Re: Re: Constructing 32-bit hexadecimal integers

I didn't explain this very clearly, sorry.

A set of 1, 2, 3 or 4 hex values are stored in one variable. If you printed the variable, you'd get whatever ASCII characters corresponded to the hex values. (There's probably a better way to explain this, but I don't really speak the language quite yet.)

Should I split() the variable, then sprintf, then concatenate?

Thanks once more...

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donfreenut

Comment on Re: Re: Constructing 32-bit hexadecimal integers

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Re:{3} Constructing 32-bit hexadecimal integers by jeroenes (Priest) on Apr 12, 2001 at 00:26 UTC
Nah, use unpack at will. You will have to pad them missing values: `$a = "\001"; #just another hex string $a = ("\000" x (4-length($a))).$a; #Zero pad that string print unpack 'L', $a; #Hey, that's one!` [download] Jeroen "We are not alone"(FZ)	[reply] [d/l]
(tye)Re: Constructing 32-bit hexadecimal integers by tye (Sage) on Apr 12, 2001 at 00:54 UTC
Well, it is either 1 or 0x01000000, depending on your platform. Use 'N' or 'V' in place of 'L' if you want to use big- or little-endian format (respectively) regardless of your current platform. - tye (but my friends call me "Tye")	[reply]