将成员函数指针传递给 C++ 中的成员对象
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Passing member function pointer to member object in c++
提问by Moomin
I have a problem with using a pointer to function in C++. Here is my example:
我在 C++ 中使用指向函数的指针时遇到问题。这是我的例子：
#include <iostream>
using namespace std;
class bar
{
public:
void (*funcP)();
};
class foo
{
public:
bar myBar;
void hello(){cout << "hello" << endl;};
};
void byebye()
{
cout << "bye" << endl;
}
int main()
{
foo testFoo;
testFoo.myBar.funcP = &byebye; //OK
testFoo.myBar.funcP = &testFoo.hello; //ERROR
return 0;
}
Compilator returns an error at testFoo.myBar.funcP = &testFoo.hello;:
编译器在testFoo.myBar.funcP = &testFoo.hello;以下位置返回错误：
ISO C++ forbids taking the address of a bound member function to form a pointer to member function. Say '&foo::hello'
cannot convert 'void (foo::)()' to 'void ()()' in assignment
ISO C++ 禁止使用绑定成员函数的地址来形成指向成员函数的指针。说 '&foo::hello'
无法在赋值中将“void (foo:: )()”转换为“void ()()”
So i tried it like this:
所以我这样试过：
class bar
{
public:
void (*foo::funcP)();
};
But now the compilator adds one more:
但是现在编译器又增加了一个：
'foo' has not been declared
'foo' 尚未声明
Is there a way make it work?
有没有办法让它工作？
Thanks in advance for suggestions
预先感谢您的建议
采纳答案by Bill
Taking everyone's suggestions together, your final solution will look like:
综合大家的建议，您的最终解决方案将如下所示：
#include <iostream>
using std::cout;
usind std::endl;
class foo; // tell the compiler there's a foo out there.
class bar
{
public:
// If you want to store a pointer to each type of function you'll
// need two different pointers here:
void (*freeFunctionPointer)();
void (foo::*memberFunctionPointer)();
};
class foo
{
public:
bar myBar;
void hello(){ cout << "hello" << endl; }
};
void byebye()
{
cout << "bye" << endl;
}
int main()
{
foo testFoo;
testFoo.myBar.freeFunctionPointer = &byebye;
testFoo.myBar.memberFunctionPointer = &foo::hello;
((testFoo).*(testFoo.myBar.memberFunctionPointer))(); // calls foo::hello()
testFoo.myBar.freeFunctionPointer(); // calls byebye()
return 0;
}
The C++ FAQ Litehas some guidance on how to simplify the syntax.
在C ++ FAQ精简版提供了如何简化的语法一些指导。
Taking Chris' idea and running with it, you could get yourself something like this:
采纳 Chris 的想法并运行它，你可以得到这样的东西：
#include <iostream>
using std::cout; using std::endl;
class foo;
typedef void (*FreeFn)();
typedef void (foo::*MemberFn)();
class bar
{
public:
bar() : freeFn(NULL), memberFn(NULL) {}
void operator()(foo* other)
{
if (freeFn != NULL) { freeFn(); }
else if (memberFn != NULL) { ((other)->*(memberFn))(); }
else { cout << "No function attached!" << endl; }
}
void setFreeFn(FreeFn value) { freeFn = value; memberFn = NULL; }
void setMemberFn(MemberFn value) { memberFn = value; freeFn = NULL; }
private:
FreeFn freeFn;
MemberFn memberFn;
};
class foo
{
public:
bar myBar;
void hello() { cout << "foo::hello()" << endl; }
void operator()() { myBar(this); }
};
void bye() { cout << "bye()" << endl; }
int main()
{
foo testFoo;
testFoo();
testFoo.myBar.setMemberFn(&foo::hello);
testFoo();
testFoo.myBar.setFreeFn(&bye);
testFoo();
return 0;
}
回答by UncleBens
As the error says, methods belong to the class, not to individual instances. For this reason pointers to free functions and pointers to non-static methods are completely different things. You'll also need an instance to call the method on.
正如错误所说，方法属于类，而不是单个实例。因此，指向自由函数的指针和指向非静态方法的指针是完全不同的东西。您还需要一个实例来调用该方法。
//declaring and taking the address of a foo's method
void (foo::*method)() = &foo::hello; //as the compiler nicely suggests
//calling a function through pointer
free_func();
//calling a method through pointer
foo instance;
(instance.*method)();
You can use libraries like Boost.Bindand Boost.Function(also in std::tr1 I think) to abstract away the difference and also bind an instance to the method:
您可以使用Boost.Bind和Boost.Function（我认为也在 std::tr1 中）之类的库来抽象差异并将实例绑定到方法：
#include <iostream>
#include <boost/bind.hpp>
#include <boost/function.hpp>
using namespace std;
class foo
{
public:
void hello(){cout << "hello" << endl;};
};
void byebye()
{
cout << "bye" << endl;
}
int main()
{
foo testFoo;
boost::function<void()> helloFunc(boost::bind(&foo::hello, testFoo));
boost::function<void()> byeFunc(byebye);
helloFunc();
byeFunc();
return 0;
}
回答by R Samuel Klatchko
To make your second option work, declare foo so the compiler knows that it is a class.
要使您的第二个选项起作用，请声明 foo 以便编译器知道它是一个类。
Also note that your function pointer syntax is incorrect. The *comes just before the name of the variable:
另请注意，您的函数指针语法不正确。该*时逢变量的名称之前：
class foo;
class bar
{
public:
void (foo::*funcP)();
};
回答by Andrey
forward foo's declaration in front of bar:
在 bar 前转发 foo 的声明：
class foo;
