Chapter 12 - Assembly Language Examples
snowhite 29: more labtest.s
ORG $1000 ; start at $1000
main MOVE.W #$00FFFE,D0 ; x:=y
ADD.W #$0001,D0 ; x:=x+1
ADD.W #$0001,D0
ADD.W #$00FFFE,D0 ; x:= x + $FFFE
ADD.W #$0002,D0 ; x:= x+ 2
LEA.L $6000,A0 ;
LEA.L adr,A2
MOVEA.W $2000,A1
MOVE.W D0,(A0)
ORG $6000
stuff DC.W $2000
adr DC.W $4000
DC.W $5000
END main ; forces debug start at main
This program has two parts:
The normal sequence of running a program is:
snowhite 24: as68k -L demoprog.s > demoprog.lisThe input file has the extension .s. The assembler creates an output file demoprog.o for use by the linker and a list file demoprog.lis from the assembler. This list file shows the program, the assembled form of each instruction, memory locations used by the program, and all symbols defined in the program and their values. In particular, note the values of stuff, main, and adr in the following listing.
snowhite 25: more demoprog.lis
Hewlett Packard AS68000 V(01.20 10Mar88) Page 1 Thu May 21 22:50:52
Cmdline - as68k -L demoprog.s
Line Address
1
2
3 ORG $1000 ;start at $1000
4
5 00001000 303C FFFE main MOVE.W #$00FFFE,D0 ;x:=y
6 00001004 5240 ADD.W #$0001,D0 ;x:=x+1
7 00001006 5240 ADD.W #$0001,D0
8 00001008 0640 FFFE ADD.W #$00FFFE,D0 ;x:= x + $FFF2
10 0000100E 41F8 6000 LEA.L $6000,A0 ;
11 00001012 45F8 6002 4E71 LEA.L adr,A2
12 00001018 3278 2000 MOVEA.W $2000,A1
13 0000101C 3080 MOVE.W D0,(A0)
14
15 ORG $6000
16 00006000 2000 stuff DC.W $2000
17 00006002 4000 adr DC.W $4000
18 00006004 5000 DC.W $5000
19 END main ;starts at main
Symbol Table
Label Value
adr 00006002
main 00001000
stuff 00006000
Note: There is an important difference between the LEA commands in the above program. The instruction LEA.L $6000,A0 puts the number $6000 into address register A0. The instruction LEA.L adr,A2 puts the symbol ŇadrÓ (which has a value of $6002) into the instruction and has the effect of putting the number $6002 into A2. The 4E71 is a filler for the possibility that the linker would change the absolute value of ŇadrÓ from a word length $6002 to a different long word value.
snowhite 26: ld68k -L -o demoprog.x demoprog > demoprog.llisThe input file is assumed to have the extension .o. The assembler creates an output file demoprog.x for use by the debugger and a list file demoprog.llis from the linker. This list file shows the location in memory of the program and its components. In particular, note that there are two components of the program: one begins at address $1000 and the other begins at address $6000. These are the code and data components of the program as defined by the ORG statements in the program. A printout of demoprog.llis is shown below:
snowhite 27: more demoprog.llis
Hewlett-Packard LD68000 V(01.20 10Mar88) Thu May 21 22:57:02
Command line: ld68k -L -o demoprog.x demoprog
LOAD demoprog
OUTPUT MODULE NAME: demoprog
OUTPUT MODULE FORMAT: IEEE
SECTION SUMMARY
---------------
SECTION ATTRIBUTE START END LENGTH ALIGN
ABSOLUTE 00001000 0000101D 0000001E 0 (BYTE
ABSOLUTE 00006000 00006005 00000006 0 (BYTE
MODULE SUMMARY
--------------
MODULE SECTION:START SECTION:END FILE
demoprog :00001000 :0000101D /users/merat/eeap282/labs
:00006000 :00006005
START ADDRESS: 00001000
Load Completed
snowhite 28: db68k demoprogIt is assumed that the input file has the extension .x. The debugger uses this file which contains the program, memory assignments, and symbol definitions and values to simulate the operation of the program on a 68000 microprocessor.
