汇编器64b师 [英] Assembler 64b division

问题描述

我需要一些简单的方法来在 x86 的汇编器中对 64b 无符号整数进行除法.我的号码保存在两个 32b 寄存器 EDX:EAX 中，我需要将结果放回 EDX:EAX.因子是 32b 整数.一些代码，好吗?

I need some easy way to divide 64b unsigned integers in assembler for x86. My number is saved in two 32b registers EDX:EAX and I need to put result back to EDX:EAX. Factor is in 32b integer. Some code, please?

推荐答案

如果我正确解释了你的问题(特别是 Factor is in 32b integer 部分)，你想要除以 64 位红利由一个 32 位的除数得到一个 64 位的商.

If I interpret your question correctly (particularly the part Factor is in 32b integer), you want to divide a 64-bit dividend by a 32-bit divisor and get a 64-bit quotient.

如果这种解释是正确的，那么在 32 位代码中实际上很容易做到.

If that interpretation is correct, then it's actually easy to do in 32-bit code.

这个想法是将被除数的两个一半"除以除数，然后将第一次除法的余数重复用于第二次除法.

The idea is that you divide both "halves" of the dividend by the divisor and reuse the remainder from the first division for the second division.

说明如何操作的 C 代码:

C code illustrating how to do it:

#include <stdio.h>
#include <limits.h>

#define C_ASSERT(expr) extern char CAssertExtern[(expr)?1:-1]

#if UINT_MAX >= 0xFFFFFFFF
typedef unsigned int uint32;
#else
typedef unsigned long uint32;
#endif
typedef unsigned long long uint64;

typedef unsigned long ulong;

// Make sure uint32=32 bits and uint64=64 bits
C_ASSERT(sizeof(uint32) * CHAR_BIT == 32);
C_ASSERT(sizeof(uint64) * CHAR_BIT == 64);

int div64by32eq64(uint64* dividend, uint32 divisor)
{
  uint32 dividendHi = (uint32)(*dividend >> 32);
  uint32 dividendLo = (uint32)*dividend;
  uint32 quotientHi;
  uint32 quotientLo;

  if (divisor == 0)
    return 0;

  // This can be done as one 32-bit DIV, e.g. "div ecx"
  quotientHi = dividendHi / divisor;
  dividendHi = dividendHi % divisor;

  // This can be done as another 32-bit DIV, e.g. "div ecx"
  quotientLo = (uint32)((((uint64)dividendHi << 32) + dividendLo) / divisor);

  *dividend = ((uint64)quotientHi << 32) + quotientLo;

  return 1;
}

int main(void)
{
  static const struct
  {
    uint64 dividend;
    uint32 divisor;
  } testData[] =
  {
    { 1 , 0 },
    { 0xFFFFFFFFFFFFFFFFULL, 1 },
    { 0xFFFFFFFFFFFFFFFFULL, 2 },
    { 0xFFFFFFFF00000000ULL, 0xFFFFFFFFUL },
    { 0xFFFFFFFFFFFFFFFFULL, 0xFFFFFFFFUL },
  };
  int i;

  for (i = 0; i < sizeof(testData)/sizeof(testData[0]); i++)
  {
    uint64 dividend = testData[i].dividend;
    uint32 divisor = testData[i].divisor;

    printf("0x%016llX / 0x%08lX = ", dividend, (ulong)divisor);

    if (div64by32eq64(&dividend, divisor))
      printf("0x%016llX
", dividend);
    else
      printf("division by 0 error
");
  }

  return 0;
}

输出(ideone):

0x0000000000000001 / 0x00000000 = division by 0 error
0xFFFFFFFFFFFFFFFF / 0x00000001 = 0xFFFFFFFFFFFFFFFF
0xFFFFFFFFFFFFFFFF / 0x00000002 = 0x7FFFFFFFFFFFFFFF
0xFFFFFFFF00000000 / 0xFFFFFFFF = 0x0000000100000000
0xFFFFFFFFFFFFFFFF / 0xFFFFFFFF = 0x0000000100000001

现在汇编中的等效除法代码(NASM 语法)不检查除以 0:

And now the equivalent division code in assembly (NASM syntax) without checking for division by 0:

; 64-bit dividend
mov edx, 0xFFFFFFFF
mov eax, 0xFFFFFFFF

; 32-bit divisor
mov ecx, 0xFFFFFFFF

push eax
mov eax, edx
xor edx, edx
div ecx ; get high 32 bits of quotient
xchg eax, [esp] ; store them on stack, get low 32 bits of dividend
div ecx ; get low 32 bits of quotient
pop edx ; 64-bit quotient in edx:eax now
; edx:eax should now be equal 0x0000000100000001

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