S-C DocuMentor — Applesoft
SAVE S.EC4A
1010 *--------------------------------
1020 * CONVERT STRING TO FP VALUE IN FAC
1030 *
1040 * STRING POINTED TO BY TXTPTR
1050 * FIRST CHAR ALREADY SCANNED BY CHRGET
1060 * (A) = FIRST CHAR, C=0 IF DIGIT.
1070 *--------------------------------
EC4A- A0 00 1080 FIN LDY #0 CLEAR WORKING AREA ($99...$A3)
EC4C- A2 0A 1090 LDX #10 TMPEXP, EXPON, DPFLG, EXPSGN, FAC, SERLEN
EC4E- 94 99 1100 .1 STY TMPEXP,X
EC50- CA 1110 DEX
EC51- 10 FB 1120 BPL .1
1130 *--------------------------------
EC53- 90 0F 1140 BCC FIN.2 FIRST CHAR IS A DIGIT
EC55- C9 2D 1150 CMP #'-' CHECK FOR LEADING SIGN
EC57- D0 04 1160 BNE .2 NOT MINUS
EC59- 86 A3 1170 STX SERLEN MINUS, SET SERLEN = $FF FOR FLAG
EC5B- F0 04 1180 BEQ FIN.1 ...ALWAYS
EC5D- C9 2B 1190 .2 CMP #'+' MIGHT BE PLUS
EC5F- D0 05 1200 BNE FIN.3 NOT PLUS EITHER, CHECK DECIMAL POINT
1210 *--------------------------------
EC61- 20 B1 00 1220 FIN.1 JSR CHRGET GET NEXT CHAR OF STRING
1230 *--------------------------------
EC64- 90 5B 1240 FIN.2 BCC FIN.9 INSERT THIS DIGIT
1250 *--------------------------------
EC66- C9 2E 1260 FIN.3 CMP #'.' CHECK FOR DECIMAL POINT
EC68- F0 2E 1270 BEQ FIN.10 YES
EC6A- C9 45 1280 CMP #'E' CHECK FOR EXPONENT PART
EC6C- D0 30 1290 BNE FIN.7 NO, END OF NUMBER
EC6E- 20 B1 00 1300 JSR CHRGET YES, START CONVERTING EXPONENT
EC71- 90 17 1310 BCC FIN.5 EXPONENT DIGIT
EC73- C9 C9 1320 CMP #TOKEN.MINUS NEGATIVE EXPONENT?
EC75- F0 0E 1330 BEQ .1 YES
EC77- C9 2D 1340 CMP #'-' MIGHT NOT BE TOKENIZED YET
EC79- F0 0A 1350 BEQ .1 YES, IT IS NEGATIVE
EC7B- C9 C8 1360 CMP #TOKEN.PLUS OPTIONAL "+"
EC7D- F0 08 1370 BEQ FIN.4 YES
EC7F- C9 2B 1380 CMP #'+' MIGHT NOT BE TOKENIZED YET
EC81- F0 04 1390 BEQ FIN.4 YES, FOUND "+"
EC83- D0 07 1400 BNE FIN.6 ...ALWAYS, NUMBER COMPLETED
EC85- 66 9C 1410 .1 ROR EXPSGN C=1, SET FLAG NEGATIVE
1420 *--------------------------------
EC87- 20 B1 00 1430 FIN.4 JSR CHRGET GET NEXT DIGIT OF EXPONENT
1440 *--------------------------------
EC8A- 90 5C 1450 FIN.5 BCC GETEXP CHAR IS A DIGIT OF EXPONENT
1460 *--------------------------------
EC8C- 24 9C 1470 FIN.6 BIT EXPSGN END OF NUMBER, CHECK EXP SIGN
EC8E- 10 0E 1480 BPL FIN.7 POSITIVE EXPONENT
EC90- A9 00 1490 LDA #0 NEGATIVE EXPONENT
EC92- 38 1500 SEC MAKE 2'S COMPLEMENT OF EXPONENT
EC93- E5 9A 1510 SBC EXPON
EC95- 4C A0 EC 1520 JMP FIN.8
1530 *--------------------------------
1540 * FOUND A DECIMAL POINT
1550 *--------------------------------
EC98- 66 9B 1560 FIN.10 ROR DPFLG C=1, SET DPFLG FOR DECIMAL POINT
EC9A- 24 9B 1570 BIT DPFLG CHECK IF PREVIOUS DEC. PT.
EC9C- 50 C3 1580 BVC FIN.1 NO PREVIOUS DECIMAL POINT
1590 * A SECOND DECIMAL POINT IS TAKEN AS A TERMINATOR
1600 * TO THE NUMERIC STRING.
1610 * "A=11..22" WILL GIVE A SYNTAX ERROR, BECAUSE
1620 * IT IS TWO NUMBERS WITH NO OPERATOR BETWEEN.
1630 * "PRINT 11..22" GIVES NO ERROR, BECAUSE IT IS
1640 * JUST THE CONCATENATION OF TWO NUMBERS.
