检查两个整数是否在0的同一侧的最快方法 [英] Fastest way to check if two integers are on the same side of 0

问题描述

我需要检查两个整数是否在零的同一侧多次。我不在乎它是积极的还是消极的，只是它是同一面......并且表现非常重要。

I need to check if two integers are on the same side of zero many times. I don't care if it's positive or negative, just that it's the same side... and performance is very important.

目前我这样做：

if (int1 == 0 || int2 == 0) {
    // handle zero
} else if ((int1 ^ int2) > 0) {
    // different side
} else {
    // same side
}

速度提高了30％（使用 caliper ）更明显：

This is a 30% improvement in speed (tested with caliper) over the more obvious:

if ((int1 > 0 && int2 > 0) || (int1 < 0 && int2 < 0)) {

可以更快地完成吗？

如果有人想要看到我用于30％的测试框架，就在这里。我使用了卡尺0.5-rc1

注意：所有这些解决方案都检查了第一位，基本上，零是与正数相同。因此，如果这适用于您的应用程序，则无需进行零检查。

NOTE: All of these solutions check the first bit, basically, which for zero is the same as a positive number. So if that works for your application, you don't need to do a zero check.

基准列表：

• XOR：错误修正的原始答案


• Ifs：明显（ （&&）||（&&））解决方案


• 比特： @ hatchet的解决方案（>> 31）==（>> 31）


• BitAndXor： @ greedybuddha的解决方案（0x80000000）


• BitAndEquals： @ greedybuddha的解决方案已修改为使用 == 不是 ^


• XorShift： @ aaronman的解决方案（ ^）>> 31 == 0

• XOR: Original answer with bugfix
• Ifs: Obvious ((&&)||(&&)) solution
• Bits: @hatchet's solution (>>31) == (>>31)
• BitAndXor: @greedybuddha's solution (0x80000000)
• BitAndEquals: @greedybuddha's solution modified to use == not ^
• XorShift: @aaronman's solution (^)>>31 == 0

0% Scenario{vm=java, trial=0, benchmark=XOR} 1372.83 ns; ?=7.16 ns @ 3 trials
17% Scenario{vm=java, trial=0, benchmark=Ifs} 2397.32 ns; ?=16.81 ns @ 3 trials
33% Scenario{vm=java, trial=0, benchmark=Bits} 1311.75 ns; ?=3.04 ns @ 3 trials
50% Scenario{vm=java, trial=0, benchmark=XorShift} 1231.24 ns; ?=12.11 ns @ 5 trials
67% Scenario{vm=java, trial=0, benchmark=BitAndXor} 1446.60 ns; ?=2.28 ns @ 3 trials
83% Scenario{vm=java, trial=0, benchmark=BitAndEquals} 1492.37 ns; ?=14.62 ns @ 3 trials

  benchmark   us linear runtime
        XOR 1.37 =================
        Ifs 2.40 ==============================
       Bits 1.31 ================
   XorShift 1.23 ===============
  BitAndXor 1.45 ==================
BitAndEquals 1.49 ==================

vm: java
trial: 0

看起来@aaronman是赢家

Looks like @aaronman is the winner

推荐答案

（int1 ^ int2）>> 31 == 0？ / *在同一侧* /：/ *不同的一方* /; 这不一定正确处理0我不确定在这种情况下你想做什么。

编辑：还想指出，如果这是在c而不是java，它可以通过摆脱 == 0 进一步优化，因为布尔值的方式在c中工作，案例将被切换

(int1 ^ int2) >> 31 == 0 ? /*on same side*/ : /*different side*/ ; This doesn't necessarily handle 0 correctly I'm not sure what you wanted to do in that case.
also wanted to point out that if this was in c instead of java, it could be optimized further by getting rid of the == 0 because of the way that booleans work in c, the cases would be switched though

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