Quote from Forum Archives on December 8, 1997, 11:30 pm

Posted by: tatewake <tatewake@...>

>This is probably a basic question. So many people may know the answer.

Yep, this is a cinch.

>Is there a way in FB to convert Hex numbers into base-10 numbers, and
>vice versa. Or is there a formula that I should be smart enough to
>figure out on my own?

Sure, hex is base 16, and base 10 is a normal number set.

The easiest way to do this:
myhex$=HEX$(base10num%)
base10num%=VAL("&H"+myhex$)

I've been using those two lines(well, at least pretty similar ones) since
the Tandy CoCo days(or is that "daze"? 🙂

In C/Java, you represent Hexadecimal numbers like so:
0xFFFF
and in basic:
&HFFFF
This would yield the number 65535...

Off memory, you just separate the numbers:
FF FF
|| ||
255 255
* *
16^2 16^0

consider this:
Dec(decimal/base 10)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Hex(hexidecimal/base 16)
0 1 2 3 4 5 6 7 8 9 A B C D E F

It's actually pretty versatile, considering you can fit an 8 bit number
into 2 characters. Can't beat that for compression...

Then there's octal(base 8). Something i never really learned, same
principle, though:

0 1 2 3 4 5 6 7
////////
8 characters
then repeat. Like hexidecimal. Too easy!

And binary(base 2)
0 1
//
two characters. boolean...

Using hexadecimal in everyday life? I do...!

Tim allen: "More power!"
1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536!!
/ __/ _____-----====___/
/ / / / /
2^8 = 65536.: 65536bytes/1024bytes(~1k) = 64k exactly!
And it is my belief that all computers depend on those numbers directly!
128+512 = 640
32+64+128+256 = 480
the resolution of your screen!
1G ~ 1024M.: 1M ~ 1024k.: 1k ~ 1024bits
Your hard drive?
32M = _2_ _16_bit Simms/Dimms? or _8_ _4_bit Simms/Dimms?
Amount of Ram you have?

Even in your modems; but that's a bit more complicated.

All numbers used in the calculations above come from powers of 2.
(Remember, 2^0 = 1; but where do we pull 0 from? It's not on our list of
powers of 2. A loophole? My theory is that 1 is a bad number, and all
numbers should be able to be subtracted by it to complete a range...)

So I drifted a bit.

Hope it helps...

-- Terence Jordan (tatewake@usa.net)
Inspired Software. inspired.netstreet.net/
Macintosh, Dos, Windows, and BeOS software solutions.
Phone: 407-728-7563
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