[英]Is there a function to remove an element from a vector without shifting it in the c++ stdlib?
If I use vector.erase()
such as如果我使用
vector.erase()
例如
std::vector<int> n = { 3, 5, 6, 7 };
n.erase(n.begin() + 1);
vector
will shift all the elements after the element removed down. vector
将删除元素后的所有元素向下移动。
Is there a function in the C++ standard library that will not shift the elements? C++ 标准库中是否存在不会移动元素的 function? Like putting the back element at the removed element and pop back?
就像把后元素放在被移除的元素上然后弹回来?
There isn't anything built in, but it's trivial to do yourself.没有任何内置的东西,但自己做是微不足道的。
std::vector<int> n = { 3, 5, 6, 7 }; // create vector
n[1] = std::move(n.back()); // move last element to removal location
n.pop_back(); // remove unneeded element.
If you don't want to shift any elements, then in C++20 you can model this with a vector of std::optional
s and the ranges library如果您不想移动任何元素,那么在 C++20 中,您可以 model 使用
std::optional
的向量和范围库
#include <cstddef>
#include <optional>
#include <ranges>
#include <vector>
template<typename T>
auto remove_element(std::vector<std::optional<T>> &v, size_t i) {
v[i] = std::nullopt;
return v
| std::views::filter(&std::optional<T>::has_value)
| std::views::transform([](auto &&o) {return *o;});
}
#include <iostream>
int main() {
std::vector<std::optional<int>> v{1, 2, 3, 4, 5, 6};
for (int i : remove_element(v, 3))
cout << i << ','; // 1,2,3,5,6,
cout << '\n';
}
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