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Oct 18th, 2009

divide overflow problem

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Assembly Syntax (Toggle Plain Text)
MOV AH,DAY
DIV DIVISOR
CMP Al,'0'
JE WEEK_
JMP NOT_WEEK_
MOV AH,DAY
DIV DIVISOR
CMP Al,'0'
JE WEEK_
JMP NOT_WEEK_

that is my code that tries to find out if the "DAY" is divisible by "7", which is the DIVISOR..

i initialized them as..

Assembly Syntax (Toggle Plain Text)
DAY         DB 1
DIVISOR  DB 7
DAY         DB 1
DIVISOR  DB 7

when the "day" is divisible by 7, a 'divide oveflow' occurs..
how do i get this right? thanks

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urbancalli

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Oct 18th, 2009

Re: divide overflow problem

Try this:

Assembly Syntax (Toggle Plain Text)
MOV AL, DAY
CBW
DIV DIVISOR
MOV AL, DAY
CBW
DIV DIVISOR

Evenbit

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Junior Poster

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Oct 19th, 2009

Re: divide overflow problem

This may make your issue more obvious...

Assembly Syntax (Toggle Plain Text)
Quotient : Remainder = Dividend / Divisor
AL : AH =  AX / (8-bit)
AX : DX = DX:AX / (16-bit)
EAX : EDX = EDX:EAX /(32-bit)
Quotient : Remainder = Dividend / Divisor
AL : AH =  AX / (8-bit)
AX : DX = DX:AX / (16-bit)
EAX : EDX = EDX:EAX /(32-bit)

You needed to specfy a data type snce you're using a memory reference

Assembly Syntax (Toggle Plain Text)
MOV AH,DAY
CBW                       ;  AX = Day
DIV DIVISOR          ; Byte ptr Divisor
;;;; WRONG      CMP Al,'0'
MOV AH,DAY
CBW                       ;  AX = Day
DIV DIVISOR          ; Byte ptr Divisor
;;;; WRONG      CMP Al,'0'

; It's a decimal response not an ASCII '0'
; Note: AH is day of week (0...6)
; AL is week # (0..N)

Last edited by wildgoose; Oct 19th, 2009 at 12:08 am.

wildgoose

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Practically a Posting Shark

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Nov 2nd, 2009

Re: divide overflow problem

I just would like to ask, what is the function of the 'CBW'?
Does it replace the
CMP al,'0' part?

I am really confused, since it does not seem to pass through the WEEK_:

Here's my code;

Assembly Syntax (Toggle Plain Text)
;increment the day (since it has to go first, 
;to agree with my code I first made)
	INC DAY
 
	MOV AH,DAY
 
;I have to subtract 1 so that I will get the current day number
	SUB AH, 1
	CBW
	DIV DIVISOR
 
;jumps to the WEEK_ function 
;which summarizes the inventory for the week
	JE WEEK_
 
;jumps to the NOT_WEEK_ function 
;which makes it program loop until the sales reach zero
	JMP NOT_WEEK_
;increment the day (since it has to go first, 
;to agree with my code I first made)
	INC DAY
	
	MOV AH,DAY
		
;I have to subtract 1 so that I will get the current day number
	SUB AH, 1
	CBW
	DIV DIVISOR
		
;jumps to the WEEK_ function 
;which summarizes the inventory for the week
	JE WEEK_
	
;jumps to the NOT_WEEK_ function 
;which makes it program loop until the sales reach zero
	JMP NOT_WEEK_

really, thanks a lot. Still hoping for your help. Thanks again.

urbancalli

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Nov 5th, 2009

Re: divide overflow problem

CBW - Convert Byte to Word
Sign extends the byte in AL into the word in AX.
Sign extension is merely copying the sign bit into the rest
of a datatype when a signed value is moved into a larger
datatype.
When it is a byte being copied into a word, the value of bit 7
is copied into the 8 MSBs of the high byte of the word.
so 80 becomes FF80,
and 7F becomes 007F.

NotNull

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