如何获取析构函数的成员函数指针？ [英] How do I get the member function pointer of a destructor?

问题描述

假设我有

  struct X {
〜X（）{} 
}; 
  

什么是类型，如何获取 X的成员函数指针::〜X（）在C ++ 03中？

我不想实际调用它，只是在SFINAE中使用如果存在给定类型的析构函数。

解决方案

您不能获取析构函数的函数指针或构造函数。然而，一个析构函数总是存在一个类型，你不能检测其 private 作为访问说明符不被 SFINAE 考虑。 p>

关于调用标量类型的析构函数的问题，标准表示 [class.dtor] / 16

：

[注意：显式调用析构函数的符号可用于任何标量类型名称（5.2.4）。允许这样做使得可以编写代码，而不必知道给定类型是否存在析构函数。例如，

typedef int I;

I * p;

p-> I ::〜I（）;

-end note]

Assume I have

struct X {
  ~X() {}
};

What's the type of and how do I get the member function pointer of X::~X() in C++03?

I don't want to actually call it, just use in SFINAE to figure if there exists a destructor for a given type.

解决方案

You can't get the function pointer of a destructor nor a constructor. Nevertheless a destructor always exist for a type, and you can't detect if its private with as access specifiers are not considered by SFINAE.

On the subject of invoking what would be the destructor of a scalar type, the standard says [class.dtor]/16:

[Note:the notation for explicit call of a destructor can be used for any scalar type name (5.2.4). Allowing this makes it possible to write code without having to know if a destructor exists for a given type. For example,

typedef int I;

I* p;

p->I::~I();

—end note]

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