在表达式left()= right()中，为什么right()首先被排序? [英] In the expression left() = right(), why is right() sequenced first?

问题描述

在C ++中，表达式left() = right()计算

In C++, the expression left() = right() evaluates

1. right()
2. left()

1. right()
2. left()

按此顺序. right()首先出现，如此处所述.

in that sequence. The right() goes first, as has been discussed here.

我想不出right()首先走的原因.你能?我认为这是有原因的.否则，该标准几乎不会说出它的意思，而是要考虑:right()将返回一些结果.在机器代码级别上，CPU是否不需要在要求right()返回结果之前就知道将结果right()放回何处?

I cannot think of a reason for right() to go first. Can you? I presume that there exists a reason. Otherwise, the standard would hardly say what it says, but consider: right() will return some result. At the machine-code level, does the CPU not need to know where to put the result right() will return before asking right() to return it?

如果您碰巧知道标准委员会在想什么(因为您在会议室或已阅读备忘录)，那就太好了:我想阅读您的答案.但是，我的实际问题较为温和.我只想知道是否存在合理的原因以及该原因可能是什么.

If you happen to know what the standard committee was thinking (because you were in the room or have read the memo), that's great: I'd like to read your answer. However, my actual question is more modest. All I want to know is whether there exists a plausible reason and what that reason might be.

推荐答案

除了执行Brian所示操作时的不直观结果之外，

In addition to the unintuitive result when doing what Brian showed:

#include <map>
int main() {
    std::map<int, int> m;
    m[0] = m.size(); // before C++17 m[0] could be 0 or 1 - it was implementation defined
}

如果我们使用相同的地图，但要这样做:

If we take a the same map but do:

#include <map>
int main() {
    std::map<int, int> m;
    m[0] = Right(); // Right() may throw
}

如果Right()抛出:

在C ++ 17之前，您可以在m[0](从左到右)中获取默认的构造元素，或者根本不会创建m[0](从右到左).在C ++ 17中，根本不会创建m[0].

Before C++17 you could get a default constructed element in m[0] (left to right) or m[0] wouldn't be created at all (right to left). In C++17 m[0] will not get created at all.

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