Re: A86: Re: mod(op1,op2) == 0


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Re: A86: Re: mod(op1,op2) == 0




he's saying to use _BINOPEXEC.  to do op1 mod op2 with that, push op2
onto the op stack (with _push_OP2 = $480f) load the code for mod to a
(should be $15) and call _BINOPEXEC (= $4b1b).  the result will be in
op1.
i'm not absolutely about the order for the input there, you might have to
swap op1&op2 before all that.

-josh

On Thu, 2 Dec 1999 01:17:12 EST Jeanne9005@aol.com writes:
>
>Sorry, I guess I need a better Assembly to English translator.  Would 
>you 
>like to try that again for me, a little slower this time?
>
>In a message dated 12/1/99 6:51:16 PM Eastern Standard Time, 
>kirkmeyer@bigfoot.com writes:
>
>> 
>>  There is a call table to ROM page 5, and on it are contained 
>commands that
>>  take two arguments, such as GCF, and mod. One of the arguments is 
>in OP1 
>and
>>  the other is pushed onto the OP stack. Since there's no 
>documentation,
>>  you'll just have to keep trying numbers until you find the correct 
># for
>>  mod...
>>  
>>  ----- Original Message -----
>>  From: <Jeanne9005@aol.com>
>>  To: <assembly-86@lists.ticalc.org>
>>  Sent: Wednesday, December 01, 1999 8:48 AM
>>  Subject: A86: mod(op1,op2) == 0
>>  
>>  
>>  >
>>  > How do I do mod(op1,op2) == 0?  I have tried many ways, but 
>cannot
>>  remember
>>  > how it's done.
>>  >
>>  > Glen
>>  > tutorial1.cjb.net
>>  >
>
>
>Glen
>tutorial1.cjb.net
>

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