是否有一个强大的保证交换在C ++的设施 [英] Is there facility for a strong guaranteed exchange in C++

问题描述

我一直在寻找一个设施来交换两个具有强大异常保证的项目。那是;使交易完全进行，或者在面临异常的情况下将目标置于初始状态。目前的标准中有什么可以允许的，我还是找不到任何东西，虽然看起来很容易编写。

I have been looking for a facility to exchange two items with a strong exception guarantee. That is; to have the exchange proceed totally, or leave the target in the initial state in the face of exceptions. Is there anything included in the current standard that allows this, I have not been able to find anything, although it appears easy to write.

我以下是一个版本我把我想要的东西放在一起，但是这是不公平的，这超出了我对强的要求。出现强力保证无法测试，但无法保证可以。

What I have below is a version I put together to try out what I am looking for, however it is noexcept, which is more than my requirement of "the strong" guarantee. Is appears the "strong guarantee" cannot be tested, but the noexcept guarantee can.

#include <iostream>
#include <type_traits>

// Guarantee to exchange l and r fully
// or (not compile) leave them in the initial state
template<typename T>
void strong_exchange(T & l, T &r) noexcept
{
    using std::swap;
    static_assert( noexcept( swap(l, r) ), "Types must be noexcept swappable");
    swap(l, r);
}

struct X
{
    X()
    {
        throw "fish";
    }

    X(X &&) = delete;
};

int main(void)
{
    int a, b;
    strong_exchange(a, b);

    X i, j;
    strong_exchange(i, j);
}

推荐答案

如果副本分配不是 noexcept （或其他方式执行副本），这是不可能的。如果它是 noexcept ， std :: swap（）应该做的事情。否则，可能没有什么可以做的。

It is impossible if copy assignment is not noexcept (or other way to perform the copy). In case it is noexcept, std::swap() should do the thing. Otherwise, there is probably nothing one can do about it.

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