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elRey
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« Reply #45 on: June 27, 2013, 12:40:27 PM »
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wow. The whole time I'm reading r14 not rl4. Thanks to both of you.
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elRey
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« Reply #46 on: October 28, 2013, 12:52:57 PM »
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I need help changing referenced mem location. mov [-r0], r9
mov [-r0], r7
mov [-r0], r6
sub r0, #2
extp #0E1h, #1 ; 'ß'
movb rl4, fnwue_byte_384905
cmpb rl4, #0FFh
jmpr cc_NZ, loc_8A1922
in hex: 88 90
88 70
88 60
28 02
D7 40 E1 00
F3 F8 05 09
47 F8 FF 00
3D 10
I'd like to change it so instead of comparing fnwue_byte_384905 with #0FFh, it compares newvar_byte_380ACC. I tried changing F3 F8 05 09 to F3 F8 CC 8A but there seems to be a segment issue. It then references lamfaw_w_word_384ACC. How do I get it to 380xxx instead 384xxx ? Thanks, Rey |
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elRey
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« Reply #47 on: October 28, 2013, 01:44:16 PM »
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Funny thing, this seems to point back to the original post  I nop the extp line and it worked. |
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« Last Edit: October 28, 2013, 01:50:08 PM by elRey »
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elRey
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« Reply #48 on: July 08, 2014, 08:41:55 AM »
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Can someone explain why the lines circled in yellow are needed? I think I get the third circle. It's compiling the results of the mulu above. edit: I included more of the function. Is the first circle needed because of r3 being the result of a divlu further up? Would it be needed if it were only udslsum_l_word_380E06 (move r3, udslsum_l_word_380E06) ? This is ggdsas_ggdsl Thanks, Rey  |
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« Last Edit: July 08, 2014, 11:10:17 PM by elRey »
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phila_dot
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« Reply #49 on: July 09, 2014, 02:05:31 PM »
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The first one is making input voltage and DSLGRAD like factors.
The last two are converting 32 to 16 bit.
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elRey
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« Reply #50 on: July 09, 2014, 02:38:47 PM »
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Thank you! So if I start with 'like factors' I don't need that first one. Got it.
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terminator
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« Reply #51 on: August 25, 2014, 12:18:58 PM »
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Seg0x209@824000:5C1C mov r10, r14
Seg0x209@824000:5C1E extp r13, #1
Seg0x209@824000:5C20 mov r14, [r12+]
Seg0x209@824000:5C22 mov r4, #0
Seg0x209@824000:5C24 mov r5, #0
Please give me a hint what does [r12+] mean?
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dream3R
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« Reply #52 on: August 25, 2014, 02:21:42 PM »
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Add indirect word memory to direct GPR and post-increment source pointer by 2. (From the manual) |
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terminator
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« Reply #53 on: August 25, 2014, 02:31:21 PM »
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Thanks!) Couldnt find it)
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terminator
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« Reply #54 on: September 23, 2014, 02:23:23 PM »
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I know its a noob question but extp changes only op2? Or both op1 and op2?
For example,
extp r13, #1;
mov op1, op2
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terminator
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« Reply #55 on: September 23, 2014, 04:32:22 PM »
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This is part of LDRXN subroutine:
extp r13, #3 ; I dont understand it. Because if there are 20Ah pages (r13) after subtraction the final result will be 0, I mean without 20A pages, and looks like no sense to use EXTP in that case?
sub r4, [r2] ; nmot - LDRXN RPM
mov r5, [r2+2] ; LDRXN RPM to r5
sub r5, [r2]
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« Last Edit: September 23, 2014, 04:34:45 PM by terminator »
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phila_dot
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« Reply #56 on: September 23, 2014, 05:20:52 PM »
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I know its a noob question but extp changes only op2? Or both op1 and op2?
For example,
extp r13, #1;
mov op1, op2
No op1 is the page and op2 is the number of lines affected. This is part of LDRXN subroutine:
extp r13, #3 ; I dont understand it. Because if there are 20Ah pages (r13) after subtraction the final result will be 0, I mean without 20A pages, and looks like no sense to use EXTP in that case?
sub r4, [r2] ; nmot - LDRXN RPM
mov r5, [r2+2] ; LDRXN RPM to r5
sub r5, [r2]
r13 is the page and it is applied to the address that r2 points to for all three lines |
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terminator
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« Reply #57 on: September 24, 2014, 12:56:42 AM »
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Thank you! I got it.
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terminator
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« Reply #58 on: October 11, 2014, 01:51:28 PM »
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Please help me with [-r0]. r15, byte_81ED35; the offset contains 6 mov [-r0], r15; move 6 to r0, then r0 offset = 81ED33  |
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phila_dot
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« Reply #59 on: October 11, 2014, 02:33:27 PM »
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No, that is a stack pointer
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