如何计算SPARC Assembly中的除法余数？

一尘不染

如何计算SPARC Assembly中的除法余数？

algorithm

这是计算两个正整数相除的伪代码。
HR寄存器可节省余数，而LR可节省股息。（并最终保存根）

但是我认为该算法存在一些问题。
因为此算法有时无法恢复减法（除法是减法的延续）。

例如，6 / 3 (0110 / 011)
此算法再减去-3一次。（这种情况在我们手动计算时永远不会发生），
因此我认为该算法存在一些问题。
你不同意我吗？如何计算装配中的除法余数？

for i = 1 to num_of_bits do
(HR LR) << 1
if (HR >= 0) then
   HR = HR - DIVISOR
else
   HR = HR + DIVISOR
endif
if (HR > 0) then LR(lsb) = 1 endif
endfor

阅读 288

2020-07-28

共1个答案

一尘不染

我不会说SPARC asm，但我会说C。这是16/8 = 8,8除法算法的示例实现：

#include <stdio.h>

typedef unsigned char uint8;
typedef unsigned int uint;

int u8div(uint8* dividendh, uint8* dividendl, uint8 divisor)
{
  int i;

  if (*dividendh >= divisor)
    return 0; // overflow

  for (i = 0; i < 8; i++)
  {
    if (*dividendh >= 0x80)
    {
      *dividendh = (*dividendh << 1) | (*dividendl >> (8 - 1));
      *dividendl <<= 1;

      *dividendh -= divisor;
      *dividendl |= 1;
    }
    else
    {
      *dividendh = (*dividendh << 1) | (*dividendl >> (8 - 1));
      *dividendl <<= 1;

      if (*dividendh >= divisor)
      {
        *dividendh -= divisor;
        *dividendl |= 1;
      }
    }
  }

  return 1;
}

int u8div2(uint8* dividendh, uint8* dividendl, uint8 divisor)
{
  uint dividend = (*dividendh << 8) | *dividendl;

  if (*dividendh >= divisor)
    return 0; // overflow

  *dividendl = dividend / divisor;
  *dividendh = dividend % divisor;

  return 1;
}

int main(void)
{
  uint dividendh, dividendl, divisor;

  for (dividendh = 0; dividendh <= 0xFF; dividendh++)
    for (dividendl = 0; dividendl <= 0xFF; dividendl++)
      for (divisor = 0; divisor <= 0xFF; divisor++)
      {
        uint8 divh = dividendh, divl = dividendl, divr = divisor;
        uint8 divh2 = dividendh, divl2 = dividendl;

        printf("0x%04X/0x%02X=", (divh << 8) | divl, divr);

        if (u8div(&divh, &divl, divr))
          printf("0x%02X.0x%02X", divl, divh);
        else
          printf("ovf");

        printf(" ");

        if (u8div2(&divh2, &divl2, divr))
          printf("0x%02X.0x%02X", divl2, divh2);
        else
          printf("ovf");

        if ((divl != divl2) || (divh != divh2))
          printf(" err"); // "err" will be printed if u8div() computes incorrect result

        printf("\n");
      }

  return 0;
}

2020-07-28