Join 109,336 Programmers for FREE! Ask your question and get quick answers from experts. There are 1,006 online right now! We've got more than 500 tutorials and 2,000 snippets. Join and find out why Dream.In.Code is the #1 programming help community on the internet! Registration is fast and FREE... Join Now!
I am a student from lovely sunny Queensland in Australia. I am currently studying for a Bachelor of Technology (Infomation Systems). I am the 1/2 point and am really enjoying the challenge.
This current assignment that I am doing is a little too challenging for me. It has a list of 4 requirements that progressively build on each other. I have managed to get requirement 2 working. I am not sure if it is acceptable to post the assignment specs here or not so I will paraphrase them.
requirement 1: add two digits that are entered seperated by a space and then display the answer ie 6 8 The sum is 14. That is all good, got that working.
requirement 2: modify to accept multi-digit integers with the same number of digits up to 3 almost got that bit under control, it works until there is an overflow. ie the sum results in a 4 digit answer I end up getting an odd hex answer, I think I know where the problem is but really have no idea how to solve it, any help would be greatly appreciated. Once I have that bit sorted then maybe I can move on, or ask for more help
requirement 3: modify the program to accept multi-digit integers with numbers of different lengths. This I think I can work out but any hints wouldn't be rejected
and finally requirement 4: modify the program to accept floating point number of up to 3 digits each with exactly the same format,that is , with the same number of digits to either side of the decimal. This I have absolutely no idea how to do, again any hints would be greatly appreciated.
This is the segment of code where I think I have the problem, if more code is need I am happy to add all of it, but thought just for starter this would be a good place to be.
Sorry I am not very good at this, my whole assignment is a mismatch of bits of code found here and there, I am sure there is quite a bit of it that does absolutely nothing. I really do apologise for my ignorance, but I am giving this assignment 120% effort, I am just not getting it very well.
Thanks in advance for any help given, it will be very much appreciated.
CODE
CONVERT MOVE.W D2,D0;get the number ASR.W D1,D0;shift the number first right bit AND.W #$000F,D0;discard the other bits for this exercise ;(we can only work on one digit at a time) ADD.W #'0',D0;convert to ascii by adding zero CMP.W #'9',D0; check if the number is greater than 9 (hex digit max)
BLE NEXT_BIT;if it is greater than nine move the bit left ADD.W #$07,D0; add seven to the value to take to the next bit
NEXT_BIT TRAP #14;complete the outch function and display character SUBQ.W #04,D1; subtract 4 to point to the next digit BGE CONVERT;repeat convert subroutine if there are more digits to process MOVE.W D2,D0
Can you please post all of the code you have and I can have a look at it.
thanks
Hi,
I have solved one of my problems, and I think I have requirement 3 working, but requirement 4, oh boy not a clue. Can't even find vague examples to pull apart on that one.
