I want to take iterator from const std::vector<T>, but cannot find a way to explain that to the compiler.
This is how could it look like:
(const std::vector<T>)::iterator
The workaround: typedef or using
template<typename T>
using const_vec = const std::vector<T>;
const_vec::iterator ...
Update. std::vector::iterator is example. The question is about C++ syntax.
The const qualifier does not change anything about the definition of the type. So the iterator type in const std::vector<T> is the same as in std::vector<T>. Just use
std::vector<T>::iterator
or, if T is a template parameter, making this a dependent name:
typename std::vector<T>::iterator
You probably don't want iterator though, because there is no function of std::vector<T> that could return an iterator on a const qualified instance. You probably want const_iterator instead.
Generally you don't need to access the iterator type aliases anyway. You can obtain the correct type from auto in a variable definition:
const std::vector<T> vec{/* ... */};
auto it = vec.begin();
or with decltype if you need the type itself:
const std::vector<T> vec{/* ... */};
using iterator = decltype(vec.begin());
You can use std::vector<int>::const_iterator like:
const std::vector<int> v{ 1, 2, 3, 4, 5 };
std::vector<int>::const_iterator it = v.begin();
or just take advantage of auto like:
const std::vector<int> v{ 1, 2, 3, 4, 5 };
auto it = v.begin();
