std::unordered_set<Key,Hash,KeyEqual,Allocator>::end,std::unordered_set<Key,Hash,KeyEqual,Allocator>::cend
来自 cppreference.com
iterator end() noexcept; |
(1) | (自 C++11 起) |
const_iterator end() const noexcept; |
(2) | (自 C++11 起) |
const_iterator cend() const noexcept; |
(3) | (自 C++11 起) |
返回指向unordered_set
的最后一个元素之后的元素的迭代器。
此元素充当占位符;尝试访问它会导致未定义的行为。
内容 |
[编辑] 参数
(无)
[编辑] 返回值
指向最后一个元素之后的元素的迭代器。
[编辑] 复杂度
常数。
备注
因为iterator
和const_iterator
都是常数迭代器(实际上它们可能类型相同),所以无法通过任何这些成员函数返回的迭代器修改容器的元素。
[编辑] 示例
运行这段代码
#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 << ") "; } }
可能的输出
(3, 9) (1, 1) (2, 4) (3, 19) (1, 11) (2, 14)
[编辑] 参见
返回指向开始的迭代器 (公有成员函数) | |
(C++11)(C++14) |
返回指向容器或数组末尾的迭代器 (函数模板) |