CODE
*** DEFINE CfONSTANTS *** FIRST DC.W $2000; memory for first number SECOND DC.W $2100; memory for second number SUM DC.W $2300 ; FOR FINAL SUM OF THE TWO NUMBERS
; EQUATES FOR KERNEL ROUTINES
INCH EQU 247 ;INCH, READS A KEYBOARD ASCII KEY INTO D0 (PP91) OUTCH EQU 248 ;OUT, OUTPUTS ASCII IN D0 TO THE SCREEN (PP 91) OUTPUT EQU 243 ; EXIT EQU 228 ; EXIT, EXITS BACK TO THE SIMULATOR LF EQU $0A ; LINE FEED CR EQU $0D ; CARRIAGE RETURN NULL EQU 00 ; NULL VALUE SPACE EQU $20 ; SPACE IN ASCII
LINE_BREAK DC.B CR,LF,NULL PROMPT DC.B 'Enter two digits separated by a space.',CR,LF,NULL FINAL DC.B 'The sum is ',CR,LF,NULL
ADDING DC.B $20,NULL ; to identify the end of the first number (a space) EQUALS DC.B $0D,NULL ; to identify the end of the calculation (a carriage return)
ORG $1000
MOVE.W #$7FFE,SP ; INITIALISE THE STACK
CLR.L D0 ; clear d0 for input numbers (for debugging) CLR.L D1 ; CLEAR D1 FOR THE SUM CLR.L D2 ; clear D2 for second number
START MOVE.W #PROMPT,A1 ; send first message out BSR WRITE_STRING ; go to write_string to see how
MOVE.W #ADDING,A1 ; find end of first number BSR GET_NUMBER ; get first number
MOVE D1,FIRST
MOVE.B #OUTCH,D7 ; output space that has been entered TRAP #14 ; tell system to do it
MOVE.L #EQUALS,A1 ; point to the end of number two BSR GET_NUMBER ;
MOVE D1,SECOND ;store the new number in second memory location
MOVE.L #LINE_BREAK,A1 ; put a line break in BSR WRITE_STRING ; write to screen
MOVE.W FIRST,D1 MOVE.W SECOND,A2 ADDA.W D1,A2 ;do the addition,result in A2
MOVE.L #FINAL,A1 ; call final message JSR WRITE_STRING ; write to screen
PRINT_DECIMAL ;assumes the addition result is in A2 MOVE.L #0,D3 ; prepare the print loop counter CLR D4 MOVE.L A2,D4 ; copy sum to D4 BRA PRINT_DECIMAL_LOOPBODY;jump straight into the decimalisation part of the loop, bypassing the loop test
PRINT_DECIMAL_DIV10LOOP CMP.W #0,D4 ;check if the last base 10 division returned 10 BEQ PRINT_DECIMAL_PRINTLOOP;if so we can start printing the 0...9 values on the stack
PRINT_DECIMAL_LOOPBODY DIVU #10,D4 ;divide by ten to get the 0...9 value in the remainder for SWAP D4 ;the current least significant decimal value in D4 MOVE.W D4,-(sp) ;push that remainder onto the stack MOVE.W #0,D4 ;clear the remainder SWAP D4 ;keep the reduced exponent value ADD.L #1,D3 ;increment digit counter BRA PRINT_DECIMAL_DIV10LOOP;jump to loop condition test
PRINT_DECIMAL_PRINTLOOP ;pops all the pushed base 10 denominators off the stack in a loop CMP.L #0,D3 ;and prints them to the console as it goes so that the most significat BEQ FINISHING ;quit if we've printed all digits
SUB.L #1,D3 ;decrement the digit counter index MOVE.W (sp)+,D0 ;pop the next less significant bit off the stack ADD.B #'0',D0 ;make it an ascii numeral MOVE.B #OUTCH,D7 TRAP #14 ;print it BRA PRINT_DECIMAL_PRINTLOOP;check for next digit
*** finishing routine to exit
FINISHING MOVE.B #EXIT,D7;call the exit function TRAP #14 ; tell system to do it
RTS
*** WORKING WITH THE NUMBERS ***
GET_NUMBER move.w d7,-(sp) ; pop off stack MOVE.W D3,-(SP); MOVE.W D2,-(SP);
CLR.L D1 ; clear contents of D1 for new number
MOVE.B #$03,D3 ; SET MAXIMUM OF D3 TO 3 DIGITS (this will be our counter of how many digits in a number)
READ_NUMBER MOVE.B #INCH,D7; READ CHARACTER FROM KEYBOARD TRAP #14 ; TELL SYSTEM TO DO IT MOVE.L A1,A0 ; SET ADDRESS POINTER TO THE END OF FIRST NUMBER
CHECK_ENTRY ; IS THIS THE END OF THE NUMBER MOVE.B (A0)+,D2;get the space or carriage return CMP.B #NULL,D2; is this the last ? BEQ NUMBER_ENTRY; IF IT IS go to the number entry CMP.B D0,D2 ; is it an equals (aka carriage return) BEQ LAST ; IF IT IS go to the last routin BRA CHECK_ENTRY;ELSE GO THROUGH THIS ROUTINE AGAIN
