std::unordered_set<Key,Hash,KeyEqual,Allocator>::begin, (3) - Linux Manuals

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

NAME

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

Synopsis

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

Returns an iterator to the first element of the container.

If the container 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 are constant iterators (and may in fact be
the same type), it is not possible to mutate the elements of the container through
an iterator returned by any of these member functions.

Example

// Run this code

 #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)