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Chapter 3 Introduction to the 68000

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1 Chapter 3 Introduction to the 68000
Register Set: data, address, condition code, status. Basic Instruction Set Basic addressing modes: register, absolute, immediate, register indirect, etc. Assembling and debugging a program

2 Register Set 8 general-purpose data registers. Word operation on D00-D15, byte operation on D00-D07 D31 D16 D08 D00 D0 D7 D6 PC points at the next instruction to be executed PC

3 Address Register: A0-A7 8 address registers of 32 bits.
Information in an address register represents a location in memory. Special one: A7 is used as stack pointer. Memory 1005 1004 1006 1007 4A 0E 07 57 A0 A31 A16 A00 A0 A1 A7 [M(A0)] = [M(1005)] = 57

4 Condition Code Register
Carry oVerflow C V Z N X 7 Zero Negative eXtend The CCR is updated to reflect the result of the operation. Z=1 if the result is 0 C=1 if there is carry-out from MSB V=1 if there is overflow N=1 if the result is negative

5 Instruction Set The 68000 has a large instruction set.
Move data Modify or operate the data Change execution sequence Determine the operation mode of CPU e.g. MOVE.B D3,1234 MOVE.B #25,D2 Classification of instruction set architecture CISC (Complex instruction set computer): large instruction set, powerful, but difficult to optimize code RISC (Reduced instruction set computer): smaller instruction set, easy to optimize code, but longer program

6 Data Movement Register-to-register, register-to-memory, memory-to-register, memory-to-memory, constant-to-memory/register. 8-bit, 16-bit, 32-bit correspond to MOVE.B, MOVE.W, MOVE.L Legal: Assembler Form RTL MOVE.B D1,D2 [D2]  [D1] MOVE.B D3, 1234 [M(1234)]  [D3] MOVE.B 1234,2000 [M(2000)]  [M(1234)] MOVE.B #12,1234 [M(1234)]  12 Illegal: MOVE.B D3,#12 12  [D3] ???

7 Unconditional Branch BRA: Branch BRA address ; GOTO address Example:
BRA NEXT MOVE.B D1,D2 NEXT MOVE.B #1,D4 Which instruction will be executed after BRA NEXT is executed?

8 Conditional Branch BEQ, BNE, BCC, BCS, BVS Example:
BCC Check_5 IF c=0 THEN branch to Check_5 MOVE.B D1,D2 Check_5 MOVE.B #1,D4 BCC Check_5 C=0 C=1 Check_5 MOVE.B #1,D4 MOVE.B D1,D2

9 Conditional Branches (con.)
BNE Branch on [CCR(Z)]=0 BEQ Branch on [CCR(Z)]=1 BCC Branch on [CCR(C)]=0 BCS Branch on [CCR(C)]=1 BVC Branch on [CCR(V)]=0 BVS Branch on [CCR(V)]=1 The complete set of conditional branch instructions is given in Table 5.2 on page 208. (e.g., BLT, BLE, BGT, BGE, etc.)

10 CMP and TST Useful for changing program flow CMP: Compare
Syntax: CMP src,Dn Operation: [Dn] - [src] Result of - is not saved TST: Test an operand Syntax: TST dest Compare [dest] to 0, no result saved TST D1 is the same as CMP __,D1

11 Change Program Flow IF1 Set Flag Test Opposite Condition and BR to ENDIF if TRUE THEN1 IF-Part ENDIF1 ... IF X1 = 0 THEN X1 := Y1 IF1 MOVE.B X1,D0 (or “TST.B X1”) BNE ENDIF1 THEN1 MOVE.B Y2,X1 ENDIF1 … other code

12 IF X1 = 0 THEN X1 := Y1 ELSE X1 := Y2
Change Program Flow (con.) IF X1 = 0 THEN X1 := Y1 ELSE X1 := Y2 Right IF1 MOVE.B X1,D0 BEQ THEN1 ELSE1 MOVE.B Y2,X1 BRA ENDIF1 THEN1 MOVE.B Y1,X1 ENDIF1 ... Wrong IF1 MOVE.B X1,D0 BEQ THEN1 ELSE1 MOVE.B Y2,X1 THEN1 MOVE.B Y1,X1 ENDIF1 …

13 IF X1 = 0 THEN X1 := Y1 ELSE X1 := Y2
Change Program Flow (con.) IF X1 = 0 THEN X1 := Y1 ELSE X1 := Y2 IF1 MOVE.B X1,D0 BNE ELSE1 THEN1 MOVE.B Y1,X1 BRA ENDIF1 must have this branch ELSE1 MOVE.B Y2,X1 ENDIF1 ...

