C ++使用const函数调用指向成员的指针 [英] C++ Call pointer to member with a map from a const function

问题描述

我有一个指向成员的指针的地图，声明为：

  std :: map< char，T *）（const T& const T&）> op_map; 
 

我在我的类的构造函数中直接用指针填充我的地图：

  op_map ['+'] =& Operand :: op_add; 
  

例如，op_add源代码为：

  T op_add（const T& a，const T& b）{
 return a + b; 
} 
  

而我想从const函数调用我的指针成员。这里是源代码：

  IOperand * res_int32（char op，const IOperand& rhs）const {
 IOperand * res = const_cast< IOperand *>（& rhs）; 
 Operand< int> * tmp = dynamic_cast< Operand< int> *>（res）; 
 T res_calc =（this-> * op_map [op]）（_ value，（T）tmp-> getValue（））; 
} 
  

但它让我总是一个错误：

  Operand.hpp：70：64：error：passing'const std :: map< char，double（Operand< double> ，const double&），std :: less< char>，std :: allocator< std :: pair< const char，double（Oper double& > >'as'this'argument'std :: map< _Key，_Tp，_Compare，_Alloc> :: mapped_type& std :: map< _Key，_Tp，_Compare，_Alloc> :: operator []（const key_type&）[with _Key = char，_Tp = double（Operand< double> :: *）（const double& const double& ，_Compare = std :: less< char>，_Alloc = std :: allocator< std :: pair< const char，double（Operand< double> :: *）（const double& const double& >，std :: map< _Key，_Tp，_Compare，_Alloc> :: mapped_type = double（Operand< double> :: *）（const double& const double&），std :: map <_Key，_Tp，_Compare ，_Alloc> :: key_type = char]'丢弃限定符[-fpermissive] 
 Operand.hpp：70：64：error：从'const Operand< double> * const'到'Operand& [-fpermissive] 
  

您有任何解决方案吗？

谢谢。

解决方案

operator [] t应用于 const 映射，因为如果未找到键，它将插入一个新的元素。

如果未找到密钥，则会抛出异常：

<$ p $ p> T res_calc =（this-> * op_map.at（op））（_ value，（T）tmp-> getValue（））;
^^^^^^


在C ++ 03中，使用 find ：

  map_type :: const_iterator found = op_map.find （op）。 
 if（found！= op_map.end（））{
 T res_calc =（this-> *（found-> second））（_ value，（T）tmp-> getValue ）; 
} else {
 // handle error 
} 
  

还需要将映射中的成员函数的类型更改为

  T（Operand :: *）（const T & const T&）const 
 ^^^^^ 
  

从 const 成员函数中调用 this 。

I have a map of pointer to member declared as :

std::map<char, T (Operand::*)(const T &, const T &)> op_map;

I fill my map with pointer to member directly in the constructor of my class with :

op_map['+'] = &Operand::op_add;

For example, op_add source code is :

  T op_add(const T & a, const T & b) {
    return a + b;
  }

And I want to call my pointer to member from a const function. Here is the source code :

  IOperand *res_int32(char op, const IOperand & rhs) const {
    IOperand *res = const_cast<IOperand *>(&rhs);
    Operand<int> *tmp = dynamic_cast<Operand<int>*>(res);
    T res_calc = (this->*op_map[op])(_value, (T)tmp->getValue());
  }

But it makes me always an error :

Operand.hpp:70:64: error: passing ‘const std::map<char, double (Operand<double>::*)(const double&, const double&), std::less<char>, std::allocator<std::pair<const char, double (Operand<double>::*)(const double&, const double&)> > >’ as ‘this’ argument of ‘std::map<_Key, _Tp, _Compare, _Alloc>::mapped_type& std::map<_Key, _Tp, _Compare, _Alloc>::operator[](const key_type&) [with _Key = char, _Tp = double (Operand<double>::*)(const double&, const double&), _Compare = std::less<char>, _Alloc = std::allocator<std::pair<const char, double (Operand<double>::*)(const double&, const double&)> >, std::map<_Key, _Tp, _Compare, _Alloc>::mapped_type = double (Operand<double>::*)(const double&, const double&), std::map<_Key, _Tp, _Compare, _Alloc>::key_type = char]’ discards qualifiers [-fpermissive]
Operand.hpp:70:64: error: invalid conversion from ‘const Operand<double>* const’ to ‘Operand<double>*’ [-fpermissive]

Have you got any solution ?

Thank you.

解决方案

operator[] can't be applied to a const map, since it inserts a new element if the key is not found.

In C++11, there is an at function which throws an exception if the key is not found:

T res_calc = (this->*op_map.at(op))(_value, (T)tmp->getValue());
                           ^^^^^^^

In C++03, you'll need to use find:

map_type::const_iterator found = op_map.find(op);
if (found != op_map.end()) {
    T res_calc = (this->*(found->second))(_value, (T)tmp->getValue());
} else {
    // handle error
}

You'll also need to change the type of the member functions in the map to

T (Operand::*)(const T &, const T &) const
                                     ^^^^^

in order to call them on this from a const member function.

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