=======================Monitor=====================12===========Stack=======14=
|1 || 00000010=4E724E72
|2 0000000C=4E724E72
|3 00000008=4E724E72
|4 00000004=4E724E72
|5 SP->00000000=4E724E72
===============================================================================
Code 11 =======Registers=====13=
00001000 303C FFFE MOVE.W #$FFFE,D0 |PC=00001000 pi=00000000
00001004 5240 ADDQ.W #$1,D0 D0=00000000 A0=00000000
00001006 5240 ADDQ.W #$1,D0 D1=00000000 A1=00000000
00001008 0640 FFFE ADDI.W #$FFFE,D0 D2=00000000 A2=00000000
0000100C 5440 ADDQ.W #$2,D0 D3=00000000 A3=00000000
0000100E 41F8 6000 LEA stuff,A0 D4=00000000 A4=00000000
00001012 45F8 6002 LEA adr,A2 D5=00000000 A5=00000000
00001016 4E71 NOP D6=00000000 A6=00000000
====================================Journal=================================10=
| auto halt at address 0x00001000.
|
|
STATUS: Command 68000 MODULE: demoprog BREAK #: 0 HELP=F5 >
Breakpoint Debugger Expression File Memory Program Symbol Window
Execution HP-UX_Shell Macro Option Pause Quit Level Directory ?(Help
The above screen is what you will see as a result of the db68k command. The monitor and stack areas of the display show information that will be important in future labs. The code area shows the next instruction to be executed and, approximately, the eight instructions following it. This code area displays your program in a format that is very similar to the assembler listing. Shown are the memory address (00001000), the assembled form of the instruction (303C FFFE), and the mnemonic form of the instruction (MOVE.W #$FFFE,D0). The register window shows the current contents of the data and address registers, the program counter (PC), and the status register (SR). When the debugger first starts not all registers and the SR are shown in the register area but will appear after the first instruction is executed. The PC was automatically set to $1000 by using the END main instruction in the program. If your program is not visible when you enter the debugger, you will need to set the current value of the PC to the correct starting location of your program using the command
Memory Register @PC=&&&&h <return>
where &&&& is the hexadecimal address of the first instruction to be executed.
The MOVE instruction at $00001000 can be executed using the debugger command
Program Step
which will execute the instruction pointed to by the PC (in this case, the MOVE instruction at $00001000), update all register and memory values affected by the instruction, advance the PC to the next instruction, and display the new values as shown below. Note that the register window has been enlarged to show the remaining registers and the SR.
=======================Monitor=====================12===========Stack=======14=
|1 || 00000010=4E724E72
|2 0000000C=4E724E72
|3 00000008=4E724E72
|4 00000004=4E724E72
|5 SP->00000000=4E724E72
===============================================================================
Code 11 =======Registers=====13=
00001000 303C FFFE MOVE.W #$FFFE,D0 |PC=00001004 pi=00001000
00001004 5240 ADDQ.W #$1,D0 D0=0000FFFE A0=00000000
00001006 5240 ADDQ.W #$1,D0 D1=00000000 A1=00000000
00001008 0640 FFFE ADDI.W #$FFFE,D0 D2=00000000 A2=00000000
0000100C 5440 ADDQ.W #$2,D0 D3=00000000 A3=00000000
0000100E 41F8 6000 LEA stuff,A0 D4=00000000 A4=00000000
00001012 45F8 6002 LEA adr,A2 D5=00000000 A5=00000000
00001016 4E71 NOP D6=00000000 A6=00000000
00001018 3278 2000 MOVEA.W $2000,A1 |D7=00000000 A7=00000000
| 0000101C 3080 MOVE.W D0,(A0) SR=0010011100001000
| 0000101E 4E72 4E72 STOP #$4E72 T S III XNZVC
========================S
STATUS: Command 68000 MODULE: demoprog BREAK #: 0 HELP=F5 >
> Program Step
Breakpoint Debugger Expression File Memory Program Symbol Window
Display_Source Find_Source Run Interrupt Load Context Pc_Reset Step
The X, N, Z, V, and C bits of the SR are all updated as a result of the MOVE instruction. For a MOVE instruction the XNZVC bits will be updated according to _**00 (see the Programmer's Reference Manual). This means that the X bit doesn't change (indicated by _) from the previous value of zero, the N and Z bits are set according to the number being moved (as indicated by the *, in this case $FFFE is negative so the N bit becomes 1 and the Z bit remains 0), and the V and C bits are always set to 0 (as indicated by the 0) for a MOVE instruction.
You can continue to type
Program Step
to execute the remaining program instructions one at a time. In this case the debugger will execute the ADD.W $0001,D0 instruction, which adds a word length $0001 to the contents of D0, to get (D0)=$0000FFFF.
