const qualifier removed on getter if template argument is a pointer

Question

I have the following code:

template <typename T>
struct Val
{
    T val;

    const T& get() const { return val; }
    void set(const T& newVal) { val = newVal; }
};

int& f(const Val<int>& val)
{
    return val.get();
}

int* g(const Val<int*>& val)
{
    return val.get();
}

int main()
{
    Val<int> val1;
    f(val1);

    Val<int*> val2;
    g(val2);

    return 0;
}

It fails compilation in gcc with the following message:

main.cpp: In function 'int& f(const Val<int>&)':
main.cpp:78:20: error: invalid initialization of reference of type 'int&' from expression of type 'const int'
     return val.get();
                    ^
main.cpp:79:1: warning: control reaches end of non-void function [-Wreturn-type]

That's totally fine, but why isn't the same error produced for g() ? Why, for some reason, is the const qualifier removed on const T& , when T is a pointer?

I tried to find some resources, but it seems hard to find. I know Meyers wrote something in his newest book, but I don't have access to that one. Can someone point me to resources or keywords where I could find more on that whole topic of template type deduction when T is a pointer?

Answer 1

Why, for some reason, is the const qualifier removed on const T&, when T is a pointer?

Because you're returning the pointer by value, not by reference. It is perfectly fine to initialize a pointer from a const pointer (note it is the pointers that are const , not the things they point to.) You're doing the equivalent of this:

int* const p0 = nullptr; // const pointer
int* p1 = p0;            // copy to non-const is OK

If you were to initialize a non-cost reference to poiter from a const pointer, you'd get a similar error to the first case:

// error: binding of reference to type 'int *'
// to a value of type 'int *const' drops qualifiers
int*& p1 = p;