how to overload [index] for list in C++

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how to overload [index] for list in C++.
I want to use list, like below code.

C++

list<t> T;
t O;
O=T[1];
T[1]=O

Posted 19-Jul-15 21:26pm

kave2011

Updated 19-Jul-15 21:33pm

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[no name] 20-Jul-15 3:56am

You can't. std::list does not have subscripting access. Use std::vector

kave2011 20-Jul-15 3:58am

don't exist no way??!!
because I need list instead of vector.

[no name] 20-Jul-15 4:02am

You can't use what does not exist unless you create it yourself. http://stackoverflow.com/questions/1112724/why-isnt-there-an-operator-for-a-stdlist

Philippe Mori 20-Jul-15 13:08pm

As already mentionned, it does not make much sense. You don't use a list if you want to access element by index. For large lists, it would be extremely innefficient.

kave2011 21-Jul-15 15:31pm

sure

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Stefan_Lang · Accepted Answer · 2015-07-19T22:07:00

First, lets be clear about one thing: if you need to access container elements by index, then you shouldn't use a list! There is no index operator for std::list, because it simply doesn't make sense!

Think about your use cases, i. e. what, exactly you want to do with your container: how you create it, how you modify it over time, how you access elements, etc.. Then consider whether there is really any situation where you want really just one specific element and the only suitable way to access that element would be through an index (note that there are alternate ways, e. g. you could store an iterator rather than an index!) Most of the time, you never need an index. And if you do, std::vector is very probably a much better choice.

Second, you can extend an existing class by deriving your own class from it. E. g. the following code compiles:

C++

template <class T>
class mylist : public std::list<T> {
public:
   T operator[](int i) const {
      return front();//replace with actual code
   }
   T& operator[](int i) {
      return front();//replace with actual code
   }
};
void foo_list_index() {
   mylist<int> l;
   int x = 3;
   l.push_back(2);
   l.push_back(5);
   int y = l[1];
   l[0] = x;
}

Note however that, depending on the class you derive from, your class may unintentionally hide certain features of the base class - most notably constructors. So if you are planning to go that route you better spend some time learning about the base class' constructors and how to create appropriate 'overrides' in your derived class.