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Embedded System Design Center ARM7TDMI Microprocessor Data Processing Instructions Sai Kumar Devulapalli.

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Presentation on theme: "Embedded System Design Center ARM7TDMI Microprocessor Data Processing Instructions Sai Kumar Devulapalli."— Presentation transcript:

1 Embedded System Design Center ARM7TDMI Microprocessor Data Processing Instructions Sai Kumar Devulapalli

2 2 of 35 Objectives Detailed understanding of ARM Data processing instructions Understanding the instruction encoding formats Understanding the general format with which data processing instructions can be conditionally executed and to set the status flags Use of immediate operand in the data processing operations Understand the use of shifter operations for second operand Understand the operation of multiplication instructions

3 3 of 35 Data processing Instructions Largest family of ARM instructions, all sharing the same instruction format. Contains: Arithmetic operations Comparisons (no results - just set condition codes) Logical operations Data movement between registers Remember, this is a load / store architecture NOTThese instruction only work on registers, NOT memory. They each perform a specific operation on one or two operands. First operand always a register - Rn Second operand sent to the ALU via barrel shifter. We will examine the barrel shifter shortly.

4 4 of 35 Arithmetic AND logical Instructions: General Format Opcode{Cond}{S} Rd,Rn,Operand 2 {Cond} - Conditional Execution of instruction –E.g. GT=GREATER THAN,LT = LESS THAN {S} - Set the bits in status register after execution. {Operand 2}-various form of the instruction immediate/register/shifting you can easily check all the combinations in the quick references of ARM.

5 5 of 35 Instruction format Cond 0 0 # opcode S Rn Rd operand 2 Opcode: Arithmetic / Logic Function S : Set condition code Rn : First operand (ARM Register) Rd : Destination operand (ARM Register)

6 6 of bit immediate If # (I) = 1 then operand 2 field will be #rot 8-bit immediate The actual immediate value is formed by rotating right the 8- bit immediate value by the even positions (#rot * 2) in a 32 bit word. If #rot = 0 (when immediate value is less than 0xFF – 8 bit) then carry out from the shifter will be indicated in C flag. Because of this all 32 bit immediate are not legitimate. Only those which can be formed.

7 7 of 35 Register operand # (I) = 0 indicates that the second operand is specified in register which can also be shifted. #shift Sh 0 Rm # Shift : Immediate shift length Rs : Register shift length Sh : Shift type Rm : Register used to hold second operand. Rs 0 Sh 1 Rm Ex: ADD RO, R1, R2, LSL #3 Ex: ADD RO, R1, R2, LSL R3

8 8 of 35 Shift operations Guided by “Sh” field in the format Sh = 00 Logical Shift Left : LSL Operation Sh = 01 Logical Shift Right : LSR Operation Sh = 10 Arithmetic Shift Right : ASR Operation Sh = 11 Rotate Right : ROR Operation With Sh = 11 and #shift = (similar to ROR #0) is used for RRX operation.

9 9 of 35 Immediate value 8 bit number Can be rotated right through an even number of positions. Assembler will calculate rotate for you from constant. Register, optionally with shift operation applied. Shift value can be either be: 5 bit unsigned integer Specified in bottom byte of another register. Operand 1 Result ALU Barrel Shifter Operand 2 Using the Barrel Shifter: The Second Operand

10 10 of 35 Data Processing instructions 0000AND 0001EOR 0010SUB 0011RSB 0100ADD 0101ADC 0110SBC 0111RSC 1000TST 1001TEQ 1010CMP 1011CMN 1100ORR 1101MOV 1110BIC 1111MVN

11 11 of 35 Arithmetic Operations Operations are: ADDoperand1 + operand2 ADCoperand1 + operand2 + carry SUBoperand1 - operand2 SBCoperand1 - operand2 + carry -1 RSBoperand2 - operand1 RSCoperand2 - operand1 + carry – 1 Syntax: { }{S} Rd, Rn, Operand2 Examples ADD r0, r1, r2 SUBGT r3, r3, #1 RSBLES r4, r5, #5

12 12 of 35 Logical Operations Operations are: ANDoperand1 AND operand2 EORoperand1 EOR operand2 ORRoperand1 OR operand2 BICoperand1 AND NOT operand2 [ie bit clear] Syntax: { }{S} Rd, Rn, Operand2 Examples: ANDr0, r1, r2 BICEQr2, r3, #7 EORSr1, r3, r0

13 13 of 35 Comparisons The only effect of the comparisons is to UPDATE THE CONDITION FLAGS. Thus no need to set S bit. Operations are: CMPoperand1 - operand2, but result not written CMNoperand1 + operand2, but result not written TSToperand1 AND operand2, but result not written TEQoperand1 EOR operand2, but result not written Syntax: { } Rn, Operand2

14 14 of 35 Comparisons Examples: CMPr0, r1 CMP R1,Operand2 e.g. CMP R1,R2 –[R1] - [R2] –Set the N Z C V in CPSR register. TSTEQr2, #5 TST R1, Operand2 e.g. TST R1,R2 –[R1] AND [R2]

15 15 of 35 Data Movement Operations are: MOVRd, operand2 MVNRd, (NOT) operand2 Note that these make no use of operand1. Syntax: { }{S} Rd, Operand2 Examples: MOVr0, r1 MVNr0, r1 MOVSr2, #10 MVNEQr1, #0

16 16 of 35 Quiz Start Stop r0 = r1 ? r0 > r1 ? r0 = r0 - r1r1 = r1 - r0 Yes NoYes No *Convert the GCD algorithm given in this flowchart into 1)“Normal” assembler, where only branches can be conditional. 2)ARM assembler, where all instructions are conditional, thus improving code density. *The only instructions you need are CMP, B and SUB.

