[A83] Re: Conditional bcall()ing
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[A83] Re: Conditional bcall()ing
At 02:51 2001-10-30, you wrote:
> >
> >oh, I thought the $ just lets TASM know that the value is a hex value =)
> >thanks for telling me that
> >
This can explain some confusion :)
>$ indicates a hex value when placed at the end of a number.
>When separate, it's not that at all. $ becomes the current address.
>So LD a,($+5) stores (currentaddress+5) into a.
$ is placed in the beginning of hex numbers last I looked...
but if the $ is not followed by 0-9 or a-f/A-F, then tasm knows it
obviously isn't a number, but should be recognized as "the current adress"
You can in tasm also use * as the current adress, but that is recommended
against in the docs because it looks like the multiplier operator... (imho
$ is just as bad...)
A similar dual meaning has also the % character. It can either mean "a
binary number" or mean the modulo operator. Which one is right is also here
read by context.
(This all can be read in tasm.doc)
///Olle
References: