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! * SLMR 2.1a * Press <CTRL>-<ALT>-<DEL> to continue ... --- Maximus 2.00 * Origin: Hartford Metro Area BBS (1:142/911)

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