[A83] Re: help for a routine


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[A83] Re: help for a routine




How would you do h*256? It's much more than 255 so I guess 0 comes out of it
(??), unless you'd use a two byte register.


>
> Hi !
> I need a routine that can make this :
>
> Input :
> - h, e
> Output:
> - a=(h*256)/e
>
> At this moment i'm using div_hl_de_s, with l:=0 and d:=0, but I'm sure
> there's a faster and a more clever way to do this...
>
> Anybody has an idea ?
>
>
>
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Ronald Teune
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