将类型分配给typename关键字

Question

I was looking at the following template function and was wondering what happens behind the scenes when we assign the enable_if type to the typename keyword:

template <typename T, typename = std::enable_if<std::is_pointer<T>::value>::type>
                    // ^^ What happens here?
void foo()
{
    std::cout << "T is a pointer!" << std::endl;
}

Other then the obvious SFINAE, does the compiler actually do something with it? Perhaps it generates some kind of anonymous type?

Thanks

Answer 1

I'll assume you're using a lenient compiler which doesn't complain about the missing typename keyword after = .

It is an unnamed template parameter which is otherwise ignored, along with a default template argument which will be used if no other template argument is specified.

It means that foo<void *>() resolves to foo<void *, std::enable_if<std::is_pointer<void *>::value>::type>() , which is foo<void *, void>() .

It means that foo<int>() would resolve to foo<int, std::enable_if<std::is_pointer<int>::value>::type>() , except it gets rejected because std::enable_if<std::is_pointer<int>::value> doesn't have any type member.

It means that foo<int, int>() works and prints "T is a pointer!", since the default argument is not used if a template argument is explicitly specified.

That last one means this is probably not a good idea.

Answer 2

No, we're not assigning the enable_if type to the typename keyword, the template parameter is just ommitted here, because it won't be used:

template <typename T, typename Anonymous_Template_Parameter = std::enable_if<std::is_pointer<T>::value>::type>
//                             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^

And = makes std::enable_if<std::is_pointer<T>::value>::type> as its default argument here.