=======================Monitor=====================12===========Stack=======14=
|1 || 00000010=4E724E72
|2 0000000C=4E724E72
|3 00000008=4E724E72
|4 00000004=4E724E72
|5 SP->00000000=4E724E72
===============================================================================
Code 11 =======Registers=====13=
00001000 303C FFFE MOVE.W #$FFFE,D0 |PC=00001006 pi=00001004
00001004 5240 ADDQ.W #$1,D0 D0=0000FFFF A0=00000000
00001006 5240 ADDQ.W #$1,D0 D1=00000000 A1=00000000
00001008 0640 FFFE ADDI.W #$FFFE,D0 D2=00000000 A2=00000000
0000100C 5440 ADDQ.W #$2,D0 D3=00000000 A3=00000000
0000100E 41F8 6000 LEA stuff,A0 D4=00000000 A4=00000000
00001012 45F8 6002 LEA adr,A2 D5=00000000 A5=00000000
00001016 4E71 NOP D6=00000000 A6=00000000
00001018 3278 2000 MOVEA.W $2000,A1 |D7=00000000 A7=00000000
| 0000101C 3080 MOVE.W D0,(A0) SR=0010011100001000
| 0000101E 4E72 4E72 STOP #$4E72 T S III XNZVC
========================S
STATUS: Command 68000 MODULE: demoprog BREAK #: 0 HELP=F5 >
> Program Step
Breakpoint Debugger Expression File Memory Program Symbol Window
Display_Source Find_Source Run Interrupt Load Context Pc_Reset Step
For an ADD instruction the XNZVC bits will be updated according to ***** (see the Programmer's Reference Manual). This means that these bits are set according to the results of the addition. The addition added $0001 to $FFFE to get $FFFF. There is no overflow or carry. The X, C, and V bits are accordingly set to zero. Because the result is negative, the N bit is set to 1. The result is not zero so the Z bit is set to 0.
You can continue to type
Program Step
to execute the remaining program instructions one at a time.
=======================Monitor=====================12===========Stack=======14=
|1 || 00000010=4E724E72
|2 0000000C=4E724E72
|3 00000008=4E724E72
|4 00000004=4E724E72
|5 SP->00000000=4E724E72
===============================================================================
Code 11 =======Registers=====13=
00001000 303C FFFE MOVE.W #$FFFE,D0 |PC=00001008 pi=00001006
00001004 5240 ADDQ.W #$1,D0 D0=00000000 A0=00000000
00001006 5240 ADDQ.W #$1,D0 D1=00000000 A1=00000000
00001008 0640 FFFE ADDI.W #$FFFE,D0 D2=00000000 A2=00000000
0000100C 5440 ADDQ.W #$2,D0 D3=00000000 A3=00000000
0000100E 41F8 6000 LEA stuff,A0 D4=00000000 A4=00000000
00001012 45F8 6002 LEA adr,A2 D5=00000000 A5=00000000
00001016 4E71 NOP D6=00000000 A6=00000000
00001018 3278 2000 MOVEA.W $2000,A1 |D7=00000000 A7=00000000
| 0000101C 3080 MOVE.W D0,(A0) SR=0010011100010101
| 0000101E 4E72 4E72 STOP #$4E72 T S III XNZVC
========================S
STATUS: Command 68000 MODULE: demoprog BREAK #: 0 HELP=F5 >
> Program Step
Breakpoint Debugger Expression File Memory Program Symbol Window
Display_Source Find_Source Run Interrupt Load Context Pc_Reset Step
This addition added $0001 to $FFFF to get $10000. However, because the addition is word length, the 1 cannot be placed in the lower word of D0, and a carry occurs. This sets the C and X bits to 1. Because the signs of the numbers are mixed, no overflow occurs, and V is set to 0. Because the result put into D0 is zero, the Z bit is set to 1. The N bit is set to 0.
