将C ++函数指针分配给同一对象的成员函数 [英] Assigning C++ function pointers to member functions of the same object

问题描述

如何让test.calculate中的函数指针赋值（也许是其余的）工作？

  #include < iostream> 
 
 class test {
 
 int a; 
 int b; 
 
 int add（）{
 return a + b; 
} 
 
 int multiply（）{
 return a * b; 
} 
 
 public：
 int calculate（char operatr，int operand1，int operand2）{
 int（* opPtr）（）= NULL; 
 
 a = operand1; 
 b = operand2; 
 
 if（operatr =='+'）
 opPtr = this。* add; 
 if（operatr =='*'）
 opPtr = this。* multiply; 
 
 return opPtr（）; 
} 
}; 
 
 int main（）{
 test t; 
 std :: cout<< t.calculate（'+'，2，3）; 
} 
  

解决方案

首先， int（* opPtr）（）= NULL; 不是指向成员的指针函数，它的一个指向自由函数的指针。声明一个成员函数指针如下：

int（test :: * opPtr）（）= NULL; / p>

其次，在采取成员函数的地址时，您需要指定类作用域，例如：

  if（operatr =='+'）opPtr =& test :: add; 
 if（operatr =='*'）opPtr =& test :: multiply; 
  

最后，要通过成员函数指针调用，还有一些特殊的语法：

  return（this-> * opPtr）（）; 
  

这是一个完整的工作示例：

  #include< iostream> 
 
 class test {
 
 int a; 
 int b; 
 
 int add（）{
 return a + b; 
} 
 
 int multiply（）{
 return a * b; 
} 
 
 public：
 int calculate（char operatr，int operand1，int operand2）{
 int（test :: * opPtr）（）= NULL; 
 
 a = operand1; 
 b = operand2; 
 
 if（operatr =='+'）opPtr =& test :: add; 
 if（operatr =='*'）opPtr =& test :: multiply; 
 
 return（this-> * opPtr）（）; 
} 
}; 
 
 int main（）{
 test t; 
 std :: cout<< t.calculate（'+'，2，3）; 
} 
  

How do I get the function pointer assignments (and maybe the rest) in test.calculate to work?

#include <iostream>

class test {

    int a;
    int b;

    int add (){
        return a + b;
    }

    int multiply (){
        return a*b;
    }

    public:
    int calculate (char operatr, int operand1, int operand2){
        int (*opPtr)() = NULL;

        a = operand1;
        b = operand2;

        if (operatr == '+')
            opPtr = this.*add;
        if (operatr == '*')
            opPtr = this.*multiply;

        return opPtr();
    }
};

int main(){
    test t;
    std::cout << t.calculate ('+', 2, 3);
}

解决方案

There are several problems with your code.

First, int (*opPtr)() = NULL; isn't a pointer to a member function, its a pointer to a free function. Declare a member function pointer like this:

int (test::*opPtr)() = NULL;

Second, you need to specify class scope when taking the address of a member function, like this:

if (operatr == '+') opPtr = &test::add;
if (operatr == '*') opPtr = &test::multiply;

Finally, to call through a member function pointer, there is special syntax:

return (this->*opPtr)();

Here is a complete working example:

#include <iostream>

class test {

    int a;
    int b;

    int add (){
        return a + b;
    }

    int multiply (){
        return a*b;
    }

    public:
    int calculate (char operatr, int operand1, int operand2){
        int (test::*opPtr)() = NULL;

        a = operand1;
        b = operand2;

        if (operatr == '+') opPtr = &test::add;
        if (operatr == '*') opPtr = &test::multiply;

        return (this->*opPtr)();
    }
};

int main(){
    test t;
    std::cout << t.calculate ('+', 2, 3);
}

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