BASIC Answers to BASIC Questions App Note
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1. 27 RELOAD LOWER ORDER BYTE kkxkxkxkkkkkkkkkkkkk THE CARRY BIT SAME AS THE LOWEST BIT IN THE LOWER BYTE IS SHIFTED INTO THE TOP OF THE UPPER ORDER BYTE kkkkkkkkkkkkkkkk k RORA kkxkxkxkkkkkkkkkkkkk THE CARRY BIT SAME AS THE LOWEST BIT IN THE UPPER BYTE IS SHIFTED INTO THE TOP OF THE LOWER ORDER BYTE kkxkkxkkkkkkkkkkkk k RORB STAA 26 THE NEW DECODED UPPER ORDER BYTE STAB 27 THE NEW DECODED LOWER ORDER BYTE RTS BASIC line 210 sets up the address where the USR function can find the memory address of the assembly routine in 6800 this is at 24 and 25 Line 220 sets up the memory address of the first USR assembly routine ENCODE at 6E00 Line 230 sets up the memory address of the second USR assembly routine DECODE at 6FFO Line 300 then POKEs the address where the USR function expects to find the memory address of the assembly routine Line 400 also POKEs this address In this way the user can choose between one of several assembly routines to execute The assembly code for both ENCODE and DECODE is documented we will forgo any explanation at this time An example of the 6809 USR function would look about the same Of course the POKE memory locations and assembly language would be different In 6809 Extended BASIC or BASIC the USR function expects to find the memory address of the assembly routine at MEMEND 2 and the USR function s argument can be found at MEMEND 4 If MEMEND is 7FFF then the address of the ass2. Section 6 4 PRINT USING is a very powerful formatting feature in Extended BASIC We have been asked questions about its use and feel that a few more examples may help to clear up these questions STRINGS BACK SLASH PRINT USING 23456 THE RAIN IN SPAIN FALLS THE RAI Altogether seven 7 characters counting both of the back slashes NEGATIVE NUMBERS using the POUND SIGN PRINT USING 235 235 A minus sign can be printed before a number using the pound sign FLOATING DOLLAR SIGN PRINT USING 23 05 23 05 PRINT USING 38293 4 38 293 40 PRINT USING 3 5 3 50 Note the leading spaces right justification and floating dollar sign and RESERVE ONE NUMERICAL SPACE PRINT USING WITH PROTECTED FIELD IS 12 5 12 5 12 50 WITH PROTECTED FIELD IS 12 50 Note that both and reserve a space for a numerical field In this case it is the tens position EXAMPLE CONTAINING MANY OPTIONS PRINT USING THE BALANCE OF 23456789 IS AND ttt AJEX INC 2345 7 3568 91 THE BALANCE OF AJEX INC IS 2 345 70 AND 3 568 91 11 BASIC Answers to BASIC Questions PRINTING FLOATING POINT NUMBERS Section 6 4 Many users have asked us how to print a large floating point number in a decimal format as opposed to the scientific notation This can be done by using the PRINT USING statement and giving the floating point
3. just as before scaling A PRINT statement will still yield 0 01 however 0 184 will be printed as 0 18 with SCALE BASIC Answers to BASIC Questions factor of only 2 due to the rounding of the 4 in the thousandths place Therefore if you intend on using values like 0 0001 or 0 18574 in which you desire to preserve accuracy then you need to set the SCALE factor accordingly Note that the maximum SCALE factor is six 6 Now that you know why a SCALE factor is used let s look at the proper use with SAVE and COMPILE programs As Extended BASIC converts all constants to their binary equivalent when the program is typed in or LOADed it is not possible to change the SCALE factor while the program is in memory Not only is it an error 67 to change the SCALE factor while a program resides in memory it would be worthless as the constant values have already been set Only by first SAVEing the program to disk changing the SCALE factor and then reLOADing the program thus resetting the constants to binary values using the new SCALE factor can a SCALE factor be changed for an Extended BASIC Source program BAS extension Furthermore as a BASIC Compiled program BAC extension cannot be LOADed the SCALE factor at the time of COMPILE is stored on the disk with the COMPILED program Subsequently RUNning the program brings both the program and the SCALE factor into memory 10 BASIC Answers to BASIC Questions MORE PRINT USING EXAMPLES
4. number the exact format you wish to use For example PRINT USING AAH HEE ttt 1234567890 1234 1 234 567 890 1234 PRINT 1234567890 1234 1 2345678901234E 09 12
