Mail Archives: djgpp/1996/12/18/09:16:03
Aaron,
The hex numbers 0x000FFFFF and 0xFFFFF both equal 1048575 (or 2 to the 20th
power). This is the same as 0x000F and 0xF both equal the number 15. Hex
is like base 10 in that zeros to the left do not affect the value (e.g.
1,245 and 001,245). Therefore both my mask of (__tb & 0x000FFFFF) and
Shawn's mask of (__tb & 0xFFFFF) yield the same result.
The only reason I use the more 'formal' mask is because I like to remind
myself what is really going on. We are masking off the high 12-bits and
keeping the low 20-bits -- when I look at my mask it reminds me of both
operations. It also reminds me that __tb is a 32-bit number... Basically,
I find it makes the code a little more 'self-documenting' -- a purely
personal choice.
Sorry for the confusion...
Patrick Smith,
Traveling Software
Aaron m Clemmer <aclemmer AT juno DOT com> wrote in article
<19961217 DOT 074132 DOT 12471 DOT 2 DOT aclemmer AT juno DOT com>...
> On 16 Dec 1996 17:02:54 GMT "Patrick Smith" <patricks AT travsoft DOT com>
> writes:
> >The protected mode selector for any buffer in conventional (dos) memory
> is
> >the global variable, _dos_ds. In order to use the transfer buffer
> (__tb),
> >simply use _dos_ds for the selector and the value of (__tb & 0x000FFFFF)
> as
> >the offset. Note: The reason you mask off the upper 12-bits is because
> >real-mode dos only uses 20-bit adresses...
> Well, between all these replies I've gotten, I finally understand
> what I should be doing, and what I was doing wrong.
> There may be no difference (but I believe there is), but if you
> say (__tb & 0x000fffff), and Shawn H says (__tb &0xfffff), wouldn't one
> be incorrect?
>
> aaron
>
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