在c编程中计算有符号32位整数的二进制数字 [英] Counting the binary digits of a signed 32-bit integer in c programming

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问题描述

我有以下函数来计算一个无符号 32 位整数中的二进制位数.

I have following function which counts the number of binary digits in an unsigned 32-bit integer.

uint32_t L(uint32_t in)
{
   uint32_t rc = 0;

   while (in)
   {
      rc++;
      in >>= 1;
   }

   return(rc);
}

如果是有符号的 32 位整数,谁能告诉我,我应该采取哪种方法?实现二进制补码是一种选择.如果您有更好的方法,请告诉我.

Could anyone tell me please in case of signed 32-bit integer, which approach i should take ? implementing two's complement is an option. if you have any better approach, please let me know.

推荐答案

关于:

uint32_t count_bits(int32_t in)
{
   uint32_t unsigned_in = (uint32_t) in;
   uint32_t rc = 0;

   while (unsigned_in)
   {
      rc++;
      unsigned_in >>= 1;
   }

   return(rc);
}

只需将 signed int 转换为 unsigned 并执行与以前相同的操作.

Just convert the signed int into an unsigned one and do the same thing as before.

顺便说一句:我想你知道 - 除非你的处理器有一个特殊的指令并且你可以访问它 - 计算位的最快实现之一是:

BTW: I guess you know that - unless your processor has a special instruction for it and you have access to it - one of the fastest implementation of counting the bits is:

int count_bits(unsigned x) {
   x = x - ((x >> 1) & 0xffffffff);
   x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
   x = (x + (x >> 4)) & 0x0f0f0f0f;
   x = x + (x >> 8);
   x = x + (x >> 16);
   return x & 0x0000003f;
}

虽然它不是最快的...

It's not the fastest though...

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