Dynamic array allocation

[abhishekd18]

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


int main()
{
    int n, i;
    printf("\nEnter a number: \n");
    scanf("%d",&n);
    int * a = (int*) malloc(n*sizeof(int));
    int * b = (int*) malloc(n*sizeof(int));
    for(i=0;i<n;i++)
    a[i]=i*i;

    b=a;


    for(i=0;i<n;i++)
    {
    printf("%d: \n",b[i]);
    }
free(a);
free(b);
    return 0;
}

In the above code, I define pointers "a" and "b". Then I assign some values to "a". Then I say b = a. Here, does it mean that I have assigned memory locations of "a" to "b"? or I have assigned values stored in array "a" to array "b"?

[Salem]

> b=a;
You can't assign arrays, and you can't copy one block of memory to another just through pointer assignment.

You need another loop
for ( i = 0 ; i < n ; i++ ) b[i] = a[i];

[abhishekd18]

The program works if you just comment out either of the lines free(a); and free(b);.

[rogster001]

The program works if you just comment out either of the lines free(a); and free(b);.

Well that makes it all ok then - bonus, you get two choices of lines to comment out to hack a working program - erm..

[abhishekd18]

But I want to understand what is happening here!

[nonoob]

By doing b=a you are only copying the address. This effectively destroys what 'b' contained... Therefore doing a free(a) and free(b) will attempt to free the same area twice. But the memory that the original 'b' pointed to continues to be allocated.
I think what you meant to do was to copy the data from one array to the other. Use memcpy(b, a, n * sizeof(int)).

[abhishekd18]

Thanks for the reply! That was helpful. But then I have one more confusion. Then why can't we access the array elements by "*a[i]" I mean because "a" is pointer if I want to get value of "a", I should get it by dereferencing. But this gives error and a[i] works perfect. In other words, "a[i]" should print address itself not the value at address.

[laserlight]

Then why can't we access the array elements by "*a[i]" I mean because "a" is pointer if I want to get value of "a", I should get it by dereferencing. But this gives error and a[i] works perfect. In other words, "a[i]" should print address itself not the value at address.

Indeed, a is a pointer: a is a pointer to int. Therefore *a is an int. Now, a[i] is equivalent to *(a + i), therefore, a[i] is also an int: in particular, *a == a[0]. However, *a[i] means *(a[i]), i.e., *(*(a + i)). But *(a + i) is an int, hence *a[i] attempts to dereference an int, not a pointer, which is wrong.