What is the use of const overloading in C++?

In C++, a function's signature depends partly on whether or not it's const. This means that a class can have two member functions with identical signatures except that one is const and the other is not. If you have a class like this, then the compiler will decide which function to call based on the object you call it on: if it's a const instance of the class, the const version of the function will be called; if the object isn't const, the other version will be called.

In what circumstances might you want to take advantage of this feature?

25177 次浏览
小开

It's there so you can make the compiler enforce whether you return a const object or a regular one, and still maintain the same method signature. There's an in-depth explanation at Const Correctness.

小开

You might want to use it to decide whether or not to return a const reference to an object or not. The STL's containers use a const overloaded begin() and end() function to decide whether to return a const_iterator or a normal iterator.

小开
最佳答案

This really only makes sense when the member function returns a pointer or a reference to a data member of your class (or a member of a member, or a member of a member of a member, ... etc.). Generally returning non-const pointers or references to data members is frowned upon, but sometimes it is reasonable, or simply very convenient (e.g. [] operator). In such cases, you provide a const and a non-const versions of the getter. This way the decision on whether or not the object can be modified rests with the function using it, which has a choice of declaring it const or non-const.

小开

Have a look at the behaviour of std::map::operator[]. The const version throws an error if you try to reference an invalid key, but the non-const version does an insert. The insertion behaviour is much handier than having to use std::map::insert (and will do an overwrite, moreover) but can't work for a const map.

小开 
#include <iostream>
using namespace std;
class base
{


public:
void fun() const
{
cout<<"have fun";
}
void fun()
{
cout<<"non const";
}


};
int main()
{
base b1;
b1.fun(); //does not give error
return 0;
}

Here compiler won't give any error, because in case of const functions compiler converts this pointer to const this*. this third argument separates these two functions.


提交答案