C++ identical method signature but different return type

Question

I have seen the following code:

template <class T>
class Type {
    public:
        Type() {}
        T& operator=(const T& rhs) {value() = rhs; return value();}
        T& value() {return m_value;}
        T value() const {return m_value;}
    private:
        T m_value;
};

Why does the compiler not complain about

    T& value() {return m_value;}
    T value() const {return m_value;}

and how to know which one is invoked?

Answer 1

The two functions are actually not the same. Only the second function is declared as a const member function. If the object that the member is called from is const, the latter option is used. If the object is non-const, the first option is used.

Example:

void any_func(const Type *t)
{
    something = t->value(); //second `const` version used
}

void any_func2(Type *t)
{
    something = t->value(); //first non-`const` version used
}

If both functions were declared non-const or both were declared const, the compiler would (should, anyway) complain.