why sfinae overload is not resolved

Question

This version works fine:

template<typename T>
struct Foo
{
    template<typename U = T>
        typename std::enable_if<std::is_same<U,A>::value>::type
        bar() { std::cout << "1" << std::endl; }

    template<typename U = T> 
        typename std::enable_if<std::is_same<U,B>::value>::type
        bar() { std::cout << "2" << std::endl; }
};  

This version fails:

template<typename T>
struct Foo2
{
    template<typename U = T, typename V = typename std::enable_if<std::is_same<U,A>::value>::type >
        V bar() { std::cout << "1" << std::endl; }

    template<typename U = T, typename V = typename std::enable_if<std::is_same<U,B>::value>::type >
        V bar() { std::cout << "2" << std::endl; }
};

with:

error: 'template template V Foo2::bar()' cannot be overloaded with 'template template V Foo2::bar()'

The difference between both versions is in the first I use the expression directly, in the second one I create a template default parameter and use that one as return type.

What is the reason to fail in the second example?

Arc

Default template parameter values are not part of function signature – Messenia

There is even a note about this common mistake in std::enable_if doc. – Gluttonize

Answer 1

Because in the case#2, the two bar are considered to be equivalent. When consider whether two function templates are equivalent or not, the default template parameters are ignored; they're not the part of function template signature. So they're considered as

template<typename U, typename V>
V bar() { std::cout << "1" << std::endl; }

template<typename U, typename V>
V bar() { std::cout << "2" << std::endl; }

As you can see, they're equivalent in fact.

(emphasis mine)

Two function templates are considered equivalent if

• they are declared in the same scope
• they have the same name
• they have identical template parameter lists
• the expressions involving template parameters in their return types and parameter lists are equivalent

The case#1 works because the return type is dependent on the template parameter and used with different expression; then they're considered not equivalent.

Negro

@Arc I fixed the statement. The point is different expression. – Negro

Thanks! So I keep in mind that the expression itself makes the difference. It feels strange that this expression is something like part of the signature... but ok, it is like it is;) – Arc