Presentation on theme: "Chapter 3 Introduction to the 68000"— Presentation transcript:
1 Chapter 3 Introduction to the 68000 Register Set: data, address, condition code, status.Basic Instruction SetBasic addressing modes: register, absolute, immediate, register indirect, etc.Assembling and debugging a program
2 Register Set8 general-purpose data registers. Word operation on D00-D15, byte operation on D00-D07D31D16D08D00D0D7D6PC points at the next instruction to be executedPC
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.Memory10051004100610074A0E0757A0A31A16A00A0A1A7[M(A0)] = [M(1005)] = 57
4 Condition Code Register CarryoVerflowCVZNX7ZeroNegativeeXtendThe CCR is updated to reflect the result of the operation.Z=1 if the result is 0C=1 if there is carry-out from MSBV=1 if there is overflowN=1 if the result is negative
5 Instruction Set The 68000 has a large instruction set. Move dataModify or operate the dataChange execution sequenceDetermine the operation mode of CPUe.g. MOVE.B D3,1234MOVE.B #25,D2Classification of instruction set architectureCISC (Complex instruction set computer): large instruction set, powerful, but difficult to optimize codeRISC (Reduced instruction set computer): smaller instruction set, easy to optimize code, but longer program
7 Unconditional Branch BRA: Branch BRA address ; GOTO address Example: BRA NEXTMOVE.B D1,D2NEXT MOVE.B #1,D4Which 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_5MOVE.B D1,D2…Check_5 MOVE.B #1,D4BCC Check_5C=0C=1Check_5 MOVE.B #1,D4MOVE.B D1,D2
9 Conditional Branches (con.) BNE Branch on [CCR(Z)]=0BEQ Branch on [CCR(Z)]=1BCC Branch on [CCR(C)]=0BCS Branch on [CCR(C)]=1BVC Branch on [CCR(V)]=0BVS Branch on [CCR(V)]=1The 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,DnOperation: [Dn] - [src]Result of - is not savedTST: Test an operandSyntax: TST destCompare [dest] to 0, no result savedTST D1 is the same as CMP __,D1
11 Change Program FlowIF1 Set Flag Test Opposite Condition and BR to ENDIF if TRUE THEN1 IF-Part ENDIF1 ...IF X1 = 0 THEN X1 := Y1IF1 MOVE.B X1,D0 (or “TST.B X1”)BNE ENDIF1THEN1 MOVE.B Y2,X1ENDIF1 … other code
18 Data TypingMost 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?
22 ASL (Arithmetic Shift Left) OperandFormat:ASL #n,destorASL Di,destshifts bits in dest LEFT by n or [Di] places, respectivelyThe 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,destWhat does ASL #n, dest do?[dest] [dest] x 2nHow to multiply D0 by 2 ?[D0] =After ASL.B #1,D0[D0] =Can ASR cause overflow?
24 ASR (Arithmetic Shift Right) OperandCMSBSame as ASL, butbits shifted to RIGHTMSB is duplicated back into MSB (Why?)ASR.B #1,D0 is equivalent to dividing D0 by 2How to divideD1 by 32 ?Example: [D0] = -22 =After “ASR.B #1,D0”[D0] = = -11[CCR(c)] = 0Can ASR cause overflow?
25 Effect of Arithmetic Operations on CCR Addition: 1, if carry out from MSBC = 0, otherwise 1, if operands are of same sign andV = their sum is of the opposite sign____ ____ ____V= an-1 bn-1 sn-1 + an-1 bn-1 sn-1where an-1, bn-1, sn-1 are the MSBs ofsource destination and result, respectively
26 Effect of Arithmetic Operations on CCR Subtraction: 1, if NO carry out from MSBC = 0, otherwise 1, if operands are of opposite sign andV = 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 ofsource destination and result, respectively
28 Use RegistersAccesses 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, D6GetChar MOVE.B ACIAC,D1BTST.B #RDRF,D1BEQ GetCharMOVE.B ACIAC,D0AND.B #% ,D1MOVE.B D1,D6RTS
29 Addressing ModesConcerned 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 AddressingUse the actual or absolute address of the operand, e.g. CLR.B $234MOVE.B D2, $2000The source is data register direct, a type of absolute addressing mode. $2000 is memory location 2000Symbols can also be usedExample:MOVE.B Input,D0SUB.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 25ADD.B Hours,D2Immediate Addressing:Hours EQU 25ADD.B #Hours,D2What about “ADD.B #Hours,D2What about “ADD.B Hours,D2
32 IF 7<P<25 THEN X := 6 if ((7<P) && (P<25)) X = 6; MOVE.B P,D0CMP.B #7,D0BLE OutOfRangeCMP.B #25,D0BGE OutOfRangeMOVE.B #6,XOutOfRange …IF 7<P<25 THEN X := 6 if ((7<P) && (P<25)) X = 6;CLR.B D1 D1: XMOVE.B#1,D0 D0: INEXT ADD.B D0,D1ADD.B #1,D0CMP.B #10,D0BLE NEXTMOVE.B D1,XX := 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 A7Pointer or referenceMOVEA: copy an address to address regMOVEA.L #$1000,A0CLR.B (A0)same effect asCLR.B $10001000A00FFF10011002100300
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,D0ADD.B $2001,D0ADD.B $2002,D0ADD.B $2003,D0:ADD.B $2063,D0CLR.B D0MOVEA.L #$2000,A0NEXT ADD.B (A0),D0ADDA.L #1,A0CMPA.L #$2064,A0BNE NEXT
36 Different Addressing* Program to test the different addressing modes.* [D2] [M(0)] By Mingrui Zhang3ORG $1000A ABSOL: DC.B 26A IMMED: EQU 267ORG $2000C MOVEA.L #$1000,A0MOVE.L $0,D2A D ADD.B (A0),D212C MOVE.L $0,D2D ADD.B ABSOL,D215MOVE.L $0,D2A ADD.B #IMMED,D218C 4E STOP #$2700END $2000
37 The Teesside MC68000 Cross-assembler and Simulator TOOLS FILEStext editor Test.X68X68K Test.BIN (and Test.LIS)E68KDesing the programEdit source fileCross-assemble the programRun the program
38 Debugging Commands HELP: Provide information about commands MD (.): Displays the contents of memorye.g. MD 400MD 400 -DI ;disassemble the contents ofmemory.MM: Memory modificatione.g. MM 400 -B MM WMM 400 -B -DEC MM W -DECDF: Displays the contents of all registers.PC: set PCe.g. .PC 400GO: Execute program, (ESC to escape)TR: Executes a single instruction at a timeBR 10000: Places a marker at location 10000QU: Quit
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