copy(...) 和 copy(seq, ...) 的关系

Question

std::copy 有无执行策略参数之间是否存在正式关系？ 在实践中或在标准中。

例如，会不会只是这样，

namespace std{
    template<class It>
    It copy(std::execution::sequenced_policy, It first, It last, It d_first){
        return std::copy(first, last, d_first);
    }
}

或

namespace std{
    template<class It>
    It copy(std::execution::sequenced_policy, It first, It last, It d_first){
    //    using std::copy; // may not be needed
        return copy(first, last, d_first);
    }
}

请注意，在第一个版本中，我还需要重载 copy(par::seq, ...)。

或者是这样

namespace std{
    template<class It>
    It copy(std::execution::sequenced_policy, It first, It last, It d_first){
        ... not defined at all in terms of other `copy(It, ...)` or `std::copy(It, ...)`
    }
}

原因是我想为一种特殊的迭代器重载复制算法（在自定义命名空间中）。

Answer 1

[execpol.seq]中提到的一个区别是

During the execution of a parallel algorithm with the execution::sequenced_policy policy, if the invocation of an element access function exits via an uncaught exception, terminate() shall be called.

演示：

#include <execution>
#include <iostream>
#include <stdexcept>

struct C {
    C() {}
    C& operator=(const C&) {
        throw std::runtime_error("copy failed");
    }
};

int main() {
    C a[1];
    C b[1];

    try {
        std::copy(std::begin(a), std::end(a), std::begin(b));
    } catch(const std::runtime_error& ex) {
        std::cout << "Exception: " << ex.what() << "\n";
    }

    try {
        std::copy(std::execution::seq, std::begin(a), std::end(a), std::begin(b));
    } catch(const std::runtime_error& ex) {
        std::cout << "Exception: " << ex.what() << "\n";
    }
}

可能的输出：

Exception: copy failed
terminate called after throwing an instance of 'std::runtime_error'
  what():  copy failed
Aborted (core dumped)

copy(...) 和 copy(seq, ...) 的关系

Relationship between copy(...) and copy(seq, ...)

c++

stl

copy

argument-dependent-lookup

c++11