在自己的成员函数中构造类时如何强制类模板参数推导？

Question

Consider following code:

struct A {};

template <typename T> struct B
{
    B(T) {}
    auto foo() {return B(A{});} // error: no matching function for call to 'B<int>::B(A)'
};

auto foo() {return B(A{});} // compiles

int main()
{
    foo();
    B b(0);
    b.foo();
}

Try it live

I understand why B::foo() doesn't compile: Inside of struct B<T> , B (as an injected-class-name) means B<T> unless it's explicitly used as a template. Which in this case prevents class template argument deduction.

Let's say I can't do auto foo() {return B<A>(A{});} since my actual code relies on slightly elaborate user-provided deduction guides.

The question is: How do I force class template argument deduction when constructing B inside of B::foo ?

_{I hope I'm not missing something obvious.}

Answer 1

您对其进行限定，使其不是注入的类名。

auto foo() {return ::B(A{});}

Answer 2

Another option is to use a function to do the type deduction for you.

template <typename T> B<T> make_b(T t) { return B<T>(t); }

and use

auto foo() {return make_b(A{});}