c - Are "malloc(sizeof(struct a *))" and "malloc(sizeof(struct a))" the same?

Question

This question is a continuation of Malloc call crashing, but works elsewhere

I tried the following program and I found it working (i.e. not crashing - and this was mentioned in the above mentioned link too). I May be lucky to have it working but I'm looking for a reasonable explanation from the SO experts on why this is working?!

Here are some basic understanding on allocation of memory using malloc() w.r.t structures and pointers

• malloc(sizeof(struct a) * n) allocates n number of type struct a elements. And, this memory location can be stored and accessed using a pointer-to-type-"struct a". Basically a struct a *.
• malloc(sizeof(struct a *) * n) allocates n number of type struct a * elements. Each element can then point to elements of type struct a. Basically malloc(sizeof(struct a *) * n) allocates an array(n-elements)-of-pointers-to-type-"struct a". And, the allocated memory location can be stored and accessed using a pointer-to-(pointer-to-"struct a"). Basically a struct a **.

So when we create an array(n-elements)-of-pointers-to-type-"struct a", is it

1. valid to assign that to struct a * instead of struct a ** ?
2. valid to access/de-reference the allocated array(n-elements)-of-pointers-to-type-"struct a" using pointer-to-"struct a" ?

data * array = NULL;

if ((array = (data *)malloc(sizeof(data *) * n)) == NULL) {
    printf("unable to allocate memory 
");
    return -1; 
}   

The code snippet is as follows:

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

int main(void) 
{
    typedef struct { 
        int value1;
        int value2;
    }data;

    int n = 1000;
    int i;
    int val=0;

    data * array = NULL;

    if ((array = (data *)malloc(sizeof(data *) * n)) == NULL) {
        printf("unable to allocate memory 
");
        return -1; 
    }   
    printf("allocation successful
");

    for (i=0 ; i<n ; i++) {
        array[i].value1 = val++;
        array[i].value2 = val++;
    }   

    for (i=0 ; i<n ; i++) {
        printf("%3d %3d %3d
", i, array[i].value1, array[i].value2);
    } 

    free(array);
    printf("freeing successful
");

    return 0;
}

---

EDIT: OK say if I do the following by mistake

data * array = NULL;
if ((array = (data *)malloc(sizeof(data *) * n)) == NULL) {

Is there a way to capture (during compile-time using any GCC flags) these kind of unintended programming typo's which could work at times and might blow out anytime! I compiled this using -Wall and found no warnings!

Answer 1

No.

sizeof(struct a*) is the size of a pointer.
sizeof(struct a) is the size of the entire struct.