How do I create a memory leak?

Question

I want to allocate some memory on the heap that is not reachable from any stack pointer. (This is for test purpose).

void *ptr = malloc(sizeof(int));
void *ptr2 = malloc(sizeof(int));

ptr = ptr2;

If I do this code, I think that ptr and ptr2 at the start is two pointers on the stack referring to some allocated memory on the heap, right? And then when I do the ptr = ptr2, the first mallocated memory is still on the heap but not reachable in any way from the stack. Is that so?

I have a program that is searching the stack to find all alive objects on the heap, therefore I want to test that it actually works.

Answer 1

It's even simpler to do than that, here is an example from wikipedia:

#include <stdlib.h>

void function_which_allocates(void) {
    /* allocate an array of 45 floats */
    float * a = malloc(sizeof(float) * 45);

    /* additional code making use of 'a' */

    /* return to main, having forgotten to free the memory we malloc'd */
}

int main(void) {
    function_which_allocates();

    /* the pointer 'a' no longer exists, and therefore cannot be freed,
     but the memory is still allocated. a leak has occurred. */
}

As soon as the pointer to 'a' goes out of scope, the memory is leaked, as 'a' was never freed.

Answer 2

A more subtile leak to find would be introduced by doing

malloc(0);

though.

Answer 3

Try using:

 malloc(1);

should suffice

Answer 4

if you simply put your code into a function, then you have a memory leak as soon as your function ends and ptr and ptr2 get out of scope.

Answer 5

That works. It's more complex than necessary, though:

malloc(4);

The easiest way to leak memory is to just not save a reference to it in the first place.