0100 ;MACHINE LANGUAGE SORTS, PART 1¢0110 ;BY KEVIN PECK¢0120 ;(c)1987, ANTIC PUBLISHING¢0130 FLAST = $00 ;End, outer loop¢0140 FLENKEY = $CB ;Key length¢0150 RLEN = $CC ;Record Length¢0160 OFFSETKEY = $CF ;Key offset¢0170 FIRST = $CD ;1st element ptr¢0180 SECOND = $D4 ;2nd element ptr¢0190 LAST = $D6 ;End, inner loop¢0200 FENDKEY = $D8 ;End, key field¢0210 ORDER = $E1 ;Order, sort 0,1¢0220 FLENSEC = $D0 ;2nd field lngth¢0230 OFFSETSEC = $E0 ;2nd " " offset¢0240 FENDSEC = $D1 ;End 2nd field¢0250 ;¢0260 *= $4000¢0270 ;¢0280 CLD ¢0290 PLA ¢0300 ; Retrieve arguments from BASIC¢0310 PLA ¢0320 STA FIRST+1¢0330 PLA ¢0340 STA FIRST¢0350 PLA ¢0360 STA LAST+1¢0370 PLA ¢0380 STA LAST¢0390 PLA ¢0400 PLA ¢0410 STA FLENKEY¢0420 PLA ¢0430 PLA ¢0440 STA OFFSETKEY¢0450 CLC ;Find end of ¢0460 ADC FLENKEY ;First key ¢0470 STA FENDKEY¢0480 PLA ¢0490 PLA ¢0500 STA FLENSEC¢0510 PLA ¢0520 PLA ¢0530 STA OFFSETSEC¢0540 CLC ¢0550 ADC FLENSEC ;Find end of ¢0560 STA FENDSEC ;second key¢0570 PLA ¢0580 PLA ¢0590 STA RLEN¢0600 PLA ¢0610 PLA ¢0620 STA ORDER¢0630 ;¢0640 ; All parms. now in zero page.¢0650 ; Next, set pointer to the end¢0660 ; of the outer loop¢0670 LDA LAST+1¢0680 STA FLAST+1¢0690 SEC ¢0700 LDA LAST¢0710 SBC RLEN¢0720 STA FLAST¢0730 BCS SETSECOND¢0740 ;¢0750 DEC FLAST+1¢0760 ; Start of outer loop.¢0770 ; Adjust the second pointer to¢0780 ; point to the first pointer¢0790 ; plus the record length.¢0800 ;¢0810 SETSECOND¢0820 CLC ¢0830 LDA FIRST+1¢0840 STA SECOND+1¢0850 LDA FIRST¢0860 ADC RLEN¢0870 STA SECOND¢0880 BCC SORTKEY¢0890 INC SECOND+1¢0900 ; Start of the inner loop.¢0910 ; 1. Compare the Key field of¢0920 ; the two sort elements.¢0930 ; 2. If we find a mismatch,¢0940 ; do we need to swap them?¢0950 SORTKEY¢0960 LDY OFFSETKEY¢0970 ;¢0980 KEYLOOP¢0990 ;¢1000 LDA (FIRST),Y ;get a byte¢1010 CMP (SECOND),Y ;of each.¢1020 BEQ CHKMORE ;If = Continue.¢1030 ;¢1040 BCC NOSWAP ;If F<S no swap.¢1050 ;¢1060 BCS SWAP ;If F>S swap.¢1070 ;¢1080 CHKMORE¢1090 ;¢1100 INY ;Adjust pointer¢1110 CPY FENDKEY ;All done?¢1120 BNE KEYLOOP ;No. Continue.¢1130 ;¢1140 ; At this point, all bytes in¢1150 ; the key field of both sort¢1160 ; elements are equal. Drop to¢1170 ; SORTSECOND and check the¢1180 ; secondary fields.¢1190 ; If their lengths = 0, then¢1200 ; we don't have 2ndry fields.¢1210 ; We are only doing a one-field¢1220 ; sort--goto the Noswap routine.¢1230 ; If the key fields are equal,¢1240 ; there is no need to swap them.