NUMBER_ENTRY CMP.B #$00,D3 ;SEE THAT THE LIMIT OF DIGITS ISN'T EXCEEDED BEQ READ_NUMBER;IF NOT GET NEXT KEY ENTRY
SUBQ.B #$01,D3 ;REDUCE D3 COUNTER BY 1 MOVE.B D0,D2 ; STORE THE VALUE ENTERED IN D2
MOVE.B #OUTCH,D7;ECHO KEY ENTERED TRAP #14 ; TELL SYSTEM TO DO IT
MOVE.B D2,D0 ;COPY ENTERED VALUE BACK TO D2 SUB.B #'0',D0 ; convert to a number by subtracting ascii zero value CMP.B #$09,D0 BLE OKAY SUBQ.B #$07,D0 ; IF VALUE OVER 9 THEN TAKE AWAY 7 TO ENSURE IT ISN'T RECORDED AS HEX (aka 10-16 is A-F)
OKAY MULU #10,D1 ; multiply by 10 ADD.W D0,D1 BRA READ_NUMBER;GET NEXT ENTRY
LAST MOVE.W (SP)+,D2 MOVE.W (SP)+,D3 MOVE.W (SP)+,D7
RTS
*** Other routines, taken from SAMPLES.ASM provided at the start of this subject ***
WRITE_STRING MOVE.W D7,-(SP) ;store modified registers ensuring that the MOVE.W D0,-(SP) ; subroutine is non-destructive MOVE.B #OUTCH,D7 ;ready screen output trap NEXT MOVE.B (A1)+,D0 ;get the next character in the string CMP.B #NULL,D0 ;exit if null-an explicit check allows the BEQ ENDSTRING; null constant to be defined as non-zero. TRAP #14 ; tell system to do it BRA NEXT ;get the next character ENDSTRING MOVE.W (SP)+,D0 ;restore modified registers ensuring that the MOVE.W (SP)+,D7 ; subroutine is non-destructive RTS ;return to the calling procedure
I think solving Requirement #4 is pretty easy, given how far you've already gotten. A little insight into the problem that might help is that adding two numbers with decimals is just as easy as adding two integers without decimals.
CODE
123456 +444333 ------ 567789
just as
CODE
123.456 +444.333 ------- 567.789
If you know that you've always got three digits on either side of the decimal point, you're in great shape; just ignore the decimal when reading input, and output it after the third digit when writing your answer.
If you don't know how many digits here are, it's not so hard. You just have to make a note of where you find the point, flood everything else with zeros, and keep the values aligned around their decimal points.
Everything is easier said than done in assembler, so let me know if you need help coming up with an implementation that reaches that goal.
This post has been edited by mikeblas: 22 May, 2008 - 06:58 PM
Thanks heaps, I have already handed the assignment in, I just couldn't figure out the floating point thing, but what you say makes perfect sense, I just couldn't code it to make it work. I will have to settle for a maximum of 30/40, but for curiosity sake and my own sanity, I would still like to know how to make it work. My friends tell me to let it go, but I can't, I really want to know the answer.
OKAY MULU #10,D1; multiply by 10 ADD.W D0,D1 BRA READ_NUMBER;GET NEXT ENTRY
I would like to enquire re: this part of the code. Why do you have to multiply by 10?
Well at the time of posting that it was the only way I could get my output to display as a decimal digit, without that it would display as a hex digit. I am sure that isn't the way to do it, but not having enough knowledge of assembly language this appeared to work for me. I haven't got the results back from that assignment yet so I really can't tell you if it was acceptable or not, but my theory is that the assignment specs didn't say how we were to achieve the desired outputs, just that we had to achieve them, and I did
Thanks for explaining! It sure is an interesting program to ponder on! I tried to make it reach requirement 4 but have either address error or bus trap error... But I'll still be trying on it during my free time just for fun.
