将一个空的for循环用作睡眠被优化掉？ [英] Will an empty for loop used as a sleep be optimized away?

问题描述

我找了一些code审查和所遇到的忙等待这样：

I am looking over some code to review and have come across a busy wait as such:

int loop = us*32;
int x;
for(x = 0;x<loop;x++)
{
    /*do nothing*/      
}

我似乎记得读这些空循环可以被优化掉。这是什么会发生在这里还是可以的这项工作？

I seem to recall reading that these empty loops can be optimized away. Is this what would happen here or can this work?

推荐答案

您是在编译器的摆布。事实上，如果它的智能将检测这是一个空操作。顺便说一句，尼尔·巴特沃思有好贴哪里他还谈到了这个问题。

You're at the mercy of the compiler. Indeed if it's smart it will detect it's a noop. Incidentally, Neil Butterworth has a nice post where he also touches on this subject.

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