以下哪一种技术是整数除2的最佳选择,为什么?

技巧1:

x = x >> 1;

技巧2:

x = x / 2;

这里x是整数。


当前回答

一般右移分为:

q = i >> n; is the same as: q = i / 2**n;

这有时被用来加快程序的速度,但以清晰度为代价。我觉得你不应该这么做。编译器足够智能,可以自动执行加速。这意味着,以清晰度为代价,你不会获得任何好处。

看看《实用c++编程》的这一页。

其他回答

使用除法(/),假设这样更清楚。编译器将相应地进行优化。

哪一个是最好的选择,为什么整数除以2?

这取决于你对最佳的定义。

如果你想让你的同事讨厌你,或者让你的代码难以阅读,我肯定会选择第一个选择。

如果你想把一个数除以2,就用第二个数。

这两者是不等价的,如果数字是负的或在更大的表达式中,它们的行为是不一样的——bitshift的优先级比+或-低,除法的优先级更高。

您应该编写代码来表达其意图。如果您关心的是性能,不要担心,优化器在这类微优化方面做得很好。

X/Y是正确的…和" >> "移位运算符..如果我们想要二除一个整数,我们可以使用(/)被除数运算符。移位运算符用于移位位。

x = x / 2; x / = 2;我们可以这样用…

我想说有几件事需要考虑。

Bitshift should be faster, as no special computation is really needed to shift the bits, however as pointed out, there are potential issues with negative numbers. If you are ensured to have positive numbers, and are looking for speed then I would recommend bitshift. The division operator is very easy for humans to read. So if you are looking for code readability, you could use this. Note that the field of compiler optimization has come a long way, so making code easy to read and understand is good practice. Depending on the underlying hardware, operations may have different speeds. Amdal's law is to make the common case fast. So you may have hardware that can perform different operations faster than others. For example, multiplying by 0.5 may be faster than dividing by 2. (Granted you may need to take the floor of the multiplication if you wish to enforce integer division).

如果您追求的是纯粹的性能,我建议您创建一些可以执行数百万次操作的测试。对执行进行多次采样(您的样本量),以确定哪一个在统计上最适合您的操作系统/硬件/编译器/代码。

我们有很多理由支持使用x = x / 2;以下是一些例子:

it expresses your intent more clearly (assuming you're not dealing with bit twiddling register bits or something) the compiler will reduce this to a shift operation anyway even if the compiler didn't reduce it and chose a slower operation than the shift, the likelihood that this ends up affecting your program's performance in a measurable way is itself vanishingly small (and if it does affect it measurably, then you have an actual reason to use a shift) if the division is going to be part of a larger expression, you're more likely to get the precedence right if you use the division operator: x = x / 2 + 5; x = x >> 1 + 5; // not the same as above signed arithmetic might complicate things even more than the precedence problem mentioned above to reiterate - the compiler will already do this for you anyway. In fact, it'll convert division by a constant to a series of shifts, adds, and multiplies for all sorts of numbers, not just powers of two. See this question for links to even more information about this.

简而言之,当你真正想要进行乘法或除法运算时,编写移位代码并没有什么好处,除了可能会增加引入错误的可能性。自从编译器不够聪明到在适当的时候优化这类事情到移位以来,已经过去了。