Question

我试图以更好的方式掌握指针功能概念。所以我有一个非常简单和有效的例子：

#include <iostream>

using namespace std;

int add(int first, int second)
{
    return first + second;
}

int subtract(int first, int second)
{
    return first - second;
}

int operation(int first, int second, int (*functocall)(int, int))
{
    return (*functocall)(first, second);
}

int main()
{
    int  a, b;
    int  (*plus)(int, int);
    int  (*minus)(int, int);
    plus = &add;
    minus = &subtract;
    a = operation(7, 5, add);
    b = operation(20, a, minus);
    cout << "a = " << a << " and b = " << b << endl;
    return 0;
}

到目前为止一切顺利，现在我需要将类中的函数分组，并根据我使用的函数指针选择加或减。所以我只做了一个小修改：

#include <iostream>

using namespace std;

class A
{
public:
int add(int first, int second)
{
    return first + second;
}

int subtract(int first, int second)
{
    return first - second;
}

int operation(int first, int second, int (*functocall)(int, int))
{
    return (*functocall)(first, second);
}
};

int main()
{
    int  a, b;
    A a_plus, a_minus;
    int (*plus)(int, int) = A::add;
    int (*minus)(int, int) = A::subtract;
    a = a_plus.operation(7, 5, plus);
    b = a_minus.operation(20, a, minus);
    cout << "a = " << a << " and b = " << b << endl;
    return 0;
}

，明显的错误是：

ptrFunc.cpp: In function ‘int main()’:
ptrFunc.cpp:87:29: error: invalid use of non-static member function ‘int A::add(int, int)’
ptrFunc.cpp:88:30: error: invalid use of non-static member function ‘int A::subtract(int, int)’

因为我没有指定要调用哪个对象（我现在不想使用静态方法）

修改几条评论和答案表明非静态版本（正如我所写）是不可能的。（感谢所有人）所以，以下列方式修改类也不会起作用：

#include <iostream>

using namespace std;

class A
{
    int res;
public:
    A(int choice)
    {
        int (*plus)(int, int) = A::add;
        int (*minus)(int, int) = A::subtract;
        if(choice == 1)
            res = operation(7, 5, plus);
        if(choice == 2)
            res = operation(20, 2, minus);
        cout << "result of operation = " << res;
    }
int add(int first, int second)
{
    return first + second;
}

int subtract(int first, int second)
{
    return first - second;
}

int operation(int first, int second, int (*functocall)(int, int))
{
    return (*functocall)(first, second);
}
};

int main()
{
    int  a, b;
    A a_plus(1);
    A a_minus(2);
    return 0;
}

生成了这个错误：

ptrFunc.cpp: In constructor ‘A::A(int)’:
ptrFunc.cpp:11:30: error: cannot convert ‘A::add’ from type ‘int (A::)(int, int)’ to type ‘int (*)(int, int)’
ptrFunc.cpp:12:31: error: cannot convert ‘A::subtract’ from type ‘int (A::)(int, int)’ to type ‘int (*)(int, int)’

我可以知道如何解决这个问题吗？

感谢

Answer 1

声明成员方法的函数指针的语法是：

int (A::*plus)(int, int) = &A::add;
int (A::*minus)(int, int) = &A::subtract;

要调用成员方法，请使用。*或 - ＆gt; * operator：

 (a_plus.*plus)(7, 5);

另请查看http://msdn.microsoft.com/en-us/library/b0x1aatf(v=vs.80).aspx

希望这有帮助。

完整代码：

     #include <iostream>

    using namespace std;

    class A
    {
    public:
    int add(int first, int second)
    {
        return first + second;
    }

    int subtract(int first, int second)
    {
        return first - second;
    }

    int operation(int first, int second, int (A::*functocall)(int, int))
    {
        return (this->*functocall)(first, second);
    }
    };

    int main()
    {
        int  a, b;
        A a_plus, a_minus;
        int (A::*plus)(int, int) = &A::add;
        int (A::*minus)(int, int) = &A::subtract;
        a = a_plus.operation(7, 5, plus);
        b = a_minus.operation(20, a, minus);
        cout << "a = " << a << " and b = " << b << endl;
        return 0;
    }

Answer 2

你不能将非静态成员函数作为参数传递给easy。根据您的需求，我认为最好覆盖运营商：http://www.learncpp.com/cpp-tutorial/92-overloading-the-arithmetic-operators/

但是如果你真的需要它们作为实际的成员函数 - 只需将它们设置为静态。

Answer 3

您对代码所做的编辑仍然是错误的，因为它不会使成员函数保持静态。您需要通过添加static说明符来使加法，减法等函数保持静态：

#include <iostream>

using namespace std;

class A
{
    int res;
public:
    A(int choice)
    {
        int (*plus)(int, int) = A::add;
        int (*minus)(int, int) = A::subtract;
        if(choice == 1)
            res = operation(7, 5, plus);
        if(choice == 2)
            res = operation(20, 2, minus);
        cout << "result of operation = " << res;
    }
static int add(int first, int second)
{
    return first + second;
}

static int subtract(int first, int second)
{
    return first - second;
}

static int operation(int first, int second, int (*functocall)(int, int))
{
    return (*functocall)(first, second);
}
};

Answer 4

请参阅以下代码。函数调用正在工作而不会使它们静止。

class A
{
  public:
  int add(int first, int second)
  {
      return first + second;
  }

  int subtract(int first, int second)
  {
      return first - second;
  }

  int operation(int first, int second, int(A::*functocall)(int, int))
  {
      return (this->*functocall)(first, second);
  }
};
//typedef int(A::*PFN)(int, int) ;
int main()
{
    int  a, b;
    A a_plus, a_minus;
    a = a_plus.operation(7, 5, &A::add);
    b = a_minus.operation(20, a, &A::subtract);
    cout << "a = " << a << " and b = " << b << endl;
    return 0;
}

函数指针生成'无效使用非静态成员函数'错误

4 个答案: