OpenEuphoria

Re: Hexadecimal and whatever...

• Posted by Architek <architek at GEOCITIES.COM> Apr 15, 1997
• 1290 views

James Powell wrote:

> The problem is that 1 byte (8 bits) will hold 2^8 (256) values, ie &H00-&HFF.
> If you combine 4 bytes (32 bits, and your basic integer type) you can now stor
e
> 2^32 (4,294,967,292) values, or &H00-&HFFFFFFFF.
> You want the hex value &H06 (00000006).  Since that is less than &HFF, it is
> stored in a SINGLE byte, rather than a 4 byte integer.  When you combine the
> four
> bytes that you are reading, it makes a much larger value (10105342) than you
> wanted.

Maybe you could say that H06 could be stored in ANY amount of byte... example:
In 4 bytes: #00, #00, #06
In 2 bytes: #00, #06
.. etc...

In general you can say that the value of a number stored in several bytes is:

(byte1)+(byte2*256)+(byte3*256^2)+(byte4*256^3)+ ... +(byteN*256^N-1)
where byte1 is the least significant byte and byte N the most significant byte.
In the above examples:
(6)+(0*256)+(0*65536) = 6

> Anyway, just use each byte seperately rather than making an integer out of 4
> of them.

James is right... the number you've got 101053452 is in hexadecimal 06 05 F4 0C
which i interpret as been the right result for you to get EACH BYTE alone and
not trying to get 4 byte numbers (4 byte number = 2^32).

Hope this helps you, if you need more info about binary numbers just ask me, som
e
time ago I had the same troubles trying to understand how numbers where stored ;
)

Bye,
--

            ************************************
            This message sent by Daniel Berstein
            ************************************
     email: architek at geocities.com
 homepages: http://www.geocities.com/SiliconValley/Heights/9316
            http://www.cybercity.hko.net/silicon_valley/architek