Question

I tried to wrap something similar to Qt’s shared data pointers for my purposes, and upon testing I found out that when the const function should be called, its non-const version was chosen instead.

I’m compiling with C++0x options, and here is a minimal code:

struct Data {
    int x() const {
        return 1;
    }
};

template <class T>
struct container
{
    container() {
        ptr = new T();
    }


    T & operator*() {
        puts("non const data ptr");
        return *ptr;
    }

    T * operator->() {
        puts("non const data ptr");
        return ptr;
    }

    const T & operator*() const {
        puts("const data ptr");
        return *ptr;
    }

    const T * operator->() const {
        puts("const data ptr");
        return ptr;
    }

    T* ptr;
};

typedef container<Data> testType;

void testing() {
    testType test;
    test->x();
}

As you can see, Data.x is a const function, so the operator -> called should be the const one. And when I comment out the non-const one, it compiles without errors, so it’s possible. Yet my terminal prints:

“non const data ptr”

Is it a GCC bug (I have 4.5.2), or is there something I’m missing?

Answer 1

If you have two overloads that differ only in their const-ness, then the compiler resolves the call based on whether *this is const or not. In your example code, test is not const, so the non-const overload is called.

If you did this:

testType test;
const testType &test2 = test;
test2->x();

you should see that the other overload gets called, because test2 is const.