分支预测在循环中的性能 [英] Performance of branch prediction in a loop

问题描述

这两个代码段之间会有明显的速度差异吗？原来，我认为第二个片段会更快，因为分支指令会遇到很少，但另一方面，分支预测器应该解决这个问题。或者它会有明显的开销，尽管可预测的模式？假设不使用条件移动指令。

Would there be any noticeable speed difference between these two snippets of code? Naively, I think the second snippet would be faster because branch instructions are encountered a lot less, but on the other hand the branch predictor should solve this problem. Or will it have a noticeable overhead despite the predictable pattern? Assume that no conditional move instruction is used.

代码段1：

for (int i = 0; i < 100; i++) {
    if (a == 3)
        output[i] = 1;
    else
        output[i] = 0;
}

$ b

Snippet 2:

if (a == 3) {
    for (int i = 0; i < 100; i++)
        output[i] = 1;
} else {
    for (int i = 0; i < 100; i++)
        output[i] = 0;
}

我不打算自己优化这些情况，

I'm not intending to optimise these cases myself, but I would like to know more about the overhead of branches even with a predictable pattern.

推荐答案

由于 a 保持不变一旦你进入循环，这两个代码片段之间应该没有太大的区别。

Since a remains unchanged once you enter into the loop, there shouldn't be much difference between the two code-snippet.

就个人而言，我喜欢使用前者，除非分支预测器无法预测真的不可能的分支，因为

Personally, I would prefer the former, unless branch predictor fails to predict the branch which is really unlikely, given that a remains unchanged in the loop.

此外，编译器可以执行此优化：

Moreover, the compiler may perform this optimization:

• 循环不切换

• Loop unswitching

，从而使两个代码段产生完全相同的机器指令。

thereby making both code-snippets emit exactly same machine instructions.

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