Question

我在C ++中使用函数数组的经验不多。我需要使用一个函数数组，其中数组包含来自不同对象的函数。这里有一些虚拟代码来说明我想要实现的内容。

class base_class
{
public:
  virtual int function1(int arg1, int arg2);
  virtual int function2(int arg1, int arg2);
};

class derived_class : public base_class
{
public:
  int function1(int arg1, int arg2) { /* ... */ };
  int function2(int arg1, int arg2) { /* ... */ };
  // ...
};

typedef int (*functions) (int arg1, int arg2); 

int main()
{
  derived_class object1;  
  derived_class object2;  

  functions func_instance[4];
  func_instance[0] = object1.function1;
  func_instance[1] = object1.function2;
  func_instance[2] = object2.function1;
  func_instance[3] = object2.function2;
  // ...
}

我无法让它工作，它会抛出以下错误：

error: argument of type int () (int , int) does not match int (*) (int, int)

Answer 1

最简单的方法是使用std::function和std::bind。

#include <functional>
#include <array>

derived_class object1;
derived_class object2;

std::array< std::function<int(int, int)>, 2> > 
  arr = {{ std::bind(&derived_class::function1, &object1)
         , std::bind(&derived_class::function2, &object1)}};
         // and so on

请注意，object1和object2已受地址约束。只要绑定的函数存活，您就需要保持它们的存活。如果你只是在object1表达式中写bind，那么对象的副本将存储在绑定函数中，并且不会出现范围问题。

C ++ 03 手工卷，单用途粘合剂类型的完整示例：

#include <iostream>

struct base
{
  virtual void f0() { std::cout << "f0 base" << std::endl; }
  virtual void f1() { std::cout << "f1 base" << std::endl; }
};

struct derived : base
{
  void f0() { std::cout << "f0 derived" << std::endl; }
  void f1() { std::cout << "f1 derived" << std::endl; }
};

// typedef for a Pointer to a member function 
// of base that accepts arguments and returns void
typedef void (base::*MemberPtrType)(void);

// we pack a pointer to a class and a pointer to a member function together
struct binder
{
  MemberPtrType ptr;
  base*         base;

  // shortcut to call this thing, you might also want 
  // to have an overload of operator() and a result_type typedef
  void call()
  { ((base)->*(ptr))(); }
};


int main()
{
  base b;
  derived d;

  // initialize a 4 element array of binder, the first argument is the function,
  // the second is the object to call it on
  // 
  // remember that b and d need to be alive as long as 
  // you want to call something in this array
  binder arr[4] = { {&base::f0, &b}, {&base::f1, &b}, 
                    {&base::f0, &d}, {&base::f1, &d}};

  // now walk through the array and call each
  for(binder* begin = arr; begin != arr + 4; ++begin)
  {
    begin->call();
  }


  return 0;
}

函数数组包含来自不同对象的函数

1 个答案: