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我重载了一个函数，fn如下所示：fn(int,char)fn(int&,char&)

#include <iostream>

using namespace std;


void fn(int a, char c);
void fn(int& a, char& c);

int main()
{
   int a=10;
   char c= 'c';



  cout << "Inside main()" << endl;
  cout << hex << "&a : " << &a << endl;
  cout << hex << "&c : " << (int *)&c << endl;

   static_cast<void(*) (int&, char&)> (fn)(a, c);

    return 0;
    }


void fn(int a, char c)
{
    int tempInt;
    char tempChar;
    cout << "\n\nInside Call By Value Function " << endl;
    cout << hex << "&a : " << &a << endl;
    cout << hex << "&c : " << (int *)&c << endl;
    cout << hex << "&tempInt : " << &tempInt << endl;
    cout << hex << "&tempChar : " << (int *)&tempChar << endl;
    }


void fn(int& a, char& c)
{

    cout << "\n\nInside Call By Reference Function " << endl;
    cout << hex << "*a : " << &a << endl;
    cout << hex << "*c : " << (int*) &c << endl;

    }

调用fn(int,char)或的决议fn(int&,char&)是通过演员作出的static_cast<void(*) (int&, char&)> (fn)(a, c);

它给出输出：

$ ./Overloading
Inside main()
&a : 0x22ac5c
&c : 0x22ac5b


Inside Call By Reference Function
*a : 0x22ac5c
*c : 0x22ac5b

现在，当我把它放在下面的类中时：

#include <iostream>

using namespace std;

class Test{
public:
void fn(int a, char c);
void fn(int& a, char& c);
};



int main()
{
   int a=10;
   char c= 'c';

  Test T();
  cout << "Inside main()" << endl;
  cout << hex << "&a : " << &a << endl;
  cout << hex << "&c : " << (int *)&c << endl;

   static_cast<void(*) (int&, char&)> (T.fn)(a, c);

    return 0;
    }


void Test::fn(int a, char c)
{
    int tempInt;
    char tempChar;
    cout << "\n\nInside Call By Value Function " << endl;
    cout << hex << "&a : " << &a << endl;
    cout << hex << "&c : " << (int *)&c << endl;
    cout << hex << "&tempInt : " << &tempInt << endl;
    cout << hex << "&tempChar : " << (int *)&tempChar << endl;
    }


void Test::fn(int& a, char& c)
{

    cout << "\n\nInside Call By Reference Function " << endl;
    cout << hex << "*a : " << &a << endl;
    cout << hex << "*c : " << (int*) &c << endl;

    }

我得到以下错误：

$ g++ -Wall Overloading.cpp -o Overloading
Overloading.cpp: In function ‘int main()’:
Overloading.cpp:23:42: error: request for member ‘fn’ in ‘T’, which is of non-class type ‘Test()’

我该如何解决这个问题？如何正确调用T's fn(int&,char&)

我猜我的代码中的表达式static_cast<void(*) (int&, char&)> (T.fn)(a, c);不正确。

请帮忙。

谢谢

编辑：

我的错

编辑Test T()到Test T;

给出错误

$ g++ -Wall Overloading.cpp -o Overloading
Overloading.cpp: In function ‘int main()’:
Overloading.cpp:23:44: error: invalid static_cast from type ‘&lt;unresolved overloaded function type>’ to type ‘void (*)(int&, char&)’
4

3 回答 3

2

First: T is no variable, T is function, that returns Test and receive nothing.

Second: function-pointer is not member-function-pointer. You should use this syntax

typedef void (Test::*function)(int&, char&);
function f = &Test::fn;
(T.*f)(a, c);
于 2013-10-23T10:49:12.707 回答
0

当你这样做时Test T();，你是在说这T是一个返回类型的函数Test。但是，您的代码中不存在这样的东西。

解决方案是：

Test T;
于 2013-10-23T10:51:03.443 回答
0

See error: request for member '..' in '..' which is of non-class type: the problem is with Test T();, try ommiting the parentheses.

于 2013-10-23T10:49:07.773 回答