问题描述
说明（十六进制数）：
mov ax, 0832
mov cx, 008a
idiv cl
文件说：
when operand is a byte:
AL = AX / operand
AH = remainder (modulus)
来源：http://www.electronics.dit.ie/staff/tscarff/8086_instruction_set/8086_instruction_set.html#IDIV
据我了解，我要检查，如果最左边的位是不是1：
As far as I understand, I have to check if the leftmost bit isn't 1:
0832 -> 0000.1000.0011.0010
008a -> 0000.0000.1000.1010
由于有一个在最高位的1号的，用计算器我应该得到：
Since there's no 1's in highest bits, by using calculator I should get:
0832 MOD 008A 的剩余时间和 0832 DIV 008A 为 DIV 部分。 国防部进入 AH 和 DIV 应到 AH ，但我不能得到正确的答案。正确的答案是 5CEF 。我在做什么错了？
0832 mod 008a as remainder and 0832 div 008a as div part. Mod goes to AH and div should go to AH, but I can't get the correct answer. The correct answer is 5CEF. What am I doing wrong?
推荐答案
+1阅读的文档：）
请注意，您需要根据有效的操作数的大小来检查左边的一位。既然你是通过 CL ，这是一个8位值划分，你需要检查位＃7，这是 1 。因此，在 CL 是负数，即 -118 。因此，
Note that you need to check the leftmost bit according to the effective operand size. Since you are dividing by CL, which is a 8 bit value, you need to check bit #7, which is 1. As such, the number in CL is negative, namely -118. So,
AL = 0832h / -118 = 2098 / -118 = -17 = EFh
AH = 2098 - (-17 * -118) = 92 = 5Ch
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