What encryption is this?

GeoTrail · Post by **GeoTrail** » Mon Nov 21, 2005 1:15 am

I have this hash or whatever it is

70617373776f7264

I'm trying to find out how it is encrypted, or which hash it is. I know the unencrypted text is password.

Any takers?

MrMat · Post by **MrMat** » Mon Nov 21, 2005 1:46 am

It looks like the ascii values of each letter in "password", written in hex. So $70 = 112 = 'p', $61 = 97 = 'a', etc.

GeoTrail · Post by **GeoTrail** » Mon Nov 21, 2005 1:59 am

Like this?
Hex(Asc("p"))

Think that is right, but how do I reverse it? How do I unencode it?

MrMat · Post by **MrMat** » Mon Nov 21, 2005 2:03 am

Yes that's it. So this displays the hex string above:

Code: Select all

str.s = "password"
text.s = ""
For l = 0 To Len(str) - 1
text + Hex(PeekB(@str + l))
Next
Debug(text)

GeoTrail · Post by **GeoTrail** » Mon Nov 21, 2005 2:08 am

That is very helpfull MrMat. How can I unencrypt a hex string created in that manner?

MrMat · Post by **MrMat** » Mon Nov 21, 2005 2:10 am

It's not really encrypted so you can just PeekS it: Debug(PeekS(@str)) if it's null-terminated or Debug(PeekS(@str, length)) if you know it's length.

MrMat · Post by **MrMat** » Mon Nov 21, 2005 2:23 am

That's if it is already in memory. If you are given it as a string then:

Code: Select all

Procedure.l Convert(a.s)
    ProcedureReturn FindString("0123456789abcdef", LCase(a), 1) - 1
EndProcedure

str.s = "70617373776f7264"
text.s = ""
For l = 1 To Len(str) Step 2
val.l = Convert(Mid(str, l, 1)) << 4 + Convert(Mid(str, l + 1, 1))
text + Chr(val)
Next
Debug(text)

GeoTrail · Post by **GeoTrail** » Mon Nov 21, 2005 10:11 am

Perfect MrMat, thanks alot for your assistance.
Oh, and good morning

MrMat · Post by **MrMat** » Mon Nov 21, 2005 12:10 pm

Glad to help and good morning to you