17 17 of 35 Quiz - Sample Solutions “Normal” Assembler gcd cmp r0, r1 ;reached the end? beq stop blt less ;if r0 > r1 sub r0, r0, r1 ;subtract r1 from r0 bal gcd less sub r1, r1, r0 ;subtract r0 from r1 bal gcd stop ARM Conditional Assembler gcd cmp r0, r1 ;if r0 > r1 subgt r0, r0, r1 ;subtract r1 from r0 sublt r1, r1, r0 ;else subtract r0 from r1 bne gcd ;reached the end?

18 18 of 35 The Barrel Shifter The ARM doesn’t have actual shift instructions. Instead it has a barrel shifter which provides a mechanism to carry out shifts as part of other instructions. So what operations does the barrel shifter support?

19 19 of 35 Shifts left by the specified amount (multiplies by powers of two) e.g. LSL #5 = multiply by 32 Barrel Shifter - Left Shift Logical Shift Left (LSL) Destination CF 0

20 20 of 35 Logical Shift Right Shifts right by the specified amount (divides by powers of two) e.g. LSR #5 = divide by 32 Arithmetic Shift Right Shifts right (divides by powers of two) and preserves the sign bit, for 2's complement operations. e.g. ASR #5 = divide by 32 Barrel Shifter - Right Shifts Destination CF Destination CF Logical Shift Right Arithmetic Shift Right...0 Sign bit shifted in

21 21 of 35 Barrel Shifter - Rotations Rotate Right (ROR) Similar to an ASR but the bits wrap around as they leave the LSB and appear as the MSB. e.g. ROR #5 Note the last bit rotated is also used as the Carry Out. Rotate Right Extended (RRX) This operation uses the CPSR C flag as a 33rd bit. Rotates right by 1 bit. Encoded as ROR #0. Destination CF Rotate Right Destination CF Rotate Right through Carry

22 22 of 35 Second Operand: Shifted Register The amount by which the register is to be shifted is contained in either: the immediate 5-bit field in the instruction –NO OVERHEAD –Shift is done for free - executes in single cycle. the bottom byte of a register (not PC) –Then takes extra cycle to execute –ARM doesn’t have enough read ports to read 3 registers at once. –Then same as on other processors where shift is separate instruction. If no shift is specified then a default shift is applied: LSL #0 i.e. barrel shifter has no effect on value in register.

23 23 of 35 Using a multiplication instruction to multiply by a constant means first loading the constant into a register and then waiting a number of internal cycles for the instruction to complete. A more optimum solution can often be found by using some combination of MOVs, ADDs, SUBs and RSBs with shifts. Multiplications by a constant equal to a ((power of 2) ± 1) can be done in one cycle. Example: r0 = r1 * 5 Example: r0 = r1 + (r1 * 4) ADD r0, r1, r1, LSL #2 Example: r2 = r3 * 105 Example: r2 = r3 * 15 * 7 Example: r2 = r3 * (16 - 1) * (8 - 1) RSB r2, r3, r3, LSL #4; r2 = r3 * 15 RSB r2, r2, r2, LSL #3; r2 = r2 * 7 Second Operand:Using a Shifted Register

24 24 of 35 Loading full 32 bit constants Although the MOV/MVN mechanism will load a large range of constants into a register, sometimes this mechanism will not generate the required constant. Therefore, the assembler also provides a method which will load ANY 32 bit constant: LDR rd,=numeric constant If the constant can be constructed using either a MOV or MVN then this will be the instruction actually generated. Otherwise, the assembler will produce an LDR instruction with a PC-relative address to read the constant from a literal pool. LDR r0,=0x42; generates MOV r0,#0x42 LDR r0,=0x ; generate LDR r0,[pc, offset to lit pool] As this mechanism will always generate the best instruction for a given case, it is the recommended way of loading constants.

25 25 of 35 Multiplication Instructions The Basic ARM provides two multiplication instructions. Multiply MUL{ }{S} Rd, Rm, Rs; Rd = Rm * Rs Multiply Accumulate- does addition for free MLA{ }{S} Rd, Rm, Rs, Rn; Rd = (Rm * Rs) + Rn Restrictions on use: Rd and Rm cannot be the same register –Can be avoid by swapping Rm and Rs around. This works because multiplication is commutative. Cannot use PC. These will be picked up by the assembler if overlooked. Operands can be considered signed or unsigned Up to user to interpret correctly.

26 26 of 35 Instructions are MULL which gives RdHi,RdLo:=Rm*Rs MLAL which gives RdHi,RdLo:=(Rm*Rs)+RdHi,RdLo However the full 64 bit of the result now matter (lower precision multiply instructions simply throws top 32bits away) Need to specify whether operands are signed or unsigned Warning : Unpredictable on non-M ARMs. Syntax: UMULL/UMLAL{cond} {S} RdLo, RdHi, Rm, Rs SMULL/SMLAL{cond} {S} RdLo, RdHi, Rm, Rs Multiply-Long and Multiply_Accumulate Long

27 27 of 35 Instruction Format 000 MUL Rd := (Rm * Rs) 001 MLA Rd := (Rm * Rs) + Rn 100 UMULL RdHi : RdLo := Rm * Rs 101 UMLAL RdHi : RdLo += Rm * Rs 110 SMULL RdHi : RdLo := Rm * Rs 111 SMLAL RdHi : RdLo += Rm * Rs Cond mul S Rd/RdHi Rn/RdLo Rs Rm

28 28 of 35