无效使用非静态成员函数C ++ [英] Invalid use of non-static member function c++

问题描述

我正在遵循以下示例。但是当我编译时，它返回一个错误：

I am following this example. But when I compile, it returns an error:

非静态成员函数的无效使用

Invalid use of non-static member function

在线

void(Machine:: *ptrs[])() = 
  {
    Machine::off, Machine::on
  };

我尝试将静态添加到 void on（）; 课堂上

class Machine
{
  class State *current;
  public:
    Machine();
    void setCurrent(State *s)
    {
        current = s;
    }
    static void on(); // I add static here ...
    static void off(); // and here
};

但它抱怨

无效使用成员Machine :: current的静态成员函数

Invalid use of member Machine::current in static member function

您能帮我解决这个问题吗？

Can you help me fix this?

推荐答案

与静态成员函数或自由函数不同，非静态成员函数不会隐式转换到成员函数指针。

Unlike static member functions or free functions, non-static member functions won't implicitly convert to member function pointers.

（强调我的意思）

函数类型 T 的左值可以隐式转换为指向该值的prvalue指针功能。 这不适用于非静态成员函数，因为引用非静态成员函数的左值不存在。

An lvalue of function type T can be implicitly converted to a prvalue pointer to that function. This does not apply to non-static member functions because lvalues that refer to non-static member functions do not exist.

因此，您需要显式使用& 来获取非静态成员函数的地址（即，获取非静态成员函数的指针）。例如

So you need to use & explicitly to take the address of the non-static member functions (i.e. to get non-static member function pointers). e.g.

void(Machine:: *ptrs[])() = 
  {
    &Machine::off, &Machine::on
  };

如果将它们声明为静态成员函数，则应更改的类型ptrs （指向非成员函数指针的数组）。请注意，对于静态成员函数，最好不要显式使用& 。例如

If you declare them as static member function, you should change the type of ptrs (to array of non-member function pointers). Note that for static member function it's fine to not use & explicitly. e.g.

void(*ptrs[])() = 
  {
    Machine::off, Machine::on
  };

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