CL-UTILS revision 2G released



#2

After much fiddling with Howard Owen's HEPINI function I managed to reduce its code lenght, reclaiming about 35 bytes. I also squeezed 50 more bytes from other sections, and got rid of Y1SEC - redundant with YSEC in the YFNS ROM - all in all freeing up space for four new functions, plus usability enhancements to the existing functionality.

Also as usual in this maintenance releases some minor bugs were fixed, none of them had been reported anyway.

Should be posted at TOS shortly, drop me a line if you need it before that.


#3

Me want.

#4

Me toooo!

Thanks Angel.

#5

Cool! I look forward to seeing your optimizations.


#6

Here it is in all its g(l)ory ;-)

It wasn´t easy, but the COCONUT CPU has such an extense instruction set that it´s doable. I have to say I´ve spent the unthinkable time trying to get this one shorter - but I did!


098	"X"	
019 "Y" Y: number of pages
010 "P" X: first page#
005 "E"
008 "H" Howard Owen
0B8 READ 2(Y) Get the number of pages N
38D ?NC XQ Convert it to hex
008 ->02E3 [BCDBIN]
0E6 B<>C S&X store N in B
0F8 READ 3(X) Get the first page number S
38D ?NC XQ Convert it to hex
008 ->02E3 [BCDBIN]
106 A=C S&X But store the result in A (A=S)
130 LDI S&X comparison with the start page# in A
010 CON: Results in (hex) 010 in C [1:0]
306 ?A<C S&X Valid start page # is: 7< S <16
0B5 ?NC GO Start >= 16, DATA ERROR
0A2 ->282D [ERRDE]
130 LDI S&X Now test the lower bound
007 CON: Load hex 07 in C [1:0]
0A6 A<>C S&X C=Start page
306 ?A<C S&X is 7 less than Start?
3D3 JNC -06 Nope, DATA ERROR
0E6 B<>C S&X B=Start page, S
106 A=C S&X A=# of pages, N
130 LDI S&X Valid #Pgs is 0 < N < 9
009 CON: Load 09 into C [1:0]
306 ?A<C S&X Is N < 9?
3D3 JNC -06 Nope, DATA ERROR
346 ?A#0 S&X Is N not equal to 0?
3C3 JNC -08 Nope, DATA ERROR
0C6 C=B S&X Start page in C[S&X]
070 N=C START page number in N[S&X]
0A6 A<>C S&X
106 A=C S&X copy #pgs N to C[S&X]
126 A=A+B S&X A(X)=start page + number of pages = highest page +1
0A6 A<>C S&X C(X)=hp+1, A(X)=page count
266 C=C-1 S&X C(X)=hp
158 M=C highest page number in M[S&X]
0A6 A<>C S&X A(X)=hp, C(X)=page count
0E6 B<>C S&X B(X)=page count, C(X)=START page
0C6 C=B S&X B(X) is the count down index (page count)
266 C=C-1 S&X Underflow will set the carry bit
067 ?C RTN Done!, end function.
0E6 B<>C S&X place decremented counter back in B(X)
0B0 C=N Calculate the current page from the base pg# & counter
106 A=C S&X base page number in A(X)
126 A=A+B S&X current page in A=(S+N-1)
379 PORT DEP: Initialize the current page
03C XQ pg# expected in A(X)
1ED ->A1ED [INITPG]
3B3 JNC -10d [LOOP] ,and loop
04E C=0 ALL Prepare to load ROM address into C
0A6 A<>C S&X current Page number to C
106 A=C S&X keep it in A
13C RCR 8 Rotate page in C(X) into MSN of address field (C[6:3])
0A6 A<>C S&X Fetch current page again
106 A=C S&X
040 WROM Write X to page base address
0A0 SLCTP C[6:3] is the address field
01C PT= 3 P points to the low nibble
0E0 SLCTQ
15C PT= 6 Q is the high nibble.
112 A=C P-Q A[ADR]=page base address
0B0 C=N START page number in N[S&X]
0B2 A<>C P-Q bring it back to C
112 A=C P-Q Save base address in N
070 N=C N[ADR]=page base address
010 LD@PT- 0 N[S&X]=lowest page number
3D0 LD@PT- F
390 LD@PT- E
1D0 LD@PT- 7 C[ADR]=page relative address of the last page pointer (lpp)
