C++成员函数指针

Question

考虑下一节课

class Foo
{
    typedef bool (*filter_function)(Tree* node, std::list<std::string>& arg);

    void filter(int filter, std::list<std::string>& args)
    {
        ...
        if (filter & FILTER_BY_EVENTS) {
            do_filter(events_filter, args, false, filter & FILTER_NEGATION);
        }
        ...
    }

    void do_filter(filter_function ff, std::list<std::string>& arg, 
        bool mark = false, bool negation = false, Tree* root = NULL)
    {
        ...
    }

    bool events_filter(Tree* node, std::list<std::string>& arg)
    {
        ...
    }
};

只有当events_filter是do_filter成员时，才能将events_filter作为参数传递给static。但我不想让它成为static。我是否可以将指向成员函数的指针传递给另一个函数？可能正在使用boost库(类似函数)之类。

谢谢。

Answer 1

bool (Foo::*filter_Function)(Tree* node, std::list<std::string>& arg)

将为您提供一个成员函数指针。你通过一个：

Foo f;
f.filter(&Foo::events_filter,...);

并以下列方式引用：

(this->*ff)(...); // the parenthesis around this->*ff are important

如果您希望能够传递任何遵循语法的函数/函式，请使用Boost.Function，或者如果编译器支持它，则使用std::function。

class Foo{
  typedef boost::function<bool(Tree*,std::list<std::string>&)> filter_function;

  // rest as is
};

然后传递你想要的任何东西。函子、自由函数(或静态成员函数)，甚至是带有Boost.Bind或std::bind的非静态成员函数(同样，如果编译器支持它)：

Foo f;
f.do_filter(boost::bind(&Foo::events_filter,&f,_1,_2),...);

Answer 2

//member function pointer is declared as
bool (*Foo::filter_function)(Tree* node, std::list<std::string>& arg);

//Usage

//1. using object instance!
Foo foo;
filter_function = &foo::events_filter;

(foo.*filter_function)(node, arg); //CALL : NOTE the syntax of the line!


//2. using pointer to foo

(pFoo->*filter_function)(node, arg); //CALL: using pFoo which is pointer to Foo

(this->*filter_function)(node, arg); //CALL: using this which is pointer to Foo