C++静态模板成员，每种模板类型一个实例？

发布于 2024-08-20 06:09:11 字数 358 浏览 4 评论 0原文

通常，一个类的静态成员/对象对于具有静态成员/对象的类的每个实例是相同的。无论如何，如果静态对象是模板类的一部分并且还依赖于模板参数怎么办？例如，像这样：

template<class T>
class A{
public:
  static myObject<T> obj;
}

如果我将 A 的一个对象转换为 int，将另一个对象转换为 float，我猜会有两个 obj ，每种类型一个？

如果我创建 A 的多个对象作为 int 类型和多个 float ，它仍然是两个 obj 实例，因为我只使用两种不同的类型？

原文

Usually static members/objects of one class are the same for each instance of the class having the static member/object. Anyways what about if the static object is part of a template class and also depends on the template argument? For example, like this:

template<class T>
class A{
public:
  static myObject<T> obj;
}

If I would cast one object of A as int and another one as float, I guess there would be two obj, one for each type?

If I would create multiple objects of A as type int and multiple floats, would it still be two obj instances, since I am only using two different types?

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青衫负雪 2024-08-27 06:09:12

每个不同的模板初始化的静态成员都是不同的。这是因为每个模板初始化都是一个不同的类，由编译器第一次遇到模板的特定初始化时生成。

以下代码显示了静态成员变量不同的事实

#include <iostream>

template <class T> class Foo {
  public:
    static int bar;
};

template <class T>
int Foo<T>::bar;

int main(int argc, char* argv[]) {
  Foo<int>::bar = 1;
  Foo<char>::bar = 2;

  std::cout << Foo<int>::bar  << "," << Foo<char>::bar;
}

：

1,2

Static members are different for each diffrent template initialization. This is because each template initialization is a different class that is generated by the compiler the first time it encounters that specific initialization of the template.

The fact that static member variables are different is shown by this code:

#include <iostream>

template <class T> class Foo {
  public:
    static int bar;
};

template <class T>
int Foo<T>::bar;

int main(int argc, char* argv[]) {
  Foo<int>::bar = 1;
  Foo<char>::bar = 2;

  std::cout << Foo<int>::bar  << "," << Foo<char>::bar;
}

Which results in