=======================Monitor=====================12===========Stack=======14=
|1 || 00000010=4E724E72
|2 0000000C=4E724E72
|3 00000008=4E724E72
|4 00000004=4E724E72
|5 SP->00000000=4E724E72
===============================================================================
Code 11 =======Registers=====13=
00001000 303C FFFE MOVE.W #$FFFE,D0 |PC=0000100C pi=00001008
00001004 5240 ADDQ.W #$1,D0 D0=0000FFFE A0=00000000
00001006 5240 ADDQ.W #$1,D0 D1=00000000 A1=00000000
00001008 0640 FFFE ADDI.W #$FFFE,D0 D2=00000000 A2=00000000
0000100C 5440 ADDQ.W #$2,D0 D3=00000000 A3=00000000
0000100E 41F8 6000 LEA stuff,A0 D4=00000000 A4=00000000
00001012 45F8 6002 LEA adr,A2 D5=00000000 A5=00000000
00001016 4E71 NOP D6=00000000 A6=00000000
00001018 3278 2000 MOVEA.W $2000,A1 |D7=00000000 A7=00000000
| 0000101C 3080 MOVE.W D0,(A0) SR=0010011100001000
| 0000101E 4E72 4E72 STOP #$4E72 T S III XNZVC
========================S
STATUS: Command 68000 MODULE: demoprog BREAK #: 0 HELP=F5 >
> Program Step
Breakpoint Debugger Expression File Memory Program Symbol Window
Display_Source Find_Source Run Interrupt Load Context Pc_Reset Step
This addition added $FFFE to $0000 to get $FFFE. Because there is no overflow or carry the X, C, and V bits are set to zero. Because the result is negative, the N bit is set to 1. The number is not zero so the Z bit is set to 0.
=======================Monitor=====================12===========Stack=======14=
|1 || 00000010=4E724E72
|2 0000000C=4E724E72
|3 00000008=4E724E72
|4 00000004=4E724E72
|5 SP->00000000=4E724E72
===============================================================================
Code 11 =======Registers=====13=
00001000 303C FFFE MOVE.W #$FFFE,D0 |PC=0000100E pi=0000100C
00001004 5240 ADDQ.W #$1,D0 D0=00000000 A0=00000000
00001006 5240 ADDQ.W #$1,D0 D1=00000000 A1=00000000
00001008 0640 FFFE ADDI.W #$FFFE,D0 D2=00000000 A2=00000000
0000100C 5440 ADDQ.W #$2,D0 D3=00000000 A3=00000000
0000100E 41F8 6000 LEA stuff,A0 D4=00000000 A4=00000000
00001012 45F8 6002 LEA adr,A2 D5=00000000 A5=00000000
00001016 4E71 NOP D6=00000000 A6=00000000
00001018 3278 2000 MOVEA.W $2000,A1 |D7=00000000 A7=00000000
| 0000101C 3080 MOVE.W D0,(A0) SR=0010011100010101
| 0000101E 4E72 4E72 STOP #$4E72 T S III XNZVC
========================S
STATUS: Command 68000 MODULE: demoprog BREAK #: 0 HELP=F5 >
> Program Step
Breakpoint Debugger Expression File Memory Program Symbol Window
Display_Source Find_Source Run Interrupt Load Context Pc_Reset Step
This addition added $0002 to $FFFE to get $10000. Note that additions take place in 2's complement. Because $FFFE is -2, adding +2 to to -2 should result in zero and that is what happened. Because the addition is word length, the 1 cannot be placed in the lower word of D0 and a carry occurs. This sets the C and X bits to 1. Because the signs of the numbers are mixed, no overflow occurs, and V is set to 0. Because the result put into D0 is zero, the Z bit is set to 1. The N bit is set to 0.
=======================Monitor=====================12===========Stack=======14=
|1 || 00000010=4E724E72
|2 0000000C=4E724E72
|3 00000008=4E724E72
|4 00000004=4E724E72
|5 SP->00000000=4E724E72
===============================================================================
Code 11 =======Registers=====13=
00001000 303C FFFE MOVE.W #$FFFE,D0 |PC=00001012 pi=0000100E
00001004 5240 ADDQ.W #$1,D0 D0=00000000 A0=00006000
00001006 5240 ADDQ.W #$1,D0 D1=00000000 A1=00000000
00001008 0640 FFFE ADDI.W #$FFFE,D0 D2=00000000 A2=00000000
0000100C 5440 ADDQ.W #$2,D0 D3=00000000 A3=00000000
0000100E 41F8 6000 LEA stuff,A0 D4=00000000 A4=00000000
00001012 45F8 6002 LEA adr,A2 D5=00000000 A5=00000000
00001016 4E71 NOP D6=00000000 A6=00000000
00001018 3278 2000 MOVEA.W $2000,A1 |D7=00000000 A7=00000000
| 0000101C 3080 MOVE.W D0,(A0) SR=0010011100010101
| 0000101E 4E72 4E72 STOP #$4E72 T S III XNZVC
========================S
STATUS: Command 68000 MODULE: demoprog BREAK #: 0 HELP=F5 >
> Program Step
Breakpoint Debugger Expression File Memory Program Symbol Window
Display_Source Find_Source Run Interrupt Load Context Pc_Reset Step
The instruction LEA stuff,A0 put the number $6000 into address register A0. Note that the original instruction in our program was LEA.L $6000,A0. The debugger recognized that $6000 was the value of the symbol stuff and automatically substituted it. The LEA instruction does not affect the SR, so no SR values are changed.
