如何在模板基类中调用模板成员函数? [英] How to call a template member function in a template base class?

问题描述

在基类中调用非模板成员函数时，可以使用 using 将其名称导入派生类，然后使用它.基类中的模板成员函数也可以吗?

When calling a non-templated member function in a base class one can import its name with using into the derived class and then use it. Is this also possible for template member functions in a base class?

仅使用 using 无效(使用g ++-snapshot-20110219 -std = c ++ 0x):

Just with using it does not work (with g++-snapshot-20110219 -std=c++0x):

template <typename T>
struct A {
  template <typename T2> void f() {  }
};

template <typename T>
struct B : A<T> {
  using A<T>::f;

  template <typename T2> void g() {
    // g++ throws an error for the following line: expected primary expression before `>`
    f<T2>();
  }
};

int main() {
  B<float> b;
  b.g<int>();
}

我知道像

    A<T>::template f<T2>();

可以很好地工作，但是问题是:是否可以没有使用简单的使用声明(就像 f 不是模板函数的情况一样)?

works fine, but the question is: is it possible without and with a simple using declaration (just as it does for the case where f is not a template function)?

万一这不可能，有人知道为什么吗?

In case this is not possible, does anyone know why?

推荐答案

这可行(双关语意): this-> template f< T2>();

this works (pun intended): this->template f<T2>();

也是

template <typename T>
struct B : A<T> {
  template <typename T2> void f()
  { return A<T>::template f<T2>(); }

  template <typename T2> void g() {
    f<T2>();
  }
};

---

为什么 using 在与模板相关的模板功能上不起作用很简单-语法不允许在这种情况下使用必需的关键字.

---

Why using doesn't work on template-dependent template functions is quite simple -- the grammar doesn't allow for the required keywords in that context.

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