5. BASIC Answers to BASIC Questions Since our BASIC first came on the market we have had many questions regarding some of the more complex features of both our BASIC and Extended BASIC The purpose of this document is to help clear up some of the confusion and to demonstrate some of the features which make BASIC and Extended BASIC so powerful This document can be divided into two sections clarification of features found in both BASIC and Extended BASIC and clarification of features found only in Extended BASIC Each command statement or feature references sections of the User s Manual where the initial discussion can be found In order to be complete several lines from the User s Manuals will be included in the discussions along with further explanation and examples It is recommended that the user also rereads these sections referenced since not all of the information in those sections will be duplicated here In this way the user may be able to understand the features in more depth It is our goal to try to present clear examples in both BASIC and Extended BASIC The user is advised to follow the examples carefully and to try variations of these examples to further reinforce their proper use The user is reminded that some of these examples may contain features not supported in their version of BASIC That is Extended BASIC examples should not be expected to work on a system supporting only BASIC With this in mind let s start BASIC Answe
6. E AS 3 20 FIELD 3 4 AS SQ 4 AS SR 4 AS LG 4 AS EX 30 REM set up arithmetic table 40 INPUT Size of the table SZ 50 FOR I 1 TO SZ 60 LSET SQ CVTF I I 70 LSET SR CVTF SQR I 80 LSET LG CVTF LOG I 90 LSET EX CVTF EXP I 100 PUT 3 110 NEXT I 120 REM now the table is set up 130 PRINT Enter a number between 1 and SZ 0 to exit 140 INPUT 0 NM 150 IF NM 0 THEN 200 160 IF NM gt SZ THEN 130 169 GET 3 RECORD NM 170 PRINT NM squared is 3CVT F SQ square root CVT F SR log 3CVT F LG and exp CVT F EX 190 GOTO 130 200 CLOSE 3 210 END This program will open the random file and set up the arithmetic values of numbers from one to a value selected by the user Once the random file is set up the user can randomly select number values from one to the size previously selected and display the square square root logarithm and exp of that number The user is invited to try this example and experiment with other variations BASIC Answers to BASIC Questions THE USR FUNCTION Section 11 The USR function can be employed when the user needs to perform some function not found in the instructions of BASIC or Extended BASIC For example you have a system that includes salesmen numbers For security you wish to encode these numbers and then when they are needed you may decode them A very simple way to encode the numbers is to perform a shift or rotate on the bits making up th
7. e number The next example shows how to call multiple USR functions to encode a number by rotating to the left and decode by rotating to the right 6800 Example 200 INPUT NUMBER TO ENCODE OR DECODE CD 210 UL HEX 24 UR HEX 25 220 XL HEX 6E XR HEX 00 230 YL HEX 6F YR HEX FO 240 INPUT ENCODE E OR DECODE D TY 250 IF TY E THEN 300 260 IF TY D THEN 400 270 GOTO 240 300 POKE UL XL POKE UR XR NU USR CD 400 POKE UL YL POKE UR YR NU USR CD With an assembly program like ORG 6E00 ENCODE LDAA 26 UPPER ORDER BYTE LDAB 27 LOWER ORDER BYTE CLE RESET CARRY TO 0 TSTB TEST THE SIGN BIT OF THE LOWER BYTE BPL ROTAT1 IF BIT IS O THEN CARRY BIT IS ALREADY 0 SEC ELSE THE CARRY BIT NEEDS SET TO 1 kkkkkkkkkkkkkkkkk THE CARRY BIT THE SAME AS THE SIGN BIT OF THE LOWER ORDER BYTE IS SHIFTED INTO THE BOTTOM OF THE UPPER ORDER BYTE kkkkkkkkkkkkkkkkk ROTAT1 ROLA kkxkkxkkkkkkkkkkkkk THE CARRY BIT THE SAME AS THE SIGN BIT OF THE UPPER ORDER BYTE IS SHIFTED INTO THE BOTTOM OF THE LOWER ORDER BYTE kkxkxkxkkkkkkkkkkkkk ROLB STAA 26 THE NEW ENCODED UPPER ORDER BYTE STAB 27 THE NEW ENCODED LOWER ORDER BYTE RTS ORG 6FFO DECODE LDAA 26 UPPER ORDER BYTE LDAB 27 LOWER ORDER BYTE CLC RESET CARRY TO 0 262 BASIC Answers to BASIC Questions ANDB 01 MASK OUT ALL BUT LOWEST BIT BEQ ROTAT2 IF LOWEST BIT IS O THEN CARRY IS 0 SEC ELSE THE CARRY BIT LOWEST BIT OF LOW BYTE ROTAT2 LDAB
