C++ Decltype for function template

user_4309934 0
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I want to create an overloaded template, that runs a function Foo() if a class contains it, else it does nothing.

class A
{
public:
    template <typename U>
    void Foo(U& u)
    {
        std::cout << "Has Foo()" << std::endl;
        // Modify u
    }
};

class B
{
    // Does not contain Foo()
};

I've been trying to run it as such

template <typename T, typename U>
decltype(std::declval<T>().Foo()) TriggerFoo(T* t, U& u)
{
    t->Foo(u);
}
template <typename T, typename U>
void TriggerFoo(T* t, U& u)
{
    std::cout << "Does not have Foo()" << std::endl;
}

int main()
{
    A a;
    B b;
    U u;     // Some type

    TriggerFoo<A, U>(&a, u);    // I want to print "Has Foo()".
    TriggerFoo<B, U>(&b, u);    // Prints "Does not have Foo()".

    return 0;
}

At the moment, both classes are passed to the "Does not have Foo()" instantiation. It compiles, but obviously it doesn't work, and it is most probably because I don't understand declval well enough. I have also tried with non-template functions and it still does not work.

Any help will be greatly appreciated.

templates c++
2 Answers

Extension to Sam's answer, if you aren't using pointers you can simplify the code further which makes it look a bit neater.

#include <iostream>
#include <type_traits>

class A
{
public:
    template <typename U>
    void Foo(U& u)
    {
        std::cout << "Has Foo()\n";
    }
};

class B
{
    // Does not contain Foo()
};

template <
  typename T,
  typename U,
  typename Z=decltype(std::declval<T>().Foo(std::declval<U&>()))>
void TriggerFoo(T& t, U& u)
{
    t.Foo(u);
}

template <typename... T>
void TriggerFoo(const T&...)
{
    std::cout << "Does not have Foo()\n";
}

class U {};

int main()
{
    A a;
    B b;
    U u;

    TriggerFoo<A, U>(a, u);    // I want to print "Has Foo()".
    TriggerFoo<B, U>(b, u);    // Prints "Does not have Foo()".

    return 0;
}
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