为什么我得到错误"非模板'f'作为模板"

I'm trying to understand where to use template and typename and I came across a problem I can't quite get around. I have a template-function f<T> which uses the type passed to it (which will be a class) to call the template-member-function .f<T>. I think my use of typename in the function body is correct, however, I keep getting the following error:

我想知道在哪里使用模板和typename，我遇到了一个我无法解决的问题。我有一个模板函数f ，它使用传递给它的类型(将是一个类)来调用模板成员函数。f 。我认为我在函数体中使用typename是正确的，但是，我一直得到以下错误:

source.cpp: In function 'void f()':
source.cpp:11:19: error: non-template 'f' used as template
source.cpp:11:19: note: use 'typename T::C::template f' to indicate that it is a template

源。cpp:在函数‘void f()’中:源。错误:非模板'f'用作模板源。注意:使用'typename T:::C::template f'表示它是一个模板

struct A {
  struct C {
     template <typename T> void f() {}
  };
};

template <typename T> void f() {

  typename T::C::f<int>();

}

int main() {

  f<A>();

}

Notice how on the last error it advises to use 'typename T::C::template f' instead. So I made the following change in accordance:

注意，它建议在最后一个错误上使用“typename T::C::template f”。所以我按照以下的方式做了更改:

// ...
typename T::C::template f<int>();
// ...

I did as it said, but then I received the next line of error:

我照它说的做了，但接着我收到了下一行错误信息:

error: no class template named 'f' in 'struct A::C'

错误:在“struct A::C”中没有名为“f”的类模板

I believe this error is incorrect in that there is in fact a public template function named f in struct A::C. What could I be doing wrong here?

我认为这个错误是不正确的，因为在结构a中有一个名为f的公共模板函数::C。我在这里做错了什么?

1 个解决方案

#1

Assuming we make f static so it can be called without an instance, you don't need typename because you're not creating any ambiguities with a dependent type (i.e. C can't be a variable because you use it with :: just after). Here is the correct syntax:

假设我们使f为静态的，这样就可以不使用实例来调用它，那么就不需要typename，因为您没有创建任何依赖类型的歧义(例如，C不能是变量，因为您使用它与:::just after)一起使用)。这里有正确的语法:

struct A {
  struct C {
     template <typename T>
     static void f() {}
  };
};

template <typename T> void f() {
  T::C::template f<int>();
}

If you wanted not to make it static, you'd create a A::C and use .template f<int>():

如果不希望它是静态的，可以创建a::C并使用.template f ():

struct A {
  struct C {
     template <typename T>
     static void f() {}
  };
};

template <typename T> void f() {
  typename T::C c;
//^^^^^^^^ Now we DO need typename to disambiguate.

  c.template f<int>();
}