8. embly routine should be placed at 7FFD 7FFF 2 and the argument is found at 7FFB 7FFF 4 Only the BASIC statements will be included here the 6809 assembly programmer can easily insert the proper code 6809 Example 200 INPUT NUMBER TO ENCODE OR DECODE CD 210 UL HEX 7FFD UR HEX 7FFE 220 XL HEX C1 XR HEX 00 230 YL HEX C2 YR HEX 00 240 INPUT ENCODE E OR DECODE D TY 250 IF TY E THEN 300 260 IF TY D THEN 400 270 GOTO 240 300 POKE UL XL POKE UR XR NU USR CD 400 POKE UL YL POKE UR YR NU USR CD 6 BASIC Answers to BASIC Questions And some 6809 assembly code ORG C100 START OF ENCODE ENCODE LDA 7FFB LDB 7FFC RTS ORG 200 DECODE LDA 7FFB LDB 7FFC RTS Furthermore the programmer of USR functions should also adjust the MEMEND check the user s manual for location in your version value in FLEX In this way BASIC will not use the memory space of the USR functions to run the BASIC program MEMEND should be set through the monitor to a value just under the USR functions first memory location For example set it less than 6E00 in the 6800 example above however as MEMEND is set to BFFF initially it does not need to be changed for the 6809 example above In order to load both the BASIC program and the assembled USR function the user needs to use the FLEX GET command to bring the assembled USR function into memory and the normal BASIC LOAD to load the BASIC progra
9. lues with only whole numbers the integer type will be faster and requires less space only 2 bytes Once again the user is advised to experiment with the differences in floating point and integer arithmetic and the mixing of the two BASIC Answers to BASIC Questions SCALE COMMAND Section 5 Another point of confusion lies with the SCALE command Users have asked us both why is it used and how is it used properly with SAVE and COMPILE programs Why is scaling used A computer uses the binary base 2 number system while humans use the decimal base 10 number system There are several values to the computer which cannot be accurately represented For example 1 3 base 10 represented in decimal and 0 1 base 10 and 0 01 base 10 represented in binary are very common values they are considered repeating digits and cannot be accurately represented in a finite amount of space ie 1 3 0 33333 in base 10 and 0 1 decimal 0 000110011001 to the computer in binary Because values such as 0 1 and 0 01 cannot be represented accurately in binary this inaccuracy as small as it may be is carried or propagated through successive arithmetical operations That is by adding subtracting multiplying etc the value 0 1 or any other values not accurately represented a number of times the small inaccuracy or error grows larger with each arithmetical operation Therefore to the computer 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 does no
10. m BASIC Answers to BASIC Questions Extended BASIC features INTEGERS vs FLOATING POINT Section 4 3 In Extended BASIC the user may use either integer variables constants or floating point variables constants A great deal of confusion has risen from the use of integers Integer arithmetic will always work in whole numbers no fractions are used as in floating point Therefore the results of integer arithmetic may be different than expected For example one may assume that 5 2 would yield 2 5 however in integer arithmetic the 5 is truncated to leave only the whole part 2 This means that 5 2 2 is 4 The answer is 4 because 5 2 is 2 not 2 5 in integer arithmetic and 2 2 is 4 10 3 is 3 not 3 333333 however 10 3 is 3 3333 This brings into the discussion the mixing of floating point and integer variables constants The computer calculates the result of arithmetic by grouping the operands and operators into groups of three based on the precedence of the operator moving from left to right That is in A B C D the A B is calculated first C D second and the addition of the two sub results last The type integer or floating point of the two operands or sub results determines the type of the result If one of the operands is floating point then the result of the operation will be floating point For example 5 2 5 2 is 4 5 Because 5 2 is 2 5 5 2 is 2 and 2 5 2 is 4 5 the answer is 4 5 If the user is using va
11. rs to BASIC Questions OUTPUT TO A PRINTER Section 8 6 Section 8 6 in the User s Manual contains information referenced here Channel 0 has special meaning for the PRINT statement Using channel 0 without using an OPEN statement when PRINTing is just like PRINTing to the terminal 200 PRINT 0 THIS IS A TEST will print THIS IS A TEST on the terminal exactly as if the 0 were not present Opening channel 0 allows you to send output to some other device such as a printer instead of the terminal Using OPEN on channel 0 tells BASIC to read the file name specified as a file containing a printer driver routine such as PRINT SYS in FLEX and use this new output routine whenever PRINT 0 is specified For example 10 OPEN 0 PRINT AS 0 20 PRINT 0 TABLE OF SQUARES AND SQUARE ROOTS 30 FOR I 1 TO 100 40 PRINT 0 1 1 1I SQR I 50 NEXT I 60 CLOSE 0 Line 10 tells BASIC to use file O PRINT SYS SYS is the default extension when channel 0 is referenced as the output routine on channel 0 Notice the OPEN OLD was not necessary On line 20 we print a header line which is output through the printer routines found in the PRINT SYS file Line 40 will now print on the printer the values of the number I the square and the square root of the number I The CLOSE 0 statement will now make all following PRINT 0 statements act just as in the example at the beginning all output will be displayed on the terminal until another OPEN AS 0 sta
