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将二进制中的 2 位数字拆分为二进制中的一位数

发布于 2024-12-11 06:34:32 字数 248 浏览 1 评论 0原文

我有 47 个，二进制为 0010 1111。我需要拆分 47，所以我得到 4 个二进制 0000 0100 和 7 个二进制 0000 0111。

我看到这个答案但我想知道，因为Intel 8085 没有 div 指令，我必须通过连续减法进行除法，如何获得模数？

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墨落成白 2024-12-18 06:34:32

如果您的处理器没有除法指令（或 BCD 转换指令，这将是另一种方式），您只需重复进行减法即可。其伪代码类似于：

    val = 47

    units = val
    tens = 0
loop:
    if units < 10 goto done
    units = units - 10
    tens = tens + 1
    goto loop
done:
    ;; tens = 4, units = 7.

对于相当基本的编码风格表示歉意，我只是试图使其更接近汇编语言的外观。否则，我会使用 while 循环:-)

一旦模数降至 10 以下，模数就是单位变量中剩下的值。

为了完整性（如果这是家庭作业，可能还有额外的分数，轻推轻推，眨眼眨眼），因为单个八位字节可以处理最多 255 个：

    val = 247

    units = val
    tens = 0
    hundreds = 0
loop1:
    if units < 100 goto loop2
    units = units - 100
    hundreds = hundreds + 1
    goto loop1
loop2:
    if units < 10 goto done
    units = units - 10
    tens = tens + 1
    goto loop2
done:
    ;; hundreds = 2, tens = 4, units = 7.

作为此方法有效的证明，您可以尝试以下 Python 代码：

val = 47

units = val
tens = 0
while units >= 10:
    units = units - 10
    tens = tens + 1

print "Tens = %d, Units = %d"%(tens,units)

它确实输出：

Tens = 4, Units = 7

就使用这些值来驱动 LED 器件而言，简单的查找表可能是最简单的。构造一个需要写入两个 8 位内存地址的值数组，然后使用该数字查找这些值以写入设备。

我将使用简化的七段 LED（加上小数点），因为它更容易理解：

    a
   ---
f |   | b
   -g-
e |   | c
   ---     . h
    d

假设在此示例中，您将一个字节 abcdefgh 写入内存映射位置，并且将位写入该字节控制哪些段在上。

不同数字的值（不带小数点）是（未经测试，因此可能需要一些调整）：

   abcdefgh
0  11111100
1  01100000
2  11011010
3  11110010
4  01100110
5  10110110
6  10111110
7  11100000
8  11111110
9  11110110

给定值 4，您只需在该表中查找偏移量 4 处的字节（使用基地址的简单相加）表和值），然后从内存中提取该字节并使用它来设置 LED 段。该值 01100110 将设置段 b、c、f 和 g，给出你：

|   |
 ---
    |

你的具体情况有点复杂，因为你有一个 15 段 LED，并且必须写一个字而不是一个字节，但原理是完全相同的。

准备一个需要为每个数字写入的值表，并使用根据我上面的算法计算出的值来偏移到该表中。

If your processor doesn't have a divide instruction (or BCD conversion instructions, which would be another way), you'll just have to do repeated subtraction. The pseudo-code for this would go something like:

    val = 47

    units = val
    tens = 0
loop:
    if units < 10 goto done
    units = units - 10
    tens = tens + 1
    goto loop
done:
    ;; tens = 4, units = 7.

Apologies for the rather BASIC coding style there, I just tried to make it a little closer to how it would look in assembler language. Otherwise, I would have used a while loop :-)

Tho modulus is what's left in the units variable once it drops below ten.

For completeness (and possibly extra marks if this is a homework assignment, nudge nudge, wink wink), since a single octet can handle up to 255:

    val = 247

    units = val
    tens = 0
    hundreds = 0
loop1:
    if units < 100 goto loop2
    units = units - 100
    hundreds = hundreds + 1
    goto loop1
loop2:
    if units < 10 goto done
    units = units - 10
    tens = tens + 1
    goto loop2
done:
    ;; hundreds = 2, tens = 4, units = 7.

As proof this method works, you can try the following Python code:

val = 47

units = val
tens = 0
while units >= 10:
    units = units - 10
    tens = tens + 1

print "Tens = %d, Units = %d"%(tens,units)

which does indeed output:

Tens = 4, Units = 7

In terms of using those values to drive a LED device, a simple lookup table is probably easiest. Construct an array of values that you need to write to the two 8-bit memory addresses then use the digit to lookup those values for writing to the device.

I'll use a simplified seven-segment LED (plus decimal point) since it's easier to understand:

    a
   ---
f |   | b
   -g-
e |   | c
   ---     . h
    d

Let's say in this example that you write a single byte abcdefgh to a memory-mapped location and the bits in this byte control what segments are on.

The values for the different digits (without the decimal point) are (untested, so may need some tweaking):

   abcdefgh
0  11111100
1  01100000
2  11011010
3  11110010
4  01100110
5  10110110
6  10111110
7  11100000
8  11111110
9  11110110

Given the value four, you would simply look up the byte at offset 4 in that table (using simple addition of the base address of the table and the value), then extract that byte from memory and use it to set the LED segments. That value 01100110 will set segments b, c, f and g, giving you:

|   |
 ---
    |

Your particular situation is a little more complicated since you have a fifteen-segment LED and have to write a word rather than a byte, but the theory is exactly the same.

Have a table of values that you need to write for each digit and use the values calculated from my algorithm above to offset into that table.

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