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

X = X / 2;是合适的代码使用..但是一个操作取决于你自己的程序,你想要产生怎样的输出。

其他回答

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。

第一个看起来像除法吗?不。如果你想除法,用x / 2。如果可能的话,编译器可以优化它使用位移位(它被称为强度减少),这使得它成为一个无用的微优化,如果你自己做它。

只是加了一个注释

在一些基于vm的语言中,x *= 0.5通常会更快——尤其是actionscript,因为变量不需要被检查是否除以0。

Mod 2, test for = 1。不知道c中的语法,但这可能是最快的。