More Assembly Language Multiplication

I am confused about how to handle assembly multiplication. I am working on a few problems described like so:

Given these hex values for the 8086 registers
AX = E204    BX = 30C2    CX = 5C08    DX = 38F1
What are the hex values in DX and AX after executing this instruction:
mul  cl

For mul operations, are we only multiplying using al and the multiplier? For example, would this multiplication simply be 4 * 8 = 32 (decimal) with the result stored in ax?

When imul is 开发者_如何学Goused, then we would multiply the entire value in ax and the multiplier and then store the results in dx:ax, correct?

For this problem, if I am indeed approaching it the right way, al * cl = 4 * 8 = 32. Then:

al = 20

Do I keep ah the same (E2) or do I zero it out for the answer?

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Please download complete Intel x86 CPU documentation from here.

For details on just this one instruction you can read here.

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Why didn't you try? Never mind, I did it for you:

Microsoft Windows XP [Versione 5.1.2600]
(C) Copyright 1985-2001 Microsoft Corp.

C:\Documents and Settings\db>debug
-a
15AB:0100 mov ax, e204
15AB:0103 mov bx, 30c2
15AB:0106 mov cx, 5c08
15AB:0109 mov dx, 38f1
15AB:010C mul cl
15AB:010E
-g=0100,010e

AX=0020  BX=30C2  CX=5C08  DX=38F1  SP=FFEE  BP=0000  SI=0000  DI=0000
DS=15AB  ES=15AB  SS=15AB  CS=15AB  IP=010E   NV UP EI PL NZ NA PO NC
15AB:010E 0000          ADD     [BX+SI],AL                         DS:30C2=00
-q

C:\DOCUME~1\db>