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

Knuth说:

过早的优化是万恶之源。

所以我建议用x /= 2;

这样代码很容易理解,而且我认为这种形式的操作优化,对处理器来说不会有太大的区别。

其他回答

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。

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。

我说这些是为了参加编程比赛。一般来说,他们有非常大的输入,除以2会发生很多次,已知输入是正的或负的。

X >>1比X /2好。我在ideone.com上运行了一个程序,其中发生了超过10^10除以2的运算。X /2花了将近5.5s,而X >>1花了将近2.6s。

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。