std::copy, std::copy_if
翻译:https://en.cppreference.com/w/cpp/algorithm/copy
定义在头文件 <algorithm>
函数声明
template< class InputIt, class OutputIt > OutputIt copy( InputIt first, InputIt last, OutputIt d_first ); // C++ 20 template< class InputIt, class OutputIt > constexpr OutputIt copy( InputIt first, InputIt last, OutputIt d_first ); // C++ 20 template< class ExecutionPolicy, class ForwardIt1, class ForwardIt2 > ForwardIt2 copy( ExecutionPolicy&& policy, ForwardIt1 first, ForwardIt1 last, ForwardIt2 d_first ); // C++ 17 template< class InputIt, class OutputIt, class UnaryPredicate > OutputIt copy_if( InputIt first, InputIt last, OutputIt d_first, UnaryPredicate pred ); // C++ 11 ~ C++ 20 template< class InputIt, class OutputIt, class UnaryPredicate > constexpr OutputIt copy_if( InputIt first, InputIt last, OutputIt d_first, UnaryPredicate pred ); // C++ 20 template< class ExecutionPolicy, class ForwardIt1, class ForwardIt2, class UnaryPredicate > ForwardIt2 copy_if( ExecutionPolicy&& policy, ForwardIt1 first, ForwardIt1 last, ForwardIt2 d_first, UnaryPredicate pred ); // C++ 17
函数解释:
复制 [first, last) 范围内的元素到 d_first 开始中
1) 元素复制的范围是 [first,last),是从 first 开始,但是不包括 last。如果 d_first 这个位置是包含在 [first,last)中的话,那么这个行为(指的是复制这个动作)是未定义的。在这种情况下,考虑使用 std::copy_backward。
3)当 pred 返回是真时,才会复制元素。元素的复制相对的顺序是收到保护的,如果复制的源范围和目的范围有重叠的话,那么该行为是未定义的。
2)4)和 1)3)相同。。。。
函数参数:
first,last -- 复制的元素范围
d_first:复制的目的范围的起始位置
policy:使用执行策略
pred:当返回值为真时,才进行复制动作。
返回值:
返回带有复制结果的 OutputIt iterator。
函数实现:
// first version template<class InputIt, class OutputIt> OutputIt copy(InputIt first, InputIt last, OutputIt d_first) { for (; first != last; (void)++first, (void)++d_first) { *d_first = *first; } return d_first; }
// second version template<class InputIt, class OutputIt, class UnaryPredicate> OutputIt copy_if(InputIt first, InputIt last, OutputIt d_first, UnaryPredicate pred) { for (; first != last; ++first) { if (pred(*first)) { *d_first = *first; ++d_first; } } return d_first; }
例子:
#include <algorithm> #include <iostream> #include <vector> #include <iterator> #include <numeric> int main() { std::vector<int> from_vector(10); std::iota(from_vector.begin(), from_vector.end(), 0); std::vector<int> to_vector; std::copy(from_vector.begin(), from_vector.end(), std::back_inserter(to_vector)); // or, alternatively, // std::vector<int> to_vector(from_vector.size()); // std::copy(from_vector.begin(), from_vector.end(), to_vector.begin()); // either way is equivalent to // std::vector<int> to_vector = from_vector; std::cout << "to_vector contains: "; std::copy(to_vector.begin(), to_vector.end(), std::ostream_iterator<int>(std::cout, " ")); std::cout << '\n'; std::cout << "odd numbers in to_vector are: "; std::copy_if(to_vector.begin(), to_vector.end(), std::ostream_iterator<int>(std::cout, " "), [](int x) { return x % 2 != 0; }); std::cout << '\n'; std::cout << "to_vector contains these multiples of 3:\n"; to_vector.clear(); std::copy_if(from_vector.begin(), from_vector.end(), std::back_inserter(to_vector), [](int x) { return x % 3 == 0; }); for (int x : to_vector) std::cout << x << ' '; std::cout << '\n'; }
输出:
to_vector contains: 0 1 2 3 4 5 6 7 8 9 odd numbers in to_vector are: 1 3 5 7 9 to_vector contains these multiples of 3: 0 3 6 9