NASM：计数如何在一32位号码许多位被设置为1 [英] NASM: Count how many bits in a 32 Bit number are set to 1

问题描述

我有一个32位的数字，想知道计数位多少1。

I have a 32 Bit number and want to count know how many bits are 1.

我想这个伪code的：

I'm thinking of this pseudocode:

mov eax, [number]
while(eax != 0)
{
  div eax, 2
  if(edx == 1)
  {
   ecx++;
  } 
  shr eax, 1
}

有没有更有效的方法？

Is there a more efficient way?

我使用NASM在x86处理器。

I'm using NASM on a x86 processor.

（我只是用汇编开始的，所以请不要告诉我用从图书馆的extern code，因为我甚至不知道如何将它们;））

(I'm just beginning with assembler, so please do not tell me to use code from extern libraries, because I do not even know how to include them ;) )

（我刚刚发现的http://stackoverflow.com/questions/109023/best-algorithm-to-count-the-number-of-set-bits-in-a-32-bit-integer其中还包含我的解决办法。还有其他发布的解决方案，但不幸的是我似乎无法弄清楚，我将如何把它们写在汇编语言）

(I just found http://stackoverflow.com/questions/109023/best-algorithm-to-count-the-number-of-set-bits-in-a-32-bit-integer which also contains my solution. There are other solutions posted, but unfortunatly I can't seem to figure out, how I would write them in assembler)

推荐答案

最有效的方法（在执行时间，无论如何）是有一个查找表。显然，你不会有一个四十亿项表，但你可以下破32位到8位的块，只需要256项表，或进一步分解成4位块，只需要16个条目。祝你好运！

The most efficient way (in terms of execution time, anyway) is to have a lookup table. Obviously you're not going to have a 4-billion entry table, but you could break the 32 bits down into 8-bit chunks and only need a 256-entry table, or further down into 4-bit chunks and only need 16 entries. Good luck!

这篇关于NASM：计数如何在一32位号码许多位被设置为1的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！