避免显式仿函数模板类型(Avoiding explicit functor template type)

 我正在玩弄玩家。 我正在使用下面的标准示例： 
 
class C {
public:
    template <typename Func>
    void foo(Func fun)
    {
        fun();
    }
};

struct S {
    void operator()() { printf ("in S\n"); }
};
....
C myClass;
myClass.foo (S());
 
 这很好用，我不需要在调用foo（）时显式提供S模板类型，它只是想出来。 但是假设我想将仿函数存储为成员变量并稍后调用它： 
 
class C {
public:
    template <typename Func>
    void foo(Func fun) {
        _myFunc = fun;
    }

    void someOtherThing() { 
        _myFunc();
    }
private:
    WHAT_IS_THIS_TYPE _myFunc;
};
 
 我现在需要让整个班级成为模板吗？ 如果是这样，编译器可以像使用单个仿函数一样推断模板类型，还是我必须明确提供它？ 谢谢。 

I'm playing with functors. I'm using the standard example below:
 
class C {
public:
    template <typename Func>
    void foo(Func fun)
    {
        fun();
    }
};

struct S {
    void operator()() { printf ("in S\n"); }
};
....
C myClass;
myClass.foo (S());
 
This works nicely and I don't need to explicitly provide the S template type in the call to foo(), it just figures it out. But suppose I want to store the functor as a member variable and call it later:
 
class C {
public:
    template <typename Func>
    void foo(Func fun) {
        _myFunc = fun;
    }

    void someOtherThing() { 
        _myFunc();
    }
private:
    WHAT_IS_THIS_TYPE _myFunc;
};
 
Do I now need to make the whole class a template? If so, can the compiler infer the template type as it did with the single functor, or must I provide it explicitly? Thanks.

原文：https://stackoverflow.com/questions/18870285