c ++ STL map.find（）或map.operator []不能在具有const限定符的类成员函数中使用

Question

I'm confused by the following code, why it cannot be successfully compiled?

class Test { 
public:
  int GetValue( int key ) const
  {
      return testMap[key];
  }

  map<const int, const int> testMap; 
};

There is always a compling error:

error C2678: binary '[': no ​​operator found which takes "const std :: map <_Kty,_Ty>" type of the left operand operator (or there is no acceptable conversion).

I've tried to put const qualifier everywhere but it still couldn't pass. Could you tell me why?

Answer 1

operator[] is not const , because it inserts an element if one doesn't already exist with the given key. find() does have a const overload, so you can call it with a const instance or via a const reference or pointer.

In C++11, there is std::map::at() , which adds bounds checking and raises an exception if an element with the given key is not present. So you can say

class Test { 
public:
  int GetValue( int key ) const
  {
      return testMap.at(key);
  }

  std::map<const int, const int> testMap; 
};

Otherwise, use find() :

  int GetValue( int key ) const
  {
    auto it = testMap.find(key);
    if (it != testMap.end()) {
      return it->second;
    } else {
      // key not found, do something about it
    }
  }

Answer 2

You got an excellent answer by juanchopanza

Just wanted to show a boost way to return something that's not valid

with boost::optional you can return empty type

#include<boost\optional.hpp>
...

boost::optional<int> GetValue(int key){

    auto it = testMap.find(key);
    if (it != testMap.end()) {
      return it->second;
    } else {
      return boost::optional<int>();
    }
}


boost::optional<int> val = GetValue(your_key);
if(!val) //Not empty
{

}