模板的C ++用法＆lt;＆gt;

Question

  

可能重复：
  template<> in c++

我在c ++代码中看到了template<>。

这是有效的语法吗？如果是，那是什么意思？

Answer 1

只要您明确地专门化模板（类或函数模板）或类模板的成员，就会使用它。第一组示例使用此类模板和成员：

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

  template<typename U>
  void g();

  struct B {
    void h();
  };

  template<typename U>
  struct C {
    void h();
  };
};

---

专业和定义成员

// define A<T>::f() */
template<typename T>
void A<T>::f() { 
}

// specialize member A<T>::f() for T = int */
template<>
void A<int>::f() {
}

// specialize member A<T>::g() for T = float
template<>
template<typename T>
void A<float>::g() { 
}

// specialize member A<T>::g for T = float and
// U = int
template<>
template<>
void A<float>::g<int>() {
}


// specialize A<T>::B for T = int
template<>
struct A<int>::B {
  /* different members may appear here! */
  void i();
};

/* defining A<int>::B::i. This is not a specialization, 
 * hence no template<>! */
void A<int>::B::i() {
}

/* specialize A<T>::C for T = int */
template<>
template<typename U>
struct A<int>::C {
  /* different members may appear here! */
  void i();
};

/* defining A<int>::C<U>::i. This is not a specialization, 
 * BUT WE STILL NEED template<>.
 * That's because some of the nested templates are still unspecialized.
 */
template<>
template<typename U>
void A<int>::C<U>::i() {
}

---

专门化非成员类模板的示例

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

/* explicitly specialize 'C' */
template<>
struct C<int> {
  /* different members may appear here */ 
  void h(int);
};

/* define void C<int>::h(int). Not a specialization, hence no template<>! */
void C<int>::h(int) { 
}

/* define void C<T>::h(). */
template<typename T>
void C<T>::h() {
}

Answer 2

是的，这是有效的 - 它用于课程模板专业化...请参阅http://www.cplusplus.com/doc/tutorial/templates/