2

How can I check if a member function exists and is not inherited?

I need this to resolve ambiguity for the following example:

A type either has a foo() or a bar() member function. Caller will call the one that exists for the given type. However, DerivedWithBar inherits foo() from BaseWithFoo but defines its own bar(). Thus, Caller does not know which function to call.

I'd need a way to give the non-inherited foo precedence over the inherited bar(), but I do not know how to check if a member function is inherited or not.

#include <iostream>

struct BaseWithFoo
{
    template <typename T> void foo(T&&){std::cout << "Base::foo" << std::endl;}
};

struct DerivedWithBar : public BaseWithFoo
{
    template <typename T> void bar(T&&){std::cout << "DerivedWithBar::bar" << std::endl;}
};

struct DerivedWithFoo : public BaseWithFoo
{
    template <typename T> void foo(T&&){std::cout << "DerivedWithFoo::foo" << std::endl;}
};

struct EmptyDerived : public BaseWithFoo {};

struct BaseWithBar
{
    template <typename T> void bar(T&&){std::cout << "BaseWithBar::bar" << std::endl;}
};


struct Caller
{
    template <typename T>
    auto call(T&& x) -> decltype(x.foo(*this), void())
    {
        x.foo(*this);
    }

    template <typename T>
    auto call(T&& x) -> decltype(x.bar(*this), void())
    {
        x.bar(*this);
    }
};

int main()
{
  Caller c;
  c.call(BaseWithFoo());
  c.call(DerivedWithFoo());
  c.call(DerivedWithBar());
  c.call(EmptyDerived());
  c.call(BaseWithBar());
}

live example

desired output:

Base::foo
DerivedWithFoo::foo
DerivedWithBar::bar
Base::foo
BaseWithBar::bar
m.s.
  • 16,063
  • 7
  • 53
  • 88
  • Do you know the signature of the function to test (`void U::foo(Caller&)`) ? – Jarod42 Jul 13 '16 at 08:52
  • @Jarod42 the signature is as you wrote it, however it is a template function, so its needs to be generic – m.s. Jul 13 '16 at 08:58

2 Answers2

2

I found a way to distinguish between inherited and non-inherited member functions by comparing types of member function pointers.

The following is a partial solution to my full problem ("giving non-inherited member functions precedence over inherited ones"). This will only call non-inherited foo or non-inherited bar.

struct Caller
{    
    template <typename T>
    auto call(T&& x) -> decltype(x.foo(*this),
        std::enable_if_t<
            std::is_same<
                decltype(&T::template foo<decltype(*this)>),
                void (T::*)(decltype(*this))
            >::value
        >())
    {
        x.foo(*this);
    }

    template <typename T>
    auto call(T&& x) -> decltype(x.bar(*this),
        std::enable_if_t<
            std::is_same<
                decltype(&T::template bar<decltype(*this)>),
                void (T::*)(decltype(*this))
            >::value
        >())
    {
        x.bar(*this);
    }
};

live example

m.s.
  • 16,063
  • 7
  • 53
  • 88
  • 1
    You may add *priority* to solve the other part: [Demo](http://coliru.stacked-crooked.com/a/b2e3e5512ef241c1). – Jarod42 Jul 13 '16 at 09:20
  • @Jarod42 thanks, I was just typing something similar (based on [this answer](http://stackoverflow.com/a/38283990/678093)). – m.s. Jul 13 '16 at 09:23
0

With a bunch of template, we can get it as it follows:

#include<iostream>
#include<type_traits>
#include<utility>

struct BaseWithFoo
{
    template <typename T> void foo(T&&){std::cout << "Base::foo" << std::endl;}
};

template<typename T>
struct DerivedWithBarT : public T
{
    template <typename U> void bar(U&&){std::cout << "DerivedWithBar::bar" << std::endl;}
};

using DerivedWithBar = DerivedWithBarT<BaseWithFoo>;

template<typename T>
struct DerivedWithFooT : public T
{
    template <typename U> void foo(U&&){std::cout << "DerivedWithFoo::foo" << std::endl;}
};

using DerivedWithFoo = DerivedWithFooT<BaseWithFoo>;

template<typename T>
struct EmptyDerivedT : public T {};

using EmptyDerived = EmptyDerivedT<BaseWithFoo>;

struct BaseWithBar
{
    template <typename T> void bar(T&&){std::cout << "BaseWithBar::bar" << std::endl;}
};

struct EmptyBase {};

template<typename...>
using void_t = void;

template<typename, typename = void_t<>>
struct HasFoo: std::false_type {};

template<typename T, template<typename> class U>
struct HasFoo<U<T>, void_t<decltype(&U<EmptyBase>::template foo<T>)>>: std::true_type {};

template<typename, typename = void_t<>>
struct HasBar: std::false_type {};

template<typename T, template<typename> class U>
struct HasBar<U<T>, void_t<decltype(&U<EmptyBase>::template bar<T>)>>: std::true_type {};

template<typename T>
struct BarOverFoo {
    static constexpr bool value = HasBar<T>::value && !HasFoo<T>::value;
};

template<typename T>
struct FooOverBar {
    static constexpr bool value = HasFoo<T>::value && !HasBar<T>::value;
};

template<typename T>
struct Both {
    static constexpr bool value = HasFoo<T>::value && HasBar<T>::value;
};

template<typename T>
struct None {
    static constexpr bool value = !HasFoo<T>::value && !HasBar<T>::value;
};

struct Caller
{
    template <typename T>
    std::enable_if_t<FooOverBar<T>::value>
    call(T&& x)
    {
        x.foo(*this);
    }

    template <typename T>
    std::enable_if_t<BarOverFoo<T>::value>
    call(T&& x)
    {
        x.bar(*this);
    }

    template <typename T>
    std::enable_if_t<Both<T>::value>
    call(T&& x)
    {
        x.bar(*this);
    }

    template <typename T>
    std::enable_if_t<None<T>::value>
    call(T&& x)
    {
        callInternal(std::forward<T>(x));
    }

private:
    template <typename T>
    auto callInternal(T&& x) -> decltype(x.foo(*this), void())
    {
        x.foo(*this);
    }

    template <typename T>
    auto callInternal(T&& x) -> decltype(x.bar(*this), void())
    {
        x.bar(*this);
    }
};

int main()
{
    Caller c;
    c.call(BaseWithFoo());
    c.call(DerivedWithFoo());
    c.call(DerivedWithBar());
    c.call(EmptyDerived());
    c.call(BaseWithBar());
}

Adjust the call methods according with your requirements.
I've set a few defaults I'm not sure are fine.

skypjack
  • 49,335
  • 19
  • 95
  • 187