将零、负和正映射到 0、1、2 的无分支代码

问题描述

编写一个无分支函数，如果两个有符号整数之间的差为零、负数或正数，则返回 0、1 或 2.

Write a branchless function that returns 0, 1, or 2 if the difference between two signed integers is zero, negative, or positive.

这是一个带有分支的版本:

Here's a version with branching:

int Compare(int x, int y)
{
    int diff = x - y;
    if (diff == 0)
        return 0;
    else if (diff < 0)
        return 1;
    else
        return 2;
}

这是一个可能更快的版本，具体取决于编译器和处理器:

Here's a version that may be faster depending on compiler and processor:

int Compare(int x, int y)
{
    int diff = x - y;
    return diff == 0 ? 0 : (diff < 0 ? 1 : 2);
}

你能想出一个没有分支的更快的吗?

Can you come up with a faster one without branches?

总结

我进行基准测试的 10 个解决方案具有相似的性能.实际数字和获胜者因编译器 (icc/gcc)、编译器选项(例如，-O3、-march=nocona、-fast、-xHost)和机器而异.佳能的解决方案在许多基准测试中表现良好，但性能优势同样微乎其微.我很惊讶在某些情况下，某些解决方案比带有分支的幼稚解决方案慢.

The 10 solutions I benchmarked had similar performance. The actual numbers and winner varied depending on compiler (icc/gcc), compiler options (e.g., -O3, -march=nocona, -fast, -xHost), and machine. Canon's solution performed well in many benchmark runs, but again the performance advantage was slight. I was surprised that in some cases some solutions were slower than the naive solution with branches.

推荐答案

int Compare(int x, int y) {
     return (x < y) + (y < x) << 1;
}

仅按位?猜猜并且 > 不计算，然后?

Bitwise only? Guess < and > don't count, then?

int Compare(int x, int y) {
    int diff = x - y;
    return (!!diff) | (!!(diff & 0x80000000) << 1);
}

但是有那个讨厌的-.

换个方向.

嗯，再试一次:

int Compare(int x, int y) {
    int diff = y - x;
    return (!!diff) << ((diff >> 31) & 1);
}

但我猜测 !! 没有标准的 ASM 指令.此外，<< 可以替换为 +，具体取决于哪个更快...

But I'm guessing there's no standard ASM instruction for !!. Also, the << can be replaced with +, depending on which is faster...

点点滴滴很有趣！

嗯，我刚刚了解了 setnz.

我还没有检查汇编器的输出(但这次我确实测试了一下)，如果幸运的话，它可以节省一整条指令！:

I haven't checked the assembler output (but I did test it a bit this time), and with a bit of luck it could save a whole instruction!:

subl  %edi, %esi
setnz %eax
sarl  $31, %esi
andl  $1, %esi
sarl  %eax, %esi
mov   %esi, %eax
ret

漫步很有趣.

我需要睡觉.

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