std::erase, std::erase_if(std::vector)
来自 cppreference.com
定义在头文件 <vector> 中 |
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(1) | ||
template< class T, class Alloc, class U > constexpr std::vector<T, Alloc>::size_type |
(自 C++20 起) (直到 C++26) |
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template< class T, class Alloc, class U = T > constexpr std::vector<T, Alloc>::size_type |
(自 C++26 起) | |
template< class T, class Alloc, class Pred > constexpr std::vector<T, Alloc>::size_type |
(2) | (自 C++20 起) |
1) 从容器中删除所有与 value 相等的元素。等效于
auto it = std::remove(c.begin(), c.end(), value); auto r = c.end() - it; c.erase(it, c.end()); return r;
2) 从容器中删除所有满足谓词 pred 的元素。等效于
auto it = std::remove_if(c.begin(), c.end(), pred); auto r = c.end() - it; c.erase(it, c.end()); return r;
内容 |
[编辑] 参数
c | - | 要从中删除的容器 |
value | - | 要删除的值 |
pred | - | 一元谓词,如果应删除元素则返回 true。 表达式 pred(v) 必须可转换为 bool,适用于类型为(可能为常量) |
[编辑] 返回值
已删除元素的数量。
[编辑] 复杂度
线性。
注释
功能测试 宏 | 值 | Std | 功能 |
---|---|---|---|
__cpp_lib_algorithm_default_value_type |
202403 | (C++26) | 列表初始化 针对算法 (1) |
[编辑] 示例
运行此代码
#include <complex> #include <iostream> #include <numeric> #include <string_view> #include <vector> void println(std::string_view comment, const auto& c) { std::cout << comment << '['; bool first{true}; for (const auto& x : c) std::cout << (first ? first = false, "" : ", ") << x; std::cout << "]\n"; } int main() { std::vector<char> cnt(10); std::iota(cnt.begin(), cnt.end(), '0'); println("Initially, cnt = ", cnt); std::erase(cnt, '3'); println("After erase '3', cnt = ", cnt); auto erased = std::erase_if(cnt, [](char x) { return (x - '0') % 2 == 0; }); println("After erase all even numbers, cnt = ", cnt); std::cout << "Erased even numbers: " << erased << '\n'; std::vector<std::complex<double>> nums{{2, 2}, {4, 2}, {4, 8}, {4, 2}}; #ifdef __cpp_lib_algorithm_default_value_type std::erase(nums, {4, 2}); #else std::erase(nums, std::complex<double>{4, 2}); #endif println("After erase {4, 2}, nums = ", nums); }
输出
Initially, cnt = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] After erase '3', cnt = [0, 1, 2, 4, 5, 6, 7, 8, 9] After erase all even numbers, cnt = [1, 5, 7, 9] Erased even numbers: 5 After erase {4, 2}, nums = [(2,2), (4,8)]
[编辑] 另请参阅
删除满足特定条件的元素 (函数模板) | |
(C++20)(C++20) |
删除满足特定条件的元素 (niebloid) |