14 Change Program Flow (con.)
WHILE (K > 0) DO S WHILE TST.B K BLE ENDWH test opposite condition S loop body BRA WHILE ENDWH ...

15 Change Program Flow (con.)
FOR I = N1 TO N2 DO S MOVE.B N1,D0 D0: loop counter NEXT CMP.B N2,D0 BGT ENDFOR S loop body ADD.B #1,D0 BRA NEXT ENDFOR ...

16 1 assembly language instruction = 1 machine language instruction
1 high-level language instruction  1 machine language instructions

17 Subroutine BSR ADD12 BSR ADD12 ADD12 ADD.B D1,D2 SUB.B #12,D2 RTS

18 Data Typing Most high-level language, like Pascal, Java, Ada, are said to be strongly typed. Assembly language is not strongly typed. How about C/C++? Example: A character can be multiplied by an integer. JAVA?

19 Data Typing (con.) Memory Map A 1000 12 00 ORG $1000 A DC.B $12
B DC.W $3456 C DS.B 1 D DS.L 1 B 1002 34 56 C 1004 00 00 D 1006 00 00 00 00 MOVE.B A,D0 ADD.B B,D0 [D0]  $12 + $34 MOVE.B A,D0 ADD.W B,D0 [D0]  $12 + $3456 MOVE.W A,D0 ADD.W B,D0 [D0]  $ $3456 MOVE.L A,D0 ADD.L B,D0 [D0]  ?

20 Arithmetic Operation ADD, SUB, CLR, NEG, ASL, ASR
ADD.B 1234,D3 ; [D3]  [M(1234)]+[D3] The CCR is updated accordingly. Example: V3 = V1 + V2 signed integers unsigned integers ORG $400 V1 DC.B 12 V2 DC.B 14 V3 DS.B 1 ORG $600 MOVE.B V1,D0 ADD.B V2,D0 MOVE.B D0,V3 BVS Error1 ... ORG $400 V1 DC.B 12 V2 DC.B 14 V3 DS.B 1 ORG $600 MOVE.B V1,D0 ADD.B V2,D0 MOVE.B D0,V3 BCS Error1 ... Data Are the codes correct? Program Where is the right place for BVS/BCS?

21 Arithmetic Operation (con.)
Subtraction: SUB src, dest ; [dest]  [dest] - [src] SUB.B D2,D0 ; [D0(0:7)]  [D0(0:7)] - [D2(0:7)] SUB.W D2,D0 ; [D0(0:15)]  [D0(0:15)] - [D2(0:15)] SUB.L D2,D0 ; [D0]  [D0] - [D2] Clear CLR.B D ; [D0(0:7)]  0 Negation: negative value, i.e,, 2’s complement NEG.B D4 ; 2’s complement of D4 If [D4] = , after [D4] =

22 ASL (Arithmetic Shift Left)
Operand Format: ASL #n,dest or ASL Di,dest shifts bits in dest LEFT by n or [Di] places, respectively The bit shifted out is shifted in C-bit of CCR. Example: ASL.B #3,D0 [D0] = [C]= [C]= [C]=

23 ASL (Arithmetic Shift Left)
Why is ASL useful? ASL is the fastest way to perform “multiply by 2’s power” ASL dest = ASL #1,dest What does ASL #n, dest do? [dest]  [dest] x 2n How to multiply D0 by 2 ? [D0] = After ASL.B #1,D0 [D0] = Can ASR cause overflow?

24 ASR (Arithmetic Shift Right)
Operand C MSB Same as ASL, but bits shifted to RIGHT MSB is duplicated back into MSB (Why?) ASR.B #1,D0 is equivalent to dividing D0 by 2 How to divide D1 by 32 ? Example: [D0] = -22 = After “ASR.B #1,D0” [D0] = = -11 [CCR(c)] = 0 Can ASR cause overflow?

25 Effect of Arithmetic Operations on CCR
Addition:  1, if carry out from MSB C =   0, otherwise  1, if operands are of same sign and V =  their sum is of the opposite sign ____ ____ ____ V= an-1  bn-1  sn-1 + an-1  bn-1  sn-1 where an-1, bn-1, sn-1 are the MSBs of source destination and result, respectively

26 Effect of Arithmetic Operations on CCR
Subtraction:  1, if NO carry out from MSB C =   0, otherwise  1, if operands are of opposite sign and V =  the result is of same sign as the source ____________ V= (an-1  bn-1)  (dn-1  an-1) where an-1, bn-1, dn-1 are the MSBs of source destination and result, respectively