1650 *--------------------------------
1660 * NUMBER TERMINATED, ADJUST EXPONENT NOW
1670 *--------------------------------
EC9E- A5 9A 1680 FIN.7 LDA EXPON E-VALUE
ECA0- 38 1690 FIN.8 SEC MODIFY WITH COUNT OF DIGITS
ECA1- E5 99 1700 SBC TMPEXP AFTER THE DECIMAL POINT
ECA3- 85 9A 1710 STA EXPON COMPLETE CURRENT EXPONENT
ECA5- F0 12 1720 BEQ .15 NO ADJUST NEEDED IF EXP=0
ECA7- 10 09 1730 BPL .14 EXP>0, MULTIPLY BY TEN
ECA9- 20 55 EA 1740 .13 JSR DIV10 EXP<0, DIVIDE BY TEN
ECAC- E6 9A 1750 INC EXPON UNTIL EXP=0
ECAE- D0 F9 1760 BNE .13
ECB0- F0 07 1770 BEQ .15 ...ALWAYS, WE ARE FINISHED
ECB2- 20 39 EA 1780 .14 JSR MUL10 EXP>0, MULTIPLY BKY TEN
ECB5- C6 9A 1790 DEC EXPON UNTIL EXP=0
ECB7- D0 F9 1800 BNE .14
ECB9- A5 A3 1810 .15 LDA SERLEN IS WHOLE NUMBER NEGATIVE?
ECBB- 30 01 1820 BMI .16 YES
ECBD- 60 1830 RTS NO, RETURN, WHOLE JOB DONE!
ECBE- 4C D0 EE 1840 .16 JMP NEGOP NEGATIVE NUMBER, SO NEGATE FAC
1850 *--------------------------------
1860 * ACCUMULATE A DIGIT INTO FAC
1870 *--------------------------------
ECC1- 48 1880 FIN.9 PHA SAVE DIGIT
ECC2- 24 9B 1890 BIT DPFLG SEEN A DECIMAL POINT YET?
ECC4- 10 02 1900 BPL .1 NO, STILL IN INTEGER PART
ECC6- E6 99 1910 INC TMPEXP YES, COUNT THE FRACTIONAL DIGIT
ECC8- 20 39 EA 1920 .1 JSR MUL10 FAC = FAC * 10
ECCB- 68 1930 PLA CURRENT DIGIT
ECCC- 38 1940 SEC <<<SHORTER HERE TO JUST "AND #$0F">>>
ECCD- E9 30 1950 SBC #'0' <<<TO CONVERT ASCII TO BINARY FORM>>>
ECCF- 20 D5 EC 1960 JSR ADDACC ADD THE DIGIT
ECD2- 4C 61 EC 1970 JMP FIN.1 GO BACK FOR MORE
1980 *--------------------------------
1990 * ADD (A) TO FAC
2000 *--------------------------------
ECD5- 48 2010 ADDACC PHA SAVE ADDEND
ECD6- 20 63 EB 2020 JSR COPY.FAC.TO.ARG.ROUNDED
ECD9- 68 2030 PLA GET ADDEND AGAIN
ECDA- 20 93 EB 2040 JSR FLOAT CONVERT TO FP VALUE IN FAC
ECDD- A5 AA 2050 LDA ARG.SIGN
ECDF- 45 A2 2060 EOR FAC.SIGN
ECE1- 85 AB 2070 STA SGNCPR
ECE3- A6 9D 2080 LDX FAC TO SIGNAL IF FAC=0
ECE5- 4C C1 E7 2090 JMP FADDT PERFORM THE ADDITION
2100 *--------------------------------
2110 * ACCUMULATE DIGIT OF EXPONENT
2120 *--------------------------------
ECE8- A5 9A 2130 GETEXP LDA EXPON CHECK CURRENT VALUE
ECEA- C9 0A 2140 CMP #10 FOR MORE THAN 2 DIGITS
ECEC- 90 09 2150 BCC .1 NO, THIS IS 1ST OR 2ND DIGIT
ECEE- A9 64 2160 LDA #100 EXPONENT TOO BIG
ECF0- 24 9C 2170 BIT EXPSGN UNLESS IT IS NEGATIVE
ECF2- 30 11 2180 BMI .2 LARGE NEGATIVE EXPONENT MAKES FAC=0
ECF4- 4C D5 E8 2190 JMP OVERFLOW LARGE POSITIVE EXPONENT IS ERROR
ECF7- 0A 2200 .1 ASL EXPONENT TIMES 10
ECF8- 0A 2210 ASL
ECF9- 18 2220 CLC
ECFA- 65 9A 2230 ADC EXPON
ECFC- 0A 2240 ASL
ECFD- 18 2250 CLC <<< ASL ALREADY DID THIS! >>>
ECFE- A0 00 2260 LDY #0 ADD THE NEW DIGIT
ED00- 71 B8 2270 ADC (TXTPTR),Y BUT THIS IS IN ASCII,
ED02- 38 2280 SEC SO ADJUST BACK TO BINARY
ED03- E9 30 2290 SBC #'0'
ED05- 85 9A 2300 .2 STA EXPON NEW VALUE
ED07- 4C 87 EC 2310 JMP FIN.4 BACK FOR MORE
2320 *--------------------------------
S.EB72