std::erase_if (std::unordered_map)
< cpp | container | unordered map
| 定义于头文件 <unordered_map>
|
||
| template< class Key, class T, class Hash, class KeyEqual, class Alloc, class Pred > typename std::unordered_map<Key,T,Hash,KeyEqual,Alloc>::size_type |
(C++20 起) | |
从容器中擦除所有满足谓词 pred 的元素。等价于
auto old_size = c.size(); for (auto i = c.begin(), last = c.end(); i != last; ) { if (pred(*i)) { i = c.erase(i); } else { ++i; } } return old_size - c.size();
参数
| c | - | 要从中擦除的元素 |
| pred | - | 若应该擦除元素则对它返回 true 的谓词 |
返回值
擦除的元素数。
复杂度
线性。
示例
运行此代码
#include <unordered_map> #include <iostream> template<typename Os, typename Container> inline Os& operator<<(Os& os, Container const& cont) { os << "{"; for (const auto& item : cont) { os << "{" << item.first << ", " << item.second << "}"; } return os << "}"; } int main() { std::unordered_map<int, char> data {{1, 'a'},{2, 'b'},{3, 'c'},{4, 'd'}, {5, 'e'},{4, 'f'},{5, 'g'},{5, 'g'}}; std::cout << "Original:\n" << data << '\n'; const auto count = std::erase_if(data, [](const auto& item) { auto const& [key, value] = item; return (key & 1) == 1; }); std::cout << "Erase items with odd keys:\n" << data << '\n' << count << " items removed.\n"; }
输出：
Original:
{{5, e}{4, d}{3, c}{2, b}{1, a}}
Erase items with odd keys:
{{4, d}{2, b}}
3 items removed.参阅
| 移除满足特定判别标准的元素 (函数模板) |