std::reverse_iterator<Iter>::base
From cppreference.com
< cpp | iterator | reverse iterator
iterator_type base()const; | (constexpr since C++17) | |
Returns the underlying iterator.
Contents |
[edit]Return value
[edit]Notes
The base iterator refers to the element that is next (from the iterator_type
perspective) to the element the reverse_iterator
is currently pointing to. That is &*(this->base()-1)==&*(*this).
[edit]Example
Run this code
#include <iostream>#include <iterator>#include <vector> int main(){std::vector<int> v ={0, 1, 2, 3, 4, 5}; using RevIt =std::reverse_iterator<std::vector<int>::iterator>; constauto it = v.begin()+3; RevIt r_it{it}; std::cout<<"*it == "<<*it <<'\n'<<"*r_it == "<<*r_it <<'\n'<<"*r_it.base() == "<<*r_it.base()<<'\n'<<"*(r_it.base() - 1) == "<<*(r_it.base()-1)<<'\n'; RevIt r_end{v.begin()}; RevIt r_begin{v.end()}; for(auto it = r_end.base(); it != r_begin.base();++it)std::cout<<*it <<' ';std::cout<<'\n'; for(auto it = r_begin; it != r_end;++it)std::cout<<*it <<' ';std::cout<<'\n';}
Output:
*it == 3 *r_it == 2 *r_it.base() == 3 *(r_it.base() - 1) == 2 0 1 2 3 4 5 5 4 3 2 1 0
[edit]See also
accesses the pointed-to element (public member function) |