=======================Monitor=====================12===========Stack=======14=
|1 || 00000010=4E724E72
|2 0000000C=4E724E72
|3 00000008=4E724E72
|4 00000004=4E724E72
|5 SP->00000000=4E724E72
===============================================================================
Code 11 =======Registers=====13=
00001000 303C FFFE MOVE.W #$FFFE,D0 |PC=00001018 pi=00001016
00001004 5240 ADDQ.W #$1,D0 D0=00000000 A0=00006000
00001006 5240 ADDQ.W #$1,D0 D1=00000000 A1=00000000
00001008 0640 FFFE ADDI.W #$FFFE,D0 D2=00000000 A2=00006002
0000100C 5440 ADDQ.W #$2,D0 D3=00000000 A3=00000000
0000100E 41F8 6000 LEA stuff,A0 D4=00000000 A4=00000000
00001012 45F8 6002 LEA adr,A2 D5=00000000 A5=00000000
00001016 4E71 NOP D6=00000000 A6=00000000
00001018 3278 2000 MOVEA.W $2000,A1 |D7=00000000 A7=00000000
| 0000101C 3080 MOVE.W D0,(A0) SR=0010011100010101
| 0000101E 4E72 4E72 STOP #$4E72 T S III XNZVC
========================S
STATUS: Command 68000 MODULE: demoprog BREAK #: 0 HELP=F5 >
> Program Step
Breakpoint Debugger Expression File Memory Program Symbol Window
Display_Source Find_Source Run Interrupt Load Context Pc_Reset Step
The instruction LEA adr,A2 put the number $6002 into address register A2. Note that the PC automatically advances beyond the NOP (the $4E71) inserted by the assembler. No long word address was needed, since our address $6000 is representable by a single word. The LEA instruction does not affect the SR, so no SR values are changed.
=======================Monitor=====================12===========Stack=======14=
|1 || 00000010=4E724E72
|2 0000000C=4E724E72
|3 00000008=4E724E72
|4 00000004=4E724E72
|5 SP->00000000=4E724E72
===============================================================================
Code 11 =======Registers=====13=
00001000 303C FFFE MOVE.W #$FFFE,D0 |PC=0000101C pi=00001018
00001004 5240 ADDQ.W #$1,D0 D0=00000000 A0=00006000
00001006 5240 ADDQ.W #$1,D0 D1=00000000 A1=00004E72
00001008 0640 FFFE ADDI.W #$FFFE,D0 D2=00000000 A2=00006002
0000100C 5440 ADDQ.W #$2,D0 D3=00000000 A3=00000000
0000100E 41F8 6000 LEA stuff,A0 D4=00000000 A4=00000000
00001012 45F8 6002 LEA adr,A2 D5=00000000 A5=00000000
00001016 4E71 NOP D6=00000000 A6=00000000
00001018 3278 2000 MOVEA.W $2000,A1 |D7=00000000 A7=00000000
| 0000101C 3080 MOVE.W D0,(A0) SR=0010011100010101
| 0000101E 4E72 4E72 STOP #$4E72 T S III XNZVC
========================S
STATUS: Command 68000 MODULE: demoprog BREAK #: 0 HELP=F5 >
> Program Step
Breakpoint Debugger Expression File Memory Program Symbol Window
Display_Source Find_Source Run Interrupt Load Context Pc_Reset Step
The instruction MOVEA.W $2000,A1 put the contents of memory location $2000 into address register A1. Since we did not put anything into that particular memory location, the debugger found the default value of $4E72 there. (The debugger initializes all memory to $4E72.) The MOVEA instruction does not affect the SR, so no SR values are changed.
