/ I'd like to try your method but don't know what do you mean by
/ "calculating the CRC". What's it?
Cyclic Redundancy Check... goes into some theory of binary polynomials
etc, well, in this case that's the number sys(2007, cExp) returns. Since
it returns a string containing the 16-bir CRC number, I use something
like this:
Proc CrcHex
para _st
priv _nv,i,_is
_nv=val(sys(2007,_st))
_is=""
* make just hex digits out of it
for i=1 to 4
_is=chr(mod(_nv,16)+48)+_is
_nv=int(_nv/16)
endf
retu chrt(_is,"0123456789:;<=>?","abcdefghijklmnop")
From that we get a four-letter string (lowercase, my personal
preference). For real hex, the last line would read
retu chrt(_is,":;<=>?","abcdef")
For real jumble, try this:
? CrcHex("somestring"+"my_password")
And replace the "abcdef..." with characters above 128.
