correct, although you should use offsetof instead of attempting to figure out the offset on your own. I’m not sure why you mention memset — it sets the contents of a block to a specified value, which seems quite unrelated to the question at hand.

Here’s some code to demonstrate how it works:

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

typedef struct x {
    int member_a;
    int member_b;
} x;

int main() { 
    x *s = malloc(sizeof(x));
    char *base;
    size_t offset;
    int *b;

    // initialize both members to known values
    s->member_a = 1;
    s->member_b = 2;

    // get base address
    base = (char *)s;

    // and the offset to member_b
    offset = offsetof(x, member_b);

    // Compute address of member_b
    b = (int *)(base+offset);

    // write to member_b via our pointer
    *b = 10;

    // print out via name, to show it was changed to new value.
    printf("%d\n", s->member_b);
    return 0;
}