Will free() clean up this allocated memory?

I have some legacy code that I have a question about. A pointer to allocated memory is passed as a parameter to a function. In the function, there is a declaration statement that creates a new pointer and assigns it the value of the allocated pointer. Will the free() statement actually free the allocated memory?

Thanks.

Code:

---

void test(int *intptr)
{
    int* newptr = intptr;

      .
      .
      .
      .
      free(newptr);
}

---

Yes. free() only cares about the value passed to it. It cannot possibly know which variable was holding the value, because it is passed the value.

Thanks --- that was my interpretation also, but wanted a second opinion (or even a 3rd and 4th)

At the risk of stating the obvious, the caller of your function should not use the argument it passes again, except to reassign it. For example;

Code:

---

void test(int *pointer)
{
    int *newpointer = pointer;
  free(newpointer);

}

int main()
{
    int *data = malloc(5*sizeof(int));
    test(data);
    data[2] = 42;    /* undefined behaviour as test() has effectively performed free(data) */
}

---

Creating a copy of a pointer (pointerA = pointerB) means that both pointers point at the same data. So freeing one, frees both.