Question

我将函数fn重载为fn(int,char)＆amp; fn(int&,char&)如下所示：

#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;

    }

致电fn(int,char)或fn(int&,char&)的决议是通过施法static_cast<void(*) (int&, char&)> (fn)(a, c);

进行的

它提供输出：

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


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

现在当我把它放在如下的课程中时：

#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？如何正确拨打T's fn(int&,char&)

我想在我的代码中表达式static_cast<void(*) (int&, char&)> (T.fn)(a, c);是不正确的。

请帮忙。

由于

修改

我的错误

将Test T()修改为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 ‘<unresolved overloaded function type>’ to type ‘void (*)(int&, char&)’

Answer 1

首先：T不是变量，T是函数，它返回Test并且什么都不收。

第二：函数指针不是成员函数指针。您应该使用此语法

typedef void (Test::*function)(int&, char&);
function f = &Test::fn;
(T.*f)(a, c);

Answer 2

请参阅error: request for member '..' in '..' which is of non-class type：问题出在Test T();，请尝试省略括号。

Answer 3

当您执行Test T();时，您说T是一个返回类型为Test的函数。但是你的代码中不存在这样的东西。

解决方案是：

Test T;

重载类成员函数

3 个答案: