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elRey
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« on: March 29, 2012, 04:48:07 PM »
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Kind a general thread where one can ask what one or a small set of instructions is doing. NOT for asking what a whole function or several lines is doing. In the spirit of us newbies trying to wrap our heads around assembly language itself, not so much ME7. I'll start. the middle line in: mov r4, #1000h
extp #0E1h, #1 ; 'ß'
mov flamsl_w_word_3849F0, r4
So what's going on after r4 is set to 1 and before r4 (now 1) is moved to RAMword_3849F0 (flamsl_w)? Thanks, Rey |
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« Last Edit: March 29, 2012, 04:55:12 PM by elRey »
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nyet
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« Reply #1 on: March 29, 2012, 04:54:46 PM »
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Not that I know anything about C166 ASM, but from the C166 manual:
Syntax EXTP op1, op2
Operation (count) ← (op2) [1 ≤ op2 ≤ 4]
Disable interrupts and Class A traps
Data_Page = (op1)
DO WHILE ((count) ≠ 0 AND Class_B_trap_condition ≠ TRUE)
Next Instruction
(count) ← (count) - 1
END WHILE
(count) = 0
Data_Page = (DPPx)
Enable interrupts and traps
Description Overrides the standard DPP addressing scheme of the long and
indirect addressing modes for a specified number of instructions.
During their execution both standard/PEC interrupts and class A
hardware traps are locked. The EXTP instruction becomes
immediately active such that no additional NOPs are required.
For any long (‘mem’) or indirect ([…]) address in the EXTP
instruction sequence, the 10-bit page number (address bits A23 -
A14) is not determined by the contents of a DPP register but by the
value of op1 itself. The 14-bit page offset (address bits A13 - A0)
is derived from the long or indirect address as usual. The value of
op2 defines the length of the effected instruction sequence.
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elRey
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« Reply #2 on: March 29, 2012, 04:56:49 PM »
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Yeah, I read that..... Over my head a little. I should add to 1st post 'For Dummies'
What I'm looking for in an answer is something like...
It jump to address 0E1+1000 (address 10E1) and runs one line of code (#1) then jumps back and continues.
Obviously that's not it, but explained liked that, please.
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« Last Edit: March 29, 2012, 05:01:11 PM by elRey »
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nyet
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« Reply #3 on: March 29, 2012, 04:58:23 PM »
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So
EXTP #0E1h, #1
means
"ignore dpp for one (1) instruction following this EXP, and use page 0xe1 instead"
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nyet
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« Reply #4 on: March 29, 2012, 05:00:31 PM »
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In other words, I know the exact address is
flamsl_w_word_3849F0 | (0xe1 << 14)
and not
flamsl_w_word_3849F0 | (dpp << 14)
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nyet
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« Reply #5 on: March 29, 2012, 05:09:44 PM »
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You're telling the cpu that this is a location that is definitely not in the current page pointed to by dpp (but you know exactly where it is, since its a constant absolute, not relative, location)
If you're an x86 programmer, this is basically a "far" pointer deference (update both the segment and offset).
on the C166, you can do this in one fell swoop (protected atomically) w/o have to save/restore the segment.
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elRey
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« Reply #6 on: March 29, 2012, 05:10:31 PM »
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Thank you.
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elRey
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« Reply #7 on: May 09, 2012, 07:18:33 PM »
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Trying to figure out relative locations... Starting @ address 8B41A: mov r12, #2148h
mov r13, #206h
movbz r14, byte_F89C
movbz r15, byte_380A61
calls 83h, LookupM_833f24 ; KFLDRXO Lookup What address does the first line refer to? (#2148h = what address? xx148) From another file starting @ address 8E398: mov r12, #19BAh
mov r13, #206h
movbz r14, byte_F89C
movbz r15, byte_380A5F
calls 82h, LookupM_825eac ; KFLDRXO Lookup I know the first line refers to address 199BA (#19BAh = 199BA) When I see address references like this, how can I figure out what absolute address they are pointing to? I don't understand the #1xxx vs #2xxx part. Thanks, Rey |
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« Last Edit: May 09, 2012, 07:22:56 PM by elRey »
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nyet
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« Reply #8 on: May 09, 2012, 07:32:17 PM »
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i dont understand what you mean by #1xxx and #2xxx. that makes no sense :/
#19BAh is 0x19BA
#206h is 0x0206
the upper bits of the actual address are whatever is in the appropriate DPP register.
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nyet
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« Reply #9 on: May 09, 2012, 07:39:24 PM »
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I know the first line refers to address 199BA (#19BAh = 199BA)
the bottom 14 bits of 0x199BA is 0x199BA & 0x0x3fff = 0x19ba the upper bits of 0x199BA is 0x18000>>14 = 6 So dpp is 6. so dpp | 0x206 is 6<<14 | 0x206 = 0x18000 | 0x206 = 0x18206 |
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elRey
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« Reply #10 on: May 09, 2012, 07:48:49 PM »
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i dont understand what you mean by #1xxx and #2xxx. that makes no sense :/ sorry. b/c of my lack of understanding. comparing the first line of the two files: mov r12, #2148h vs mov r12, #19BAh I know the 2nd ( #19BAh) = 199BAH because I've mapped that file. But the 1st ( #2148h) I can't figure out. Seemed to me that only the last 3 digits were part of the absolute address (148 and 9BA). I can't figure out the #1 and #2 that each begins with. the bottom 14 bits of 0x199BA is 0x199BA & 0x0x3fff = 0x19ba Can you explain this using #2148h please? |
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« Last Edit: May 09, 2012, 07:53:49 PM by elRey »
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nyet
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« Reply #11 on: May 09, 2012, 09:34:56 PM »
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#1 and #2 doesn't mean what you think it does.
ignore the #
it just means its a constant.
dpp | 0x2148 is 6<<14 | 0x2148 = 0x18000 | 0x2148 = 0x1a148
14 bits is halfway through that digit. You can't do it by digits.
8|2 = 10 = 0xa
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matchew
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« Reply #12 on: May 09, 2012, 09:39:51 PM »
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But the 1st (#2148h) I can't figure out.
0x1A148 The clue is in the 2nd line down in both examples. R12 and R13 are both used to describe an address outside of the current data page. |
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« Last Edit: May 09, 2012, 09:43:11 PM by matchew »
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rajivc666
Full Member
Karma: +23/-2
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Posts: 127
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« Reply #13 on: May 09, 2012, 11:22:19 PM »
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Ideally You will have to reverse the two functions that are called (calls 83h, LookupM_833f24 & calls 82h, LookupM_825eac) to understand which memory address they are referring to. But as matchew said it is most likely going to be 206H * 4000H + #2148h = 8, 1a148h in the first case and similar in the second.
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nyet
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« Reply #14 on: May 14, 2012, 01:27:30 PM »
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Also, you have to figure out what data segment each subroutine thinks they sit in, and keep an eye out for anything that modifies the segment register(s) (for this processor, the dpp registers)
Otherwise, you'll have no prayer at figuring out what a near pointer is actually pointing to.
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