12. t quite equal 0 8 In situations where you may have several hundreds or thousands of computations this error called round off error could result in false results or in improper testing in IF statements For example if you were testing for equality of a computed result with a constant due to round off error they may not be exactly equal For example 10 FOR 19 1 TO 1000 20 F F 0 1 30 NEXT 19 40 IF F 100 THEN PRINT EQUAL 50 END Without setting the SCALE factor EQUAL will never be printed the test of equality will fail Now where does scaling come in If we could tell the computer to do something to 0 1 or 0 01 in order that it would accurately represent these values we would no longer need to worry about round off error Scaling is the answer With the SCALE factor set to non zero Extended BASIC scales all floating point values by 10 SCALE factor and rounds to a whole number In effect all floating point numbers become integers the fractional parts disappear With a SCALE factor of 1 0 1 becomes 1 0 internally and 0 29 becomes 3 0 internally rounded to whole number 3 Using a SCALE factor of 1 the above program will now print EQUAL the equality test will succeed With a SCALE factor of 2 0 1 becomes 10 0 0 01 is 1 0 0 29 is 29 0 and 0 184 is 18 0 rounded Please note 0 01 is only 1 0 to the computer internally used in arithmetic the computer scales all values performs arithmetic and then will print them out
13. tement is executed It is very important that the file specified in the OPEN AS 0 statement is actually a printer system file such as PRINT SYS If it is not unpredictable results will occur possibly crashing BASIC BASIC Answers to BASIC Questions VIRTUAL ARRAYS AND RECORD 1 0 Section 10 2 10 14 There still exists some confusion about virtual arrays and record 1 0 An example using both of these powerful features should aid in clarifying their use 10 OPEN EMPLOYEE AS 1 20 FIELD 1 20 AS NM 22 AS Z1 2 AS DI 64 AS Z2 30 OPEN CHANGES AS 2 40 DIM 2 DV 999 50 FOR I 0 TO 999 60 GET 1 70 LSET DI DV I 80 PRINT NM division changed to DI 90 PUT 1 100 NEXT I This program is designed to mass update the employee s division number of an employee file stored in record I O form with a table of division changes stored as a virtual array Line 10 OPENs the file EMPLOYEE on channel 1 Line 20 describes the file s format Each employee record is a total of 108 bytes long however BASIC will use an entire sector 252 bytes to store it The two variables Z1 and Z2 are used to preserve spaces to position DI properly Line 30 then OPENs the virtual array on channel 2 This virtual array is dimensioned as 1000 elements in line 40 The loop lines 50 through 100 goes sequentially down the file and array making the proper change and displaying a message to the terminal Realize that with record 1 0 we could ha
14. ve done the GETs and PUTs ina random order The same program could have been written using a two dimensioned virtual array with the employee number in one field and the division change in the other field Now the GETs and PUTs depend upon the employee s number For example 10 OPEN EMPLOYEE AS 1 20 FIELD 1 20 AS NM 22 AS Z1 2 AS DI 64 AS Z2 30 OPEN CHANGES AS 2 40 DIM 2 DV 350 1 50 FOR I 0 TO 350 60 GET 1 RECORD CVT F DV 1I 0 70 LSET DI DV I 1 80 PRINT NM division changed to DI 90 PUT 1 RECORD CVT F DV I 0 100 NEXT I This example works just like the previous one except that the employee number and record number is found in the first column of the virtual array table while the division change is in the second column One more example just using record I 0 with random access will help BASIC Answers to BASIC Questions to clarify record I O This example can be typed in and run because it will set up its own data This example will use floating point numbers in a random file therefore there will be a difference between the standard BASIC and Extended BASIC version of this program Standard BASIC uses only four bytes to store a floating point number whereas Extended BASIC will need eight bytes to store a floating point number This means that in the example below the FIELD 3 4AS 4AS should have the 4 s replaced with 8 s to run on the Extended BASIC versions 10 OPEN NEW TABL
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