15C PT= 6 Set Q back from the LC3
152 A=A+C P-Q A[ADR]=absolute address of the lpp
092 B=A P-Q save xFE7 in B[ADR] for later
0B0 C=N ALL lowest page number in C[S&X]
31C PT= 1
010 LD@PT- 0
058 G=C @PT+ G= PCALC flag:lowest page number
379 PORT DEP: Set LPP and write it to ROM address
03C XQ Lowest Page Pointer = LPP
25D ->A25D [PCALC]
0D2 C=B P-Q restore previous absolute ADR
23A C=C+1 M adr: xFE8
112 A=C P-Q
198 C=M ALL highest page number in C[S&X]
31C PT= 1
050 LD@PT- 1 This flag says we will subtract from the current page to get the previous page
058 G=C @PT+ G has '1P', where 'P' is the highest page number.
379 PORT DEP: Set NPP and write it to ROM address
03C XQ Next Page Pointer = NPP
25D ->A25D [PCALC]
0D2 C=B P-Q restore previous ADR
010 LD@PT- 0 remove absolute nibble
23A C=C+1 M adr: xFE8
23A C=C+1 M adr: xFE9
3DC PT=PT+1
0F2 B<>C P-Q
130 LDI S&X
091 CON:
106 A=C S&X
379 PORT DEP: Load Character
03C XQ
250 ->A250 [LMLDL]
010 LD@PT- 0
3D0 LD@PT- F
390 LD@PT- E adr: xFED
350 LD@PT- D word: 090
15C PT= 6
0F2 B<>C P-Q
130 LDI S&X
090 CON:
106 A=C S&X
379 PORT DEP: Load Character
03C XQ
250 ->A250 [LMLDL]
0F2 B<>C P-Q
23A C=C+1 M adr: xFEE
23A C=C+1 M adr: xFEF
0F2 B<>C P-Q
130 LDI S&X
091 CON:
106 A=C S&X
379 PORT DEP: Load Character
03C XQ
250 ->A250 [LMLDL]
0F2 B<>C P-Q
23A C=C+1 M adr: xFF0
23A C=C+1 M adr: xFF1
0F2 B<>C P-Q
130 LDI S&X
0E5 CON:
106 A=C S&X
379 PORT DEP: Load Character
03C XQ
250 ->A250 [LMLDL]
0F2 B<>C P-Q
23A C=C+1 M adr: xFF2
0F2 B<>C P-Q
130 LDI S&X
00F CON:
106 A=C S&X
379 PORT DEP: Load Character
03C XQ
250 ->A250 [LMLDL]
0F2 B<>C P-Q
23A C=C+1 M adr: xFF3
0F2 B<>C P-Q
130 LDI S&X
200 CON:
106 A=C S&X
0B0 C=N ALL
09C PT= 5
010 LD@PT- 0
010 LD@PT- 0
010 LD@PT- 0
112 A=C P-Q
15C PT= 6
132 A=A+B P-Q ADR field
0B2 A<>C P-Q ADR field
112 A=C P-Q ADR field
0A6 A<>C S&X get byte value to C[S&X]
040 WROM
3E0 RTN
04E C=0 ALL
39C PT= 0
098 C=G @PT+
31C PT= 1
010 LD@PT- 0 C(X)=passed extreme (highest or lowest) page number
106 A=C S&X Save it in A[S&X]
0B0 C=N Get current address into C[ADR]
07C RCR 4 Rotate the high nibble into C[3]
3C6 RSHFC S&X Mask off the nibbles to the left
3C6 RSHFC S&X What remains is the current page number
246 C=A-C S&X Difference with current page number
2E6 ?C#0 S&X are they different?
03F JC +07 yes, go to [PNEQ]
15C PT= 6 no, go on
010 LD@PT- 0 Otherwise, we are on one of the extreme pages
010 LD@PT- 0 The pointer to the next or previous page is
010 LD@PT- 0 therefore 0.
03C RCR 3
0AB JNC +21d [PCONT]
0B0 C=N We are in a page between the extremes
07C RCR 4 Get current base address into C[ADR])
3C6 RSHFC S&X Rotate the high nibble into C[3]
3C6 RSHFC S&X Mask off the nibbles to the left
106 A=C S&X Store CURRENT page number in A(X)
39C PT= 0 Get the add/subtract flag
098 C=G @PT+
3CE RSHFC ALL
21C PT= 2
010 LD@PT- 0
010 LD@PT- 0
2E6 ?C#0 S&X Add/subtract flag
02F JC +05 ADD
0A6 A<>C S&X
106 A=C S&X Get the page number back in C[S&X]
266 C=C-1 S&X the current page minus 1 is the previous page
023 JNC +04 PCONT
0A6 A<>C S&X
106 A=C S&X C=A S&X
226 C=C+1 S&X the current page plus 1 is the next page
15C PT= 6
0B2 A<>C P-Q Load the previous pointer address
040 WROM and write the computed value
3E0 RTN


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