How does delete deal with pointer constness?

Question

I was reading this question Deleting a const pointer and wanted to know more about delete behavior. Now, as per my understanding:

delete expressio

Answer 1

operator delete accepts a void*. As part of a test program I overloaded operator delete and found that operator delete doesn't accept const pointer.

How did you try this? It certainly does accept const pointers:

#include <memory>

int main() {
    void* const px = 0;
    delete px;
    ::operator delete(px);
}

This code is correct, compiles (albeit with a justified warning) and executes.

EDIT: Reading the original article – you aren't talking about a const pointer but a pointer to const, which is something else. The reason why this has to work is described there. As for why it's working: others have said this.

Answer 2

As this answer says, delete is not a method like any other, but a part of the langage to destruct objects. const-ness has no bearing on destructability.

Answer 3

delete just makes a call to deallocate the memory the pointer points to, it doesn't change the value of the pointer nor the object. Therefore, delete has nothing to do with the const-ness of the pointer or object pointed to.

Answer 4

const_cast doesn't really do anything – it's a way to suppress compiler moaning about const-ness of the object. delete keyword is a compiler construct, the compiler knows what to do in this case and doesn't care about const-ness of the pointer.

Answer 5

delete is an operator that you can overload. It takes a pointer as an argument, and frees the memory, possibly using free. The compiler allows this whether the pointer is const or not.