std:: erase_if (std::unordered_multimap)
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定义于头文件
<unordered_map>
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template
<
class
Key,
class
T,
class
Hash,
class
KeyEqual,
class
Alloc,
class
Pred
>
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(C++20 起)
(C++26 起为 constexpr) |
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从 c 中删除所有满足谓词 pred 的元素。
等同于
auto old_size = c.size(); for (auto first = c.begin(), last = c.end(); first != last;) { if (pred(*first)) first = c.erase(first); else ++first; } return old_size - c.size();
目录 |
参数
| c | - | 要从中擦除元素的容器 |
| pred | - | 谓词,当元素应被擦除时返回 true |
返回值
被擦除的元素数量。
复杂度
线性。
示例
#include <iostream> #include <unordered_map> void println(auto rem, const auto& container) { std::cout << rem << '{'; for (char sep[]{0, ' ', 0}; const auto& [key, value] : container) std::cout << sep << '{' << key << ", " << value << '}', *sep = ','; std::cout << "}\n"; } int main() { std::unordered_multimap<int, char> data { {1, 'a'}, {2, 'b'}, {3, 'c'}, {4, 'd'}, {5, 'e'}, {4, 'f'}, {5, 'g'}, {5, 'g'}, }; println("Original:\n", data); const auto count = std::erase_if(data, [](const auto& item) { const auto& [key, value] = item; return (key & 1) == 1; }); println("Erase items with odd keys:\n", data); std::cout << count << " items removed.\n"; }
可能的输出:
Original:
{{5, g}, {5, g}, {5, e}, {4, f}, {4, d}, {3, c}, {2, b}, {1, a}}
Erase items with odd keys:
{{4, f}, {4, d}, {2, b}}
5 items removed.
参见
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移除满足特定条件的元素
(函数模板) |
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(C++20)
(C++20)
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移除满足特定条件的元素
(算法函数对象) |