=======================Monitor=====================12===========Stack=======14=
|1 || 00000010=4E724E72
|2 0000000C=4E724E72
|3 00000008=4E724E72
|4 00000004=4E724E72
|5 SP->00000000=4E724E72
===============================================================================
Code 11 =======Registers=====13=
00001000 303C FFFE MOVE.W #$FFFE,D0 |PC=0000101E pi=0000101C
00001004 5240 ADDQ.W #$1,D0 D0=00000000 A0=00006000
00001006 5240 ADDQ.W #$1,D0 D1=00000000 A1=00004E72
00001008 0640 FFFE ADDI.W #$FFFE,D0 D2=00000000 A2=00006002
0000100C 5440 ADDQ.W #$2,D0 D3=00000000 A3=00000000
0000100E 41F8 6000 LEA stuff,A0 D4=00000000 A4=00000000
00001012 45F8 6002 LEA adr,A2 D5=00000000 A5=00000000
00001016 4E71 NOP D6=00000000 A6=00000000
00001018 3278 2000 MOVEA.W $2000,A1 |D7=00000000 A7=00000000
| 0000101C 3080 MOVE.W D0,(A0) SR=0010011100010100
| 0000101E 4E72 4E72 STOP #$4E72 T S III XNZVC
========================S
STATUS: Command 68000 MODULE: demoprog BREAK #: 0 HELP=F5 >
> Program Step
Breakpoint Debugger Expression File Memory Program Symbol Window
Display_Source Find_Source Run Interrupt Load Context Pc_Reset Step
The instruction MOVE.W D0,(A0) puts the contents of D0 (in this case, $0000) into the memory location whose address is in A0 (in this case, $6000). Because the number is zero, the Z bit is set to 1. The N, V, and C bits are set to 0. The X bit remains unchanged.
=======================Monitor=====================12===========Stack=======14=
|1 || 00000010=4E724E72
|2 0000000C=4E724E72
|3 00000008=4E724E72
|4 00000004=4E724E72
|5 SP->00000000=4E724E72
===============================================================================
Code 11 =======Registers=====13=
00001000 303C FFFE MOVE.W #$FFFE,D0 |PC=0000101E pi=0000101E
00001004 5240 ADDQ.W #$1,D0 D0=00000000 A0=00006000
00001006 5240 ADDQ.W #$1,D0 D1=00000000 A1=00004E72
00001008 0640 FFFE ADDI.W #$FFFE,D0 D2=00000000 A2=00006002
0000100C 5440 ADDQ.W #$2,D0 D3=00000000 A3=00000000
0000100E 41F8 6000 LEA stuff,A0 D4=00000000 A4=00000000
00001012 45F8 6002 LEA adr,A2 D5=00000000 A5=00000000
00001016 4E71 NOP D6=00000000 A6=00000000
====================================Journal=================================10=|
| Exception vector 8 at 101E: privilege violation
|
| Stopped due to exception interrupt S
STATUS: Command 68000 MODULE: demoprog BREAK #: 0 HELP=F5 >
> Program Step
Breakpoint Debugger Expression File Memory Program Symbol Window
Display_Source Find_Source Run Interrupt Load Context Pc_Reset Step
When we attempt to execute the next instruction, we get an error. This is because our program did not extend to this memory location and it contained the initial value of $4E72 put there by the debugger. There is no elegant way to end your program in the debugger (that we have shown you yet).
=======================Monitor=====================12===========Stack=======14=
|1 || 00000010=4E724E72
|2 0000000C=4E724E72
|3 00000008=4E724E72
|4 00000004=4E724E72
|5 SP->00000000=4E724E72
===============================================================================
Code 11 =======Registers=====13=
00001000 303C FFFE MOVE.W #$FFFE,D0 |PC=0000101E pi=0000101E
00001004 5240 ADDQ.W #$1,D0 D0=00000000 A0=00006000
00001006 5240 ADDQ.W #$1,D0 D1=00000000 A1=00004E72
00001008 0640 FFFE ADDI.W #$FFFE,D0 D2=00000000 A2=00006002
0000100C 5440 ADDQ.W #$2,D0 D3=00000000 A3=00000000
0000100E 41F8 6000 LEA stuff,A0 D4=00000000 A4=00000000
00001012 45F8 6002 LEA adr,A2 D5=00000000 A5=00000000
00001016 4E71 NOP D6=00000000 A6=00000000
====================================Journal=================================10=|
| Exception vector 8 at 101E: privilege violation
|
| Stopped due to exception interrupt S
STATUS: Command 68000 MODULE: demoprog BREAK #: 0 HELP=F5 >
> Debugger Quit Yes
We can exit the debugger with the command
Debugger Quit Yes
Next Section: Chapter 13a - Addressing Modes Table
Previous Section: Chapter 11 - Running Programs
EEAP 282 Class Notes Table of Contents
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