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32bit to long?

BlitzMax Forums/BlitzMax Programming/32bit to long?

kronholm(Posted 2008) [#1]
What's the easiest way you know to convert a string with 4 characters in it to a long? There's probably many ways, I just can't find any of them :(

I mean from e.g. "7ab4" to an unsigned long?
For instance the num equivs of a string is 34,0,58,241 and they become 570440433.

Hope it's clear enough, otherwise say something and I'll try to explain further :)


kronholm(Posted 2008) [#2]
Here's what I came up with :/
Anything quicker out there?

' from 4 bytes (8 bit * 4 bytes = 32 bit) to 32 bit unsigned long (blitz long is 64 bit though, but whatever)
Function bytes2long:Long(in$)		
	If Len(in) <> 32/8 err("invalid bytes2long input, need 4x8 bit bytes input.") ; Return
	Local first = in[0] Shl 24'  																	
	Local second = (in[2] * 256) + in[3]																
	Local result:Long = first + second																
	Return result																					
EndFunction



Ked(Posted 2008) [#3]
Local s:String="7ab4"
Print s.ToLong()


Is that what you mean?


Floyd(Posted 2008) [#4]
He wants to treat each character as a digit in a "base 256" number.

Here's my attempt. It assumes the string is four characters and each character code is in the range 0-255.
BlitzMax uses 16-bit unicode characters so this may not be true if your string was made in China.

Note the $FFFFFFFF:Long, just in case the result was a negative integer and got sign-extended to 64 bits.

BTL:Long = BytesToLong( Chr($FE) + Chr($DC) + Chr($BA) + Chr($98) )

Print
Print BTL
Print LongHex( BTL )

Function BytesToLong:Long( in$ )

	Local n:Int
	
	n = in[0]
	n = ( n Shl 8 ) | in[1]
	n = ( n Shl 8 ) | in[2]
	n = ( n Shl 8 ) | in[3]
			
	Return n & $FFFFFFFF:Long

End Function



kronholm(Posted 2008) [#5]
Thank you very much Floyd, that is definitely better than mine :) I forgot n[1] as well in mine, hah!


kronholm(Posted 2008) [#6]
Can anyone, maybe Floyd, even, show me how to make the opposite conversion? I don't even know what | does..


plash(Posted 2008) [#7]
I don't even know what | does..
'Bitwise Or' operation?


kronholm(Posted 2008) [#8]
According to Blitzmax documentation, :| is bitwise or.


kronholm(Posted 2008) [#9]
Looks like this did the trick, for future references.

Function long2bytes$(in:Long)	
  	Local b1 = (in Shr 24) Mod 256
  	Local b2 = (in Shr 16) Mod 256
  	Local b3 = (in Shr 8) Mod 256
  	Local b4 = (in Shr 0) Mod 256
	Print "hej "+b1
	Return Chr(b1)+Chr(b2)+Chr(b3)+Chr(b4)
EndFunction



Jesse(Posted 2008) [#10]
this is simpler:
function long2bytes$(in:long)
     return hex(in)
end function

Untested, Might not be faster.


Floyd(Posted 2008) [#11]
He wants base 256, not base 16.

It could be speeded up a little with bitwise And. "Mod 256" means discard everything except the last eight bits.
So "n Mod 256" has the same effect as "n & $FF", but uses integer division to get the result.