How do I remove code duplication between similar c

Question

Let's say I have the following class X where I want to return access to an internal member:

class Z
{
    // details
};

class X
{
    std::vector<Z> vecZ;

public:
    Z& Z(size_t index)
    {
        // massive amounts of code for validating index

        Z& ret = vecZ[index];

        // even more code for determining that the Z instance
        // at index is *exactly* the right sort of Z (a process
        // which involves calculating leap years in which
        // religious holidays fall on Tuesdays for
        // the next thousand years or so)

        return ret;
    }
    const Z& Z(size_t index) const
    {
        // identical to non-const X::Z(), except printed in
        // a lighter shade of gray since
        // we're running low on toner by this point
    }
};

The two member functions X::Z() and X::Z() const have identical code inside the braces. This is duplicate code and can cause maintenance problems for long functions with complex logic.

Is there a way to avoid this code duplication?

Answer 1

How about moving the logic into a private method, and only doing the "get the reference and return" stuff inside the getters? Actually, I would be fairly confused about the static and const casts inside a simple getter function, and I'd consider that ugly except for extremely rare circumstances!

Answer 2

C++17 has updated the best answer for this question:

T const & f() const {
    return something_complicated();
}
T & f() {
    return const_cast<T &>(std::as_const(*this).f());
}

This has the advantages that it:

• Is obvious what is going on
• Has minimal code overhead -- it fits in a single line
• Is hard to get wrong (can only cast away volatile by accident, but volatile is a rare qualifier)

If you want to go the full deduction route then that can be accomplished by having a helper function

template<typename T>
constexpr T & as_mutable(T const & value) noexcept {
    return const_cast<T &>(value);
}
template<typename T>
void as_mutable(T const &&) = delete;

Now you can't even mess up volatile, and the usage looks like

T & f() {
    return as_mutable(std::as_const(*this).f());
}

Answer 3

I did this for a friend who rightfully justified the use of const_cast... not knowing about it I probably would have done something like this (not really elegant) :

#include <iostream>

class MyClass
{

public:

    int getI()
    {
        std::cout << "non-const getter" << std::endl;
        return privateGetI<MyClass, int>(*this);
    }

    const int getI() const
    {
        std::cout << "const getter" << std::endl;
        return privateGetI<const MyClass, const int>(*this);
    }

private:

    template <class C, typename T>
    static T privateGetI(C c)
    {
        //do my stuff
        return c._i;
    }

    int _i;
};

int main()
{
    const MyClass myConstClass = MyClass();
    myConstClass.getI();

    MyClass myNonConstClass;
    myNonConstClass.getI();

    return 0;
}