What happens when I delete dynamic array?

Question

I made a dynamic array (example):

int *a;
a = new int[3];
a[0] = 10; a[1] = 20; a[2] = 30;

Than I create a vector which stores pointers:

vector<int*> pa;
pa.push_back(&a[0]);

After I deleted (freed) the memory with "delete[] a;", I can still access to the element, that I push_backed. (cout << *pa[0]; output: 10)

Why is this happen? When I "delete[] a", it only deletes the pointer to the elements, but the elements are still accessable?

Don't miss out on this wonderful answer about memory and undefined behavior. http://stackoverflow.com/questions/6441218/can-a-local-variables-memory-be-accessed-outside-its-scope/6445794#6445794. — R Sahu, Mar 01 '17 at 19:37

score 1 · Answer 1 · answered Mar 01 '17 at 19:28

1

Your statement about how delete[] works is backwards. It deletes the elements but not the pointer. Continuing to use the pointer after its contents have been deleted, as you're doing, is undefined behavior. This means that literally anything is allowed to happen if you do it, so you shouldn't do it. It's by pure luck that you can still access the elements now, and bad things will happen in the future if you do it.

answered Mar 01 '17 at 19:28

Joseph Sible-Reinstate Monica

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Let's imagine that I remove the elements from the vector which I want to delete (pa.pop_back(); delete[] a). Is it totally safe? – Mar 01 '17 at 19:32
That makes `pa` completely safe, but you still have to know that you can't dereference `a` anymore without reassigning it. – Joseph Sible-Reinstate Monica Mar 01 '17 at 19:34

What happens when I delete dynamic array?

1 Answers1