which amongst these two cases would result in memory leak

Question

using delete:

int** a=new int*;
**a=5;
delete a;

using delete[]:

int** a=new int*;
**a=5;
delete[] a;

To find out, I tried doing this-

int** a=new int*;
**a=5;
int* b=new int;
b=*(a);
delete[] a;
cout<<"*b: "<<*b;

getting an output- 5

but I suspect that integer stored in b might have been deleted but not changed (no good reason to get a garbage value right ahead).

What should I do to find out (or what is the answer).

Answer 1

int** a=new int*;
**a=5;
delete a;

This is wrong, because you allocate an int* and don't point it to an int afterwards. Accessing the int* itself via *a is OK, but accessing an int via the extra * is not OK. This is Undefined Behavior.

You need to add that additional int, eg:

int** a=new int*;
*a = new int;  // <-- here
**a = 5;       // <-- OK!
delete *a;     // <-- and don't forget this!
delete a;

---

int** a=new int*;
**a=5;
delete[] a;

Same problem as above - assigning to a non-existent int via **a - but worse because now you are also mismatching new and delete[]. More Undefined Behavior. You can use delete only with new, and delete[] only with new[].

---

int** a=new int*;
**a=5;
int* b=new int;
b=*(a);
delete[] a;
cout<<"*b: "<<*b;

Now, that is just a culmination of bad behaviors. Not only are you:

• assigning to a non-existent int via **a

• mismatching new and delete[]

But you are also:

• leaking the int that you originally allocated for b to point to. You are re-assigning b afterwards with a new (invalid) int* value, losing access to the int you allocated. You have no hope of ever using a valid delete statement to free that int.

• worse, you are dereferencing b to access an int after delete'ing the memory address stored in the int* that was assigned to b (and that int* was not valid to dereference to begin with).

This code is just all kinds of Undefined Behavior.