std::unordered_set<Key,Hash,KeyEqual,Allocator>::begin, std::unordered_set<Key,Hash,KeyEqual,Allocator>::

From cppreference.com

iterator begin() noexcept;
(1) (since C++11)
const_iterator begin() const noexcept;
(2) (since C++11)
const_iterator cbegin() const noexcept;
(3) (since C++11)

Returns an iterator to the first element of the unordered_set.

If the unordered_set is empty, the returned iterator will be equal to end().

range-begin-end.svg

Parameters

(none)

Return value

Iterator to the first element.

Complexity

Constant.

Notes

Because both iterator and const_iterator

Example

#include <iostream>
#include <unordered_set>
 
struct Point { double x, y; };
 
int main()
{
    Point pts[3] = {{1, 0}, {2, 0}, {3, 0}};
 
    // points is a set containing the addresses of points
    std::unordered_set<Point*> points = { pts, pts + 1, pts + 2 };
 
    // Change each y-coordinate of (i, 0) from 0 into i^2 and print the point
    for (auto iter = points.begin(); iter != points.end(); ++iter)
    {
        (*iter)->y = ((*iter)->x) * ((*iter)->x); // iter is a pointer-to-Point*
        std::cout << "(" << (*iter)->x << ", " << (*iter)->y << ") ";
    }
    std::cout << '\n';
 
    // Now using the range-based for loop, we increase each y-coordinate by 10
    for (Point* i : points)
    {
        i->y += 10;
        std::cout << "(" << i->x << ", " << i->y << ") ";
    }
}

Possible output:

(3, 9) (1, 1) (2, 4) 
(3, 19) (1, 11) (2, 14)

See also

returns an iterator to the end
(public member function)
(C++11)(C++14)
returns an iterator to the beginning of a container or array
(function template)