¢1250 ;¢1260 SORTSEC¢1270 LDA FLENSEC ;A second field?¢1280 BEQ NOSWAPBD ;No.¢1290 ;¢1300 LDY OFFSETSEC¢1310 SECLOOP¢1320 LDA (FIRST),Y ;Compare byte¢1330 CMP (SECOND),Y ;by byte...¢1340 BEQ CHKMORE2 ;If = do more.¢1350 ;¢1360 BCC NOSWAP ;If F<S Noswap.¢1370 ;¢1380 BCS SWAP ;If F>s Swap.¢1390 ;¢1400 CHKMORE2¢1410 INY ;Point to next.¢1420 CPY FENDSEC ;End of 2nd?¢1430 BNE SECLOOP ;No, do more.¢1440 ;¢1450 ; Now, both the key and the¢1460 ; secondary fields of both sort¢1470 ; elemtents are equal. Goto¢1480 ; the noswap routine through the¢1490 ; "back door." No need to check¢1500 ; the order, no need to swap.¢1510 BEQ NOSWAPBD¢1520 ;¢1530 ; This is where the swapping¢1540 ; occurs. First, check the¢1550 ; swapping order.¢1560 ; (Assume swapping in ascending¢1570 ; order.) If order<>0, then¢1580 ; sort in descending order.¢1590 ;¢1600 SWAP¢1610 ;¢1620 LDA ORDER ;Get order¢1630 BNE NOSWAPBD ;Not 0, No swap¢1640 ;¢1650 ; Swap routine's back door.¢1660 ; If NOSWAP decides we need to¢1670 ; swap by checking the order,¢1680 ; we need to come here (instead¢1690 ; of SWAP) or we would go into¢1700 ; a continuous loop.¢1710 ;¢1720 SWAPBD¢1730 LDY #0¢1740 SWAPLOOP¢1750 LDA (FIRST),Y ;Key byte¢1760 PHA ; to stack.¢1770 LDA (SECOND),Y ; 2ndry byte¢1780 STA (FIRST),Y ; to key.¢1790 PLA ;Key from stack¢1800 STA (SECOND),Y ;to 2ndry.¢1810 INY ;Next byte.¢1820 CPY RLEN ; More?¢1830 BNE SWAPLOOP ;Yes. Continue.¢1840 ;¢1850 ; All bytes have been swapped.¢1860 ; Now adjust pointers to the¢1870 ; next elements for the sort.¢1880 ; Goto the noswap back door.¢1890 ;¢1900 BEQ NOSWAPBD¢1910 ;¢1920 NOSWAP¢1930 ;¢1940 LDA ORDER ;Is ORDER=1?¢1950 BNE SWAPBD ;Yes. Swap them¢1960 ;¢1970 NOSWAPBD¢1980 CLC ¢1990 LDA SECOND¢2000 ADC RLEN¢2010 STA SECOND¢2020 LDA SECOND+1¢2030 ADC #0¢2040 STA SECOND+1¢2050 CMP LAST+1¢2060 BNE SORTKEY¢2070 ;¢2080 LDA SECOND¢2090 CMP LAST¢2100 BNE SORTKEY¢2110 ;¢2120 ; We've made one pass through¢2130 ; the sort's inner loop. Now,¢2140 ; adjust the outer loop and¢2150 ; check if we're done with it.¢2160 ; If not, readjust the inner¢2170 ; loop pointer to the outer loop¢2180 ; pointer + the record length.¢2190 ;¢2200 CLC ¢2210 LDA FIRST¢2220 ADC RLEN¢2230 STA FIRST¢2240 LDA FIRST+1¢2250 ADC #0¢2260 STA FIRST+1¢2270 CMP FLAST+1¢2280 BNE SETSECOND¢2290 ;¢2300 LDA FIRST¢2310 CMP FLAST¢2320 BNE SETSECOND¢2330 ;¢2340 RTS ;Goto BASIC¢2350 .END ¢
