Why 'enable_if' cannot be used to disable this declaration here

Question

#include<string>
#include<type_traits>

template<typename... Args>
class C {
public:
    void foo(Args&&... args) {      

    }

    template<typename = std::enable_if_t<(0 < sizeof...(Args))>>
    void foo(const Args&... args) {     

    }
};

int main() {
    C<> c;
    c.foo();
    return 0;
}

Above code works as expacted (by me :)) and calls void foo(Args&&... args) at run-time in msvc 2015 but same code fails to even compile in both gcc 7.3 and clang 6.0.0 with error:

error: no type named 'type' in 'std::enable_if'; 'enable_if' cannot be used to disable this declaration

I want to understand what is wrong with the above code and how can it be fixed?

Answer 1

SFINAE only works for deduced template arguments. In your case your method did not depend on any parameter from the method call so it is not in a deduced context. Everything is already known while instantiate the class itself.

MSVC is simply wrong in that case.

Workaround:

template<typename... Args>
class C
{
    public:
        template< typename U = std::tuple<Args...>>
            std::enable_if_t< (std::tuple_size<U>::value > 0 ) > foo(const Args&...)
            {
                std::cout << "Args  > 0 type " << std::endl;
            }

        template< typename U = std::tuple<Args...>>
            std::enable_if_t< (std::tuple_size<U>::value == 0)> foo(const Args&...)
            {
                std::cout << "Args 0 type " << std::endl;
            }
};

int main()
{
    C<>{}.foo();
    C<int>{}.foo(1);
}

I don't know why you need such a overload, because if the parameter list is empty, you simply should write an overload for such without any SFINAE stuff at all.

if your compiler is not outdated ( c++14 only ) it is much easier with using constexpr if:

template <typename... Args>
struct C
{
    void foo (const Args&... args)
    {
        if constexpr ( sizeof...(args) == 0)
        {
            std::cout << "0" << std::endl;
        }
        else
        {
            std::cout << ">0" << std::endl;
        }
    }
};

int main ()
{
    C<>    c0;
    C<int> c1;
    c0.foo();
    c1.foo(42);
}

EDIT after comment:

To avoid SFINAE you can also use specialized template classes like this:

// provide common stuff here
template <typename ... ARGS>
class CAll { protected: void DoSomeThing(){ std::cout << "Do some thing" << std::endl; } };

template<typename ... ARGS>
class C;

// special for no args
template<>
class C<>: public CAll<>
{   
    public:
        void foo() 
        {
            std::cout << "none" << std::endl; 
            this->DoSomeThing();
        }   
};  

//special for at minimum one arg
template<typename FIRST, typename ... REST>
class C<FIRST, REST...>: public CAll<FIRST, REST...>
{   
    public:
        void foo( FIRST&, REST&... )
        {   
            std::cout << "lvalue" << std::endl;
            this->DoSomeThing();
        }

        void foo( FIRST&&, REST&&... )
        {   
            std::cout << "rvalue" << std::endl;
            this->DoSomeThing();
        }   
};  

int main()
{   
    int a;
    C<>{}.foo();
    C<int>{}.foo(1);
    C<int>{}.foo(a);
}

Answer 2

As better explained by Klaus, your original code doesn't works because std::enable_if_t need to check a template of the method itself and isn't enough the template list of the class.

I propose a simplified alternative of the Klaus's solution.

First of all, you need a template parameter to check; you can use one with a default value deduced from the template paramenter of the class (Args...).

Surely you can use a type taking a std::tuple of the Args... but, taking in count that you're only interested in the number of Args... parameters, I find it's simpler to use a std::size_t template parameter initialized with the number of Args...

template <std::size_t N = sizeof...(Args)>
std::enable_if_t<N> foo (Args const & ... args)
 { std::cout << "N args" << std::endl; }  

Regarding the zero args version, there is no need to make it a template version; you can simply write it with zero parameters

void foo ()
 { std::cout << "zero args" << std::endl; }

In case of zero Args..., the not-template version take the precedence over the template one.

The following is a full compiling example

#include <iostream>
#include <type_traits>

template <typename... Args>
struct C
 {
   void foo ()
    { std::cout << "zero args" << std::endl; }

   template <std::size_t N = sizeof...(Args)>
   std::enable_if_t<N> foo(const Args&... args)
    { std::cout << "N args" << std::endl; }
};

int main ()
 {
   C<>    c0;
   C<int> c1;
   c0.foo();
   c1.foo(42);
 }