了解取消引用的类型-const_iterator

Question

I have this declaration:

list<string*>::const_iterator iter;

I am trying to understand whether the type of *iter is: string* , const string* or something else.

I read that cost_iterator returns reference. On the one hand the list contains string* and the iterator points to those. on the other hand it is const_iterator so the value it points to should be const and it acts as if the data in the list is const ? despite it is not really? not sure what am I missing and what is the right answer.

Answer 1

std::list<T>::const_iterator::operator* returns a const T& . Since your list stores a string* that means you'll get a reference to a string * const as the const applies to T , not to what T points to if it is a pointer.

Answer 2

The general answer is const T & , which I prefer to write T const & because it composes correctly in the next paragraph.

For T=string* , we get string * const & , or a reference to a constant pointer to a string .

This means you can't mutate the list element (via a const_iterator) by pointing it to a different string, but you can indirectly mutate the list by altering the string to which it points.

You can read the definitions of std::list<T> , but it isn't as clear as it could be about the iterator behaviour - it works out like so:

• std::list<T>::value_type is the same as T

  • so std::list<string*>::value_type is string* .

• std::list<T>::const_iterator models a const LegacyBidirectionalIterator

• LegacyBidirectionalIterator is a LegacyForwardIterator which must satisfy LegacyInputIterator

• LegacyInputIterator must support the expression *i returning type reference

• and finally, from the LegacyForwardIterator we rather skipped over

  • The type std::iterator_traits::reference must be exactly

    ... const T& otherwise (It is constant),

  (where T is the type denoted by std::iterator_traits::value_type)

  which is how we get from the type named reference to the const T & we started with.