This post has been edited by Joolia: 9 Jun, 2008 - 10:07 PM
PRINT_DECIMAL_LOOPBODY DIVU #10,D4 ;divide by ten to get the 0...9 value in the remainder for SWAP D4 ;the current least significant decimal value in D4 MOVE.W D4,-(sp) ;push that remainder onto the stack MOVE.W #0,D4 ;clear the remainder SWAP D4 ;keep the reduced exponent value ADD.L #1,D3 ;increment digit counter BRA PRINT_DECIMAL_DIV10LOOP;jump to loop condition test
I'm wondering abt this part too. Does it mean:
For e.g.: 111 + 2
This part is doing the (11 + 2 = 13) 13 / 10 to get the remainder 3 and swap the remainder (03) => 30 store it using parameter passing => clear this part of D4 and the rest I'm not really sure... How does this part of coding actually work?
PRINT_DECIMAL_LOOPBODY DIVU #10,D4 ;divide by ten to get the 0...9 value in the remainder for SWAP D4 ;the current least significant decimal value in D4 MOVE.W D4,-(sp) ;push that remainder onto the stack MOVE.W #0,D4 ;clear the remainder SWAP D4 ;keep the reduced exponent value ADD.L #1,D3 ;increment digit counter BRA PRINT_DECIMAL_DIV10LOOP;jump to loop condition test
I'm wondering abt this part too. Does it mean:
For e.g.: 111 + 2
This part is doing the (11 + 2 = 13) 13 / 10 to get the remainder 3 and swap the remainder (03) => 30 store it using parameter passing => clear this part of D4 and the rest I'm not really sure... How does this part of coding actually work?
To be honest I really don't know, I was grabbing bits of code from where ever I could find it trying to work out a solution, and when this did as I wanted I just accepted it and moved on. I am definitely not as adventurous as you are, I figure if it is working I am not going to question it! I had my exam for that subject last Wednesday, it wasn't as hard as I thought, luckily a large component of it was maths, ie, adding and subtracting binary, hex and floating point and error correcting with hamming code which I am good at, and I think I nailed the coding part too so I was very happy with that. I might even be looking at a distinction for that subject, now that would be funny.
When you do work it out I would love to know the answer out of curiosity, I am doing eCommerce and Managerial Communication this semester, so not quite as challenging or entertaining as assembly language.
To be honest I really don't know, I was grabbing bits of code from where ever I could find it trying to work out a solution, and when this did as I wanted I just accepted it and moved on. I am definitely not as adventurous as you are, I figure if it is working I am not going to question it! I had my exam for that subject last Wednesday, it wasn't as hard as I thought, luckily a large component of it was maths, ie, adding and subtracting binary, hex and floating point and error correcting with hamming code which I am good at, and I think I nailed the coding part too so I was very happy with that. I might even be looking at a distinction for that subject, now that would be funny.
When you do work it out I would love to know the answer out of curiosity, I am doing eCommerce and Managerial Communication this semester, so not quite as challenging or entertaining as assembly language.
Good luck with it. cheers giddyupgirl
It's ok then. I'll try to figure it out and continue to ponder on the question. I'll let you know when I've got it figured out?
16 May, 2008 - 03:44 AM
But I'll still be trying on it during my free time just for fun. 
I am definitely not as adventurous as you are, I figure if it is working I am not going to question it! I had my exam for that subject last Wednesday, it wasn't as hard as I thought, luckily a large component of it was maths, ie, adding and subtracting binary, hex and floating point and error correcting with hamming code which I am good at, and I think I nailed the coding part too so I was very happy with that. I might even be looking at a distinction for that subject, now that would be funny. 
I'll try to figure it out and continue to ponder on the question. I'll let you know when I've got it figured out? 
|