Is out-of-line sfinae on template member functions possible?

Question

Demo

A in class declaration of A::foo.

struct A {
    template 
    void foo(T a); 
};

A::foo is now split by sfi

Answer 1

One way to achieve this is to internally tag-dispatch:

#include 
#include 

struct A {
    template 
    void foo(T a); 

    private:

    template 
    auto implement_foo(T value, std::true_type) -> void;

    template 
    auto implement_foo(T value, std::false_type) -> void;
};

template 
void A::foo(T a ) {
    implement_foo(a, std::integral_constant4)>());
}

template 
auto A::implement_foo(T value, std::true_type) -> void
{
    std::cout << "> 4 \n";
}

template 
auto A::implement_foo(T value, std::false_type) -> void
{
    std::cout << "not > 4 \n";
}


main()
{
    A a;
    a.foo(char(1));
    a.foo(double(1));
}