Datamaker

 BBS: Inland Empire Archive
Date: 07-06-92 (07:06)             Number: 292
From: AARON WEST                   Refer#: NONE
  To: MARK BUTLER                   Recvd: NO  
Subj: Datamaker                      Conf: (2) Quik_Bas

MB> JV> The trick is in bit shifting..  3 bytes become 4 and........
MB> JV> .............<post partially abridged>.......................
MB> JV> To split 24 bits into 4 six bit numbers, I think you'd divide by
MB> JV> 2^18, then 2^12, then 2^6 , then.............................

MB> Hey, you're on the case now buddy! If you get that conversion formula
MB> worked out it'll be all downhill from there! <grin>

From the author of the 'automagic' (not my name, but maybe I'll use it!)
self-uudecoder:

Here's a fragment from makeuue.c, in the comment section:
Given:
7654321.7654321.7654321.
54321.54321.54321.54321.
c1=byte1, c2=byte2, c3=byte3:
o1=c1>>2; o2=(c1&3)<<4+c2>>4; o3=(c2&15)<<2+c3>>6; o4=c3&63;

OK, in BASIC (binary to UUE):

l2$=""
for i=1 to len(lin$) step 3
  c1=asc(mid$(lin$,i,1))
  c2=asc(mid$(lin$,i+1,1))
  c3=asc(mid$(lin$,i+2,1))
  l2$=l2$+chr$(c1\4)+chr$((c1 and 3)*16+(c2\16))
  l2$=l2$+chr$((c2 and 15)*4+c3\64)+chr$(c3 and 63)
next i

The reverse (UUE to binary), converted to basic:

;get 4 bytes, put in msb-lsb order
;         t0      t1      t2      t3
;         al      ah      al      ah
;input..543210..543210..543210..543210
;output 765432--107654--321076--543210
;           b0        b1        b2

c1=(asc(mid$(lin$,i,1))-32)and 63
...
c4=(asc(mid$(lin$,i+3,1))-32)and 63



l2$=l2$+chr$(c1*4+c2\64)+chr$((c2*16 and &HF0)+c3\4)
l2$=l2$+chr$((c3*64 and &HC0)+c4)

Phew, took a few minutes to do in my head!

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