what is the difference between const_iterator and iterator? [duplicate]

Answer 1

There is no performance difference.

A const_iterator is an iterator that points to const value (like a const T* pointer); dereferencing it returns a reference to a constant value (const T&) and prevents modification of the referenced value: it enforces const-correctness.

When you have a const reference to the container, you can only get a const_iterator.

Edited: I mentionned “The const_iterator returns constant pointers” which is not accurate, thanks to Brandon for pointing it out.

Edit: For COW objects, getting a non-const iterator (or dereferencing it) will probably trigger the copy. (Some obsolete and now disallowed implementations of std::string use COW.)

Answer 2

if you have a list a and then following statements

list<int>::iterator it; // declare an iterator
    list<int>::const_iterator cit; // declare an const iterator 
    it=a.begin();
    cit=a.begin();

you can change the contents of the element in the list using “it” but not “cit”, that is you can use “cit” for reading the contents not for updating the elements.

*it=*it+1;//returns no error
    *cit=*cit+1;//this will return error

Answer 3

Performance wise there is no difference. The only purpose of having const_iterator over iterator is to manage the accessesibility of the container on which the respective iterator runs. You can understand it more clearly with an example:

std::vector<int> integers{ 3, 4, 56, 6, 778 };

If we were to read & write the members of a container we will use iterator:

for( std::vector<int>::iterator it = integers.begin() ; it != integers.end() ; ++it )
       {*it = 4;  std::cout << *it << std::endl; }

If we were to only read the members of the container integers you might wanna use const_iterator which doesn't allow to write or modify members of container.

for( std::vector<int>::const_iterator it = integers.begin() ; it != integers.end() ; ++it )
       { cout << *it << endl; }

NOTE: if you try to modify the content using *it in second case you will get an error because its read-only.