Comment by nathanrf

In C and C++, `const` on pointers/references is basically just a comment to programmers - it is part of the type, but doesn't "mean" anything to the abstract machine; the rules don't treat const / regular references/pointers differently, they just say that the types only let you mutate through a mutable pointer.

Obviously, good code should treat it as more than just a comment - using `const` correctly clarifies intent and makes it possible to stay sane as a C++ developer, but the abstract machine doesn't care.

In C++, you can basically always `const_cast` a `const T&` into a `T&` and then modify it without causing UB. A function that accepts a `const T&` is just pinky promising that it will be polite and probably not do that.

It is only UB if the underlying object is "actually const", and even then, it doesn't cause UB until you actually perform the mutation; creating the mutable reference itself is perfectly fine.

For example, the following is perfectly legal:

    int& upgrade_to_mut(const int& x) {
      return *const_cast<int*>(&x);
    }

    int x = 5;
    const int& x_ref = x;
    int& x_ref_mut = upgrade_to_mut(x_ref);
    x_ref_mut = 6;

it's only invalid if the object that is pointed at is const, as in:

    int& upgrade_to_mut(const int& x) {
      return *const_cast<int*>(&x);
    }

    const int y = 5;
    const int& y_ref = y;

    int& y_ref_mut = upgrade_to_mut(y_ref); // it is actually legal to produce y_ref_mut, but we cannot modify it


    y_ref_mut = 6; // this is UB: cannot modify a const object 'y'

The difference is that in Rust, "mutation capabilities" are part of references, and so you cannot create them out of nowhere, that would be UB. But in C++, mutation capabilities are part of the object being pointed at, so as long as they happen to be there when you perform the mutation (e.g. you're not modifying a string literal or a variable declared `const`) then there's no problem.