CODEWITHSUNDEEP
assemblymasmx86-16
user3334146
user3334146

Reputation: 75

AAM instruction in 8086

I am new with 8086 and i need a little help. I know the basic of AAM. that if i multiple two no.s suppose 07H and 09H

MUL AL,BL
AAM

this will store the value 06H in AH and 03H in AL. But suppose if apply AAM at a value of 3 hexadecimal bits, Suppose 

MOV AL,77H
MOV BL,0AH
MUL BL
AAM

What will be the content of AL at last ?

Upvotes: 1

Views: 20914

Answers (2)

humpty
humpty

Reputation: 71

ASCII Adjust after Multiplication(AAM):

Corrects the result of multiplication of two BCD values.

Algorithm:

AH = AL / 10

AL = remainder

Example:

MOV AL, 15 ; // AL = 0Fh

AAM ; // AH = 01, AL = 05

RET

Upvotes: 0

qwr
qwr

Reputation: 3670

AAM (BCD ADJUST AFTER MULTIPLY)

from here

use aam only after executing a mul instruction between two BCD digits (unpacked). mul stores the result in the AX register. The result is less than 100 so it can be contained in the AL register (the low byte of the AX register). aam unpacks the AL result by dividing AL by 10, stores the quotient (most-significant digit) in AH, and stores the remainder (least-significant digit) in AL.

So question is what it will do if we provide Al bigger value than 99?

It will do the same AH = AL / 10 and AL = AL mod 10 but will leave incorrect unpacked bcd values.

So Coming to your case before AAM AL will be 166 (0xA6) (AX will be 0x04A6 after multiply) after AAM

  Ah=  166/10=16 ( 0x10)
  AL=166 mod 10=6 (0x6)

As we see AX will be 0x1006 after AAM And it left incorrect unpacked bcd number. BeCause the input was not below 100

Upvotes: 5

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