How is abstraction unsafe? Additionally, raw pointers are not dangerous. If you'...

Jweb_Guru · on Dec 16, 2014

Creating a dangling const pointer is pretty trivial. Hell, make it a const reference.

    #include <iostream>

    struct Foo {
        const int &dangling;

        Foo(const int &dangling): dangling(dangling) { }
    };

    Foo foo() {
        int dangling = 0;
        Foo foo(dangling);
        return foo;
    }

    int main() {
        std::cout << foo().dangling << std::endl;
        return 0;
    }

Neither g++ nor clang warn on this, either. I am not sure why you think const pointers are safe. They aren't. Flat out.

72deluxe · on Dec 16, 2014

Well I was assuming for pointers that they'd be initialised to something sensible. That would be a coding standard problem if they're not.

The example you give is also a problem with coding standards, I'd argue. It isn't the languages fault that you're using a reference that's going out of scope - that's just bad coding. (A pointer going out of scope isn't so bad, no?)

Animats · on Dec 16, 2014

How is abstraction (hiding) unsafe?

Objects hide their implementation details. An object is only a valid abstraction if it correctly hides its implementation details and the user can ignore them. If there are hidden constraints on what a user can do with an object, but those are not enforced by the object, the object is an unsuccessful abstraction and a potential source of bugs.

Additionally, raw pointers are not dangerous.

Two words: "buffer overflow". C pointers lose size information.