在c++中如何将成员函数指针传递给成员对象?

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我在C++中使用指针函数有个问题。以下是我的例子:

#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;
}

编译器返回一个错误:

ISO C++ forbids taking the address of a bound member function to form a pointer to member function. Say '&foo::hello'

所以我就这样做了:

class bar
{
public:
    void (*foo::funcP)();
};

但是现在编译器又添加了一个:

'foo' has not been declared

有办法让它起作用吗?

提问于
用户回答回答于

你的最终解决方案如下:

#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; 
} 

关于如何简化语法

你就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;
}
用户回答回答于

如错误所述,方法属于类,而不属于单个实例。因为这个原因指向自由函数和指向非静态方法的指针是完全不同的东西。你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)();

你可以使用像Boost.BindBoost.Function一个实例的方法:

#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;
}

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