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15C programming question - Eric Zoob - 03-27-2012
Routine A dimensions matrices A,B, and sets result to C.
Routine E multiplies AxB.
Without the second (repeated) RTN on line 22, the program does not return to 000, but will continue through and multiply AxB. What am I doing wrong here?
001- 42,21,11 LBL A
002- 2
003- 36 ENTER
004- 2
005- 42,23,11 DIM A
006- 2
007- 36 ENTER
008- 1
009- 42,23,12 DIM B
010- 42,26,13 RESULT C
011- 42,16,1 MATRIX 1
012- 0
013-u 44 11 STO A
014- 1
015-u 44 11 STO A
016- 1
017- 16 CHS
018-u 44 11 STO A
019- 0
020-u 44 11 STO A
021- 43 32 RTN
022- 43 32 RTN
023- 42,21,15 LBL E
024- 45,16,11 RCL MATRIX A
025- 36 ENTER
026- 45,16,12 RCL MATRIX B
027- 20 Multiply
028- 43 32 RTN
Edited: 27 Mar 2012, 2:08 p.m.
Re: 15C programming question - Kiyoshi Akima - 03-27-2012
Step 20 follows the "do if true" paradigm of the conditional and looping commands. It does the STO, increments the counters, and skips the next step if it had gone past the matrix boundary.
Re: 15C programming question - Tom Sauntry - 03-27-2012
Pardon my ignorance, but what are the lower-case "u"s preceding the 44 11 STO commands in the listing.
Re: 15C programming question - Gene Wright - 03-27-2012
means they were keyed in while in USER mode.
Changes the behavior of some commands related to matrix, etc.
Re: 15C programming question - Eric Zoob - 03-27-2012
Thanks. I found it in the manual. That's a very nice feature, actually.
Re: 15C programming question - C.Ret - 03-28-2012
Bonjour,
This topic is a good reason, actually a good excuse, to practice and remind me some of my HP-15C's code.
That's how I read the proposed code and interprets its progress. Feel free to comment and report any oversight or misinterpretation.
Key Code MNEMOMICS Merge Commands Remarks001-42,21,11 LBL A "A" Sub-program "A"
002- 2 2
003- 36 ENTER
004- 2 2
005-42,23,11 DIM A (2,2) DIM A Dimension Data Matrix A to 2x2
006- 2 2
007- 36 ENTER
008- 1 1
009-42,23,12 DIM B (2,1) DIM B Dimension Data Matrix B to 2x1
010-42,26,13 RESULT C Result in C Designates Matrix C for storing result
011-42,16, 1 MATRIX 1 Matrix Entry Restart entry from home position (1,1)
012- 0 0
013-u 44 11 STO A A(1,1) <-- 0 Matrix A Entry
014- 1 1
015-u 44 11 STO A A(1,2) <-- 1
016- 1 1
017- 16 CHS
018-u 44 11 STO A A(2,1) <-- -1
019- 0 0
020-u 44 11 STO A A(2,2) <-- 0
021- 43 32 RTN ............ step skiped since 'End of Matrix reached' (roll over)
022- 43 32 RTN End of "A"023-42,21,15 LBL E "E" Sub-Program "E"
024-45,16,11 RCL MATRIX A
025- 36 ENTER ? ? ?
026-45,16,12 RCL MATRIX B
027- 20 × C <-- A × B Matrix Product
028- 43 32 RTN End of "E"
Re: 15C programming question - Luiz C. Vieira (Brazil) - 03-28-2012
Hi;
I'd just add a few remarks, if you allow me to:
Key Code MNEMOMICS Merge Commands Remarks001-42,21,11 LBL A "A" Sub-program "A"
002- 2 2
003- 36 ENTER
004- 2 2 (unnecessary)
005-42,23,11 DIM A (2,2) DIM A Dimension Data Matrix A to 2x2
006- 2 2 (unnecessary)
007- 36 ENTER (unnecessary)
008- 1 1
009-42,23,12 DIM B (2,1) DIM B Dimension Data Matrix B to 2x1
010-42,26,13 RESULT C Result in C Designates Matrix C for storing result
011-42,16, 1 MATRIX 1 Matrix Entry Restart entry from home position (1,1)
012- 0 0
013-u 44 11 STO A A(1,1) <-- 0 Matrix A Entry
014- 1 1
015-u 44 11 STO A A(1,2) <-- 1
016- 1 1 (unnecessary)
017- 16 CHS
018-u 44 11 STO A A(2,1) <-- -1
019- 0 0
020-u 44 11 STO A A(2,2) <-- 0
021- 43 32 RTN ............ step skiped since 'End of Matrix reached' (roll over)
022- 43 32 RTN End of "A"023-42,21,15 LBL E "E" Sub-Program "E"
024-45,16,11 RCL MATRIX A
025- 36 ENTER ? ? ?
026-45,16,12 RCL MATRIX B
027- 20 × C <-- A × B Matrix Product
028- 43 32 RTN End of "E"
Step #4: after [ENTER] both X and Y registers already filled with 2.00, no need to enter again;
Steps #6 and #7: Same as above
Step #16: there is a copy of 1 still in the X-register, a single CHS suffices to let -1 in the X-register
And you are correct: step #25 is not needed.
Since the elements and the indexes for the matrices are not used after their contents are set, reusing them instead of introducing them back once they are still in the memory stack reduces the number of program steps. And allow the extra few bytes to be used in other programs.
Cheers.
Luiz (Brazil)
Edited: 28 Mar 2012, 5:52 a.m.
Re: 15C programming question - C.Ret - 03-28-2012
You are welcome ! :-)