这是来自 ISO C++ 标准 14.6.4.1 实例化点的声明

1. For a function template specialization, a member function template specialization, or a specialization for a member function or static data member of a class template, if the specialization is implicitly instantiated because it is referenced from within another template specialization and the context from which it is referenced depends on a template parameter, the point of instantiation of the specialization is the point of instantiation of the enclosing specialization. Otherwise, the point of instantiation for such a specialization immediately follows the namespace scope declaration or definition that refers to the specialization.

2. If a function template or member function of a class template is called in a way which uses the definition of a default argument of that function template or member function, the point of instantiation of the default argument is the point of instantiation of the function template or member function specialization.

3. For a class template specialization, a class member template specialization, or a specialization for a class member of a class template, if the specialization is implicitly instantiated because it is referenced from within another template specialization, if the context from which the specialization is referenced depends on a template parameter, and if the
   specialization is not instantiated previous to the instantiation of the enclosing template, the point of instantiation is immediately before the point of instantiation of the enclosing template. Otherwise, the point of
   instantiation for such a specialization immediately precedes the namespace scope declaration or definition that refers to the specialization.

我无法为整个部分编写程序。从昨天开始，我正在尝试为这一部分编写一个程序。

任何人都可以为我提供这些部分的代码以供理解。

拜托，通常..我试着问了1个或多个点。在任何部分。但是我无法理解本节中的任何一点。

所以，任何人都可以提供一个代码(程序)让我理解这些部分。

最佳答案

我觉得这很令人费解，委员会有 more such fun .所以我认为我在下面可能有一些错误。所以请仔细阅读:)

第三段

For a class template specialization, a class member template specialization, or a specialization for a class member of a class template, if the specialization is implicitly instantiated because it is referenced from within another template specialization, if the context from which the specialization is referenced depends on a template parameter, and if the specialization is not instantiated previous to the instantiation of the enclosing template, the point of instantiation is immediately before the point of instantiation of the enclosing template.

换句话说，如果类模板或类模板的嵌套类被实例化，并且导致实例化的上下文依赖于模板参数，则模板/嵌套类会在类模板的实例化点之前立即实例化引用它的模板。

其他特化中的上下文可以依赖于模板参数，这是主模板、部分特化和类模板成员的情况，或者它不依赖于模板参数，这来自显式特化中的引用就是这种情况。

Otherwise [i.e. the context is nondependent], the point of instantiation for such a specialization immediately precedes the namespace scope declaration or definition that refers to the specialization.

这个区别很重要。考虑一下如果来自依赖上下文的特化的实例化点会立即出现在引用它的 namespace 范围声明或定义之前会发生什么

template<typename T, int N> 
struct A {
  typedef typename A<T, N-1>::type *type;
};

template<typename T>
struct A<T, 0> {
  typedef T type;
};

typedef A<int, 2>::type ptr;

这个模板应该添加N指针声明符。所以A<int, 2>是int**例如。

• typedef A<int, 2>::type 的背景是非依赖的，所以 A<int, 2>在 typedef 声明之前实例化。

• 在 A<int, 2> , 我们有 A<int, N-1>::type , 它出现在依赖上下文中并且引用了 A<int, 1>::type .所以标准要求我们实例化A<int, 1>同时我们实例化了A<int, 2> .

  如果我们将在引用它的命名空间作用域声明之前立即实例化它(在主模板定义之前)，我们将不会注意到偏特化 A<T, 0>加工时`A<int, N-1>::type在 A<int, 1> 内因为我们会实例化 A<int, 1> 之前那个特化。

第二段

这只是为了让在默认参数中查找的名称与在它们用于的其余函数中查找的名称一致(即它们的 POI 与其函数模板/类的成员函数的 POI 相同模板)。

第一段

这与第三段基本相同。但是，函数模板在引用它们的实体之后实例化，以便可以递归使用，如下例所示。相比之下，类模板在引用它们的实体之前被实例化，因为实体需要类类型是完整的。如果类类型的 POI 位于该实体之后，则类类型仍然不存在。

template<typename T>
void f(T);

template<typename T>
struct A {
  void g() {
    f(0);
  }

  void h() { /* do a dance */ }
};

template<typename T>
void f(T t) {
  A<T> a;
  a.h();
}

void g() {
  A<int> a; 
  a.g();
}

如果f将被实例化之前 A<int> , 那么它就无法访问 a.h()因为那时它还不存在。因此，函数模板在引用它们的实体之后实例化，而类模板在引用它们的实体之前实例化。

关于c++ - 模板 :Name resolution:Point of instantiation: -->can any one tell some more examples for this statement?，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/3810269/