27 Logical Operation AND, OR, EOR, NOT If [D0] = 11110000
AND.B #% ,D0 ;[D0]= OR.B #% ,D0 ;[D0]= EOR.B #% ,D0 ;[D0]=

28 Use Registers Accesses to data registers are faster than accesses to memory. Shorter instruction, 3 bits to indicate either one of 8 general-purpose registers. Use comments to indicate how registers are used in program. * GetChar: Input an ASCII-coded character into D0 * Input Parameters: None * Output parameters: ASCII character in D0, Error code in D6 * Registers modified: D0, D1, D6 GetChar MOVE.B ACIAC,D1 BTST.B #RDRF,D1 BEQ GetChar MOVE.B ACIAC,D0 AND.B #% ,D1 MOVE.B D1,D6 RTS

29 Addressing Modes Concerned with the way in which data is accessed (where operand can be found) Data register direct, absolute, immediate, and address register indirect. Looking for a house in a familiar neighborhood, “The house next to Tim’s” (relative location) is enough. Looking for a house in a new environment, “61 William Street” (actual address) is necessary, even with a city name.

30 Absolute Addressing Use the actual or absolute address of the operand, e.g. CLR.B $234 MOVE.B D2, $2000 The source is data register direct, a type of absolute addressing mode. $2000 is memory location 2000 Symbols can also be used Example: MOVE.B Input,D0 SUB.B Time,D0

31 Immediate Addressing MOVE.B #25,D2 ; [D2]  25
Immediate addressing is faster than the absolute addressing. Data 25 is part of the instruction  stored in IR. Absolute Addressing: Hours DC.B 25 ADD.B Hours,D2 Immediate Addressing: Hours EQU 25 ADD.B #Hours,D2 What about “ADD.B #Hours,D2 What about “ADD.B Hours,D2

32 IF 7<P<25 THEN X := 6 if ((7<P) && (P<25)) X = 6;
MOVE.B P,D0 CMP.B #7,D0 BLE OutOfRange CMP.B #25,D0 BGE OutOfRange MOVE.B #6,X OutOfRange … IF 7<P<25 THEN X := 6 if ((7<P) && (P<25)) X = 6; CLR.B D1 D1: X MOVE.B #1,D0 D0: I NEXT ADD.B D0,D1 ADD.B #1,D0 CMP.B #10,D0 BLE NEXT MOVE.B D1,X X := 0 FOR I := 1 TO X = X + I X = 0; for (I = 1; I <= 10; I++) X = X + I;

33 Address Register Indirect
The address of an operand is found in an address register, A0 to A7 Pointer or reference MOVEA: copy an address to address reg MOVEA.L #$1000,A0 CLR.B (A0) same effect as CLR.B $1000 1000 A0 0FFF 1001 1002 1003 00

34 Indirect Addressing Two accesses:
1) to the address register A0 to find the actual address of operand, 1000. 2) to the memory location 1000 to get the operand. Why is address register indirect addressing useful?

35 Example: Add 100 numbers together, the data starts at memory location 200016
MOVE.B $2000,D0 ADD.B $2001,D0 ADD.B $2002,D0 ADD.B $2003,D0 : ADD.B $2063,D0 CLR.B D0 MOVEA.L #$2000,A0 NEXT ADD.B (A0),D0 ADDA.L #1,A0 CMPA.L #$2064,A0 BNE NEXT

36 Different Addressing * Program to test the different addressing modes. * [D2]  [M(0)] By Mingrui Zhang 3 ORG $1000 A ABSOL: DC.B 26 A IMMED: EQU 26 7 ORG $2000 C MOVEA.L #$1000,A0 MOVE.L $0,D2 A D ADD.B (A0),D2 12 C MOVE.L $0,D2 D ADD.B ABSOL,D2 15 MOVE.L $0,D2 A ADD.B #IMMED,D2 18 C 4E STOP #$2700 END $2000

37 The Teesside MC68000 Cross-assembler and Simulator
TOOLS FILES text editor Test.X68 X68K Test.BIN (and Test.LIS) E68K Desing the program Edit source file Cross-assemble the program Run the program

38 Debugging Commands HELP: Provide information about commands
MD (.): Displays the contents of memory e.g. MD 400 MD 400 -DI ;disassemble the contents of memory. MM: Memory modification e.g. MM 400 -B MM W MM 400 -B -DEC MM W -DEC DF: Displays the contents of all registers .PC: set PC e.g. .PC 400 GO: Execute program, (ESC to escape) TR: Executes a single instruction at a time BR 10000: Places a marker at location 10000 QU: Quit

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