8086 asm 中 64 位数字乘以 32 位数字 [英] Multiplying 64-bit number by a 32-bit number in 8086 asm

问题描述

我什至在启动问题的解决方案时遇到了问题.我曾尝试考虑乘法是重复加法算法，但无论我考虑什么算法，我似乎都专注于一个问题——8086 中的最大寄存器大小是 16 位.

I have been having problems even initiating the solution for the problem. I have tried considering the multiplication is repeated addition algorithm but whatever algorithm I consider, I seem to be fixated over a single issue- the max register size in 8086 is 16-bit.

data segment
num1 dw 0102h,0304h,0506h,0708h
num2 dw 0102h,0304h
res dw ?,?,?,?,?,?
data ends

code segment
assume CS:CODE, DS:DATA
start:
mov ax,DATA
mov DS,ax

.........填写代码.......

..........fill code..........

此时我被卡住了.即使是轻微的代码提示或算法也将不胜感激.

At this point I am stuck. Even a slight hint of code or just the algorithm would be appreciated.

推荐答案

对于 64 位数字乘以 16 位 CPU 的 32 位数字:

For multiplying a 64-bit number by a 32-bit number with a 16-bit CPU:

第 1 步:假设数字在基数 65536"中.例如，一个 64 位数字有 16 位十六进制数字(例如 0x0123456789ABCDEF)，并且在base 65536"中有 4 位数字(例如{0123}{4567}{89AB}{CDEF});以同样的方式，一个 32 位的数字在基数 65536"中有 2 位数字.

Step 1: Assume the numbers are in "base 65536". For example, a 64-bit number has 16 digits in hex (e.g. 0x0123456789ABCDEF) and would have 4 digits in "base 65536" (e.g. {0123}{4567}{89AB}{CDEF}); and in the same way a 32-bit number would have 2 digits in "base 65536".

第 2 步:要将一对数字相乘，将第一个数字的每个数字与第二个数字的每个数字相乘，并将其添加到结果中的正确位置.正确的位置取决于每个数字在原始数字中的位置.例如(十进制) 90 * 30 你会做9 * 3 = 27"并将它放在数百"的地方(例如2700)，因为第一个数字右边有一个数字加上另一个数字第二个数字的右边，这意味着它需要转到结果右边有 2 位数字的位置.

Step 2: To multiply a pair of numbers, multiply each digit from the first number with each digit from the second number, and add this into the right place in the result. The right place depends on the position of each digit in the original numbers. For example (in decimal) for 90 * 30 you'd do "9*3 = 27" and put it in the "hundreds" place (e.g. 2700) because there was one digit to the right in the first number plus another digit to the right in the second number, and that means it needs to go where there's 2 digits to the right in the result.

示例:

    0x0123456789ABCDEF * 0x87654321
  = {0123}{4567}{89AB}{CDEF} * {8765}{4321}

                      {0123}{4567}{89AB}{CDEF}
    *                             {8765}{4321}
    ------------------------------------------
    =                             {3600}{18CF}  (from 4321 * CDEF)
    +                       {6CEA}{484B}{0000}  (from 8765 * CDEF)
    +                       {2419}{800B}{0000}  (from 4321 * 89AB)
    +                 {48CF}{7D77}{0000}{0000}  (from 8765 * 89AB)
    +                 {1232}{E747}{0000}{0000}  (from 4321 * 4567)
    +           {24B4}{B2A3}{0000}{0000}{0000}  (from 8765 * 4567)
    +           {004C}{4E83}{0000}{0000}{0000}  (from 4321 * 0123)
    +     {0099}{E7CF}{0000}{0000}{0000}{0000}  (from 8765 * 0123)
    ------------------------------------------
    =     {009A}{0CD0}{5C28}{F5C1}{FE56}{18CF}
    ------------------------------------------
    = 0x009A0CD05C28F5C1FE5618CF

注意8086有16位乘以16位=32位结果"指令(MUL)；通过使用一个 ADD 指令后跟任意多条您需要的 ADC 指令，可以一次完成 16 位加法.

Note that 8086 has a "16 bit multiplied by 16 bit = 32-bit result" instruction (MUL); and addition can be done 16 bits at a time by using one ADD instruction followed by however many ADC instructions you need.

另请注意，您可以通过合并来避免一些添加.例如:

Also note that you can avoid some additions by merging. For example:

                      {0123}{4567}{89AB}{CDEF}
    *                             {8765}{4321}
    ------------------------------------------
    =                             {3600}{18CF}  (from 4321 * CDEF)
    +                       {6CEA}{484B}{0000}  (from 8765 * CDEF)
    +                       {2419}{800B}{0000}  (from 4321 * 89AB)
    +                 {48CF}{7D77}{0000}{0000}  (from 8765 * 89AB)
    +                 {1232}{E747}{0000}{0000}  (from 4321 * 4567)
    +           {24B4}{B2A3}{0000}{0000}{0000}  (from 8765 * 4567)
    +           {004C}{4E83}{0000}{0000}{0000}  (from 4321 * 0123)
    +     {0099}{E7CF}{0000}{0000}{0000}{0000}  (from 8765 * 0123)
    ------------------------------------------
    =                             {3600}{18CF}  (from 4321 * CDEF)
    +                 {48CF}{7D77}{0000}{0000}  (from 8765 * 89AB)
    +     {0099}{E7CF}{0000}{0000}{0000}{0000}  (from 8765 * 0123)
    +                       {6CEA}{484B}{0000}  (from 8765 * CDEF)
    +           {24B4}{B2A3}{0000}{0000}{0000}  (from 8765 * 4567)
    +                       {2419}{800B}{0000}  (from 4321 * 89AB)
    +           {004C}{4E83}{0000}{0000}{0000}  (from 4321 * 0123)
    +                 {1232}{E747}{0000}{0000}  (from 4321 * 4567)
    ------------------------------------------
    =     {0099}{E7CF}{48CF}{7D77}{3600}{18CF}  (THESE WERE MERGED WITHOUT ADDITION)
    +           {24B4}{B2A3}{6CEA}{484B}{0000}  (THESE WERE MERGED WITHOUT ADDITION)
    +           {004C}{4E83}{2419}{800B}{0000}  (THESE WERE MERGED WITHOUT ADDITION)
    +                 {1232}{E747}{0000}{0000}  (from 4321 * 4567)
    ------------------------------------------
    =     {009A}{0CD0}{5C28}{F5C1}{FE56}{18CF}
    ------------------------------------------
    = 0x009A0CD05C28F5C1FE5618CF

当然，因为您对(base 65536")数字对进行乘法的顺序无关紧要；您可以按照合并的最佳顺序进行所有乘法运算.

Of course, because it doesn't matter which order you do the multiplication of pairs of ("base 65536") digits; you can do all the multiplications in the optimum order for merging.

对于最终代码(带合并)；你最终会得到 8 个 MUL 指令、3 个 ADD 指令和大约 7 个 ADC 指令.我懒得写代码了.;)

For the final code (with merging); you'd end up with 8 MUL instructions, 3 ADD instructions and about 7 ADC instructions. I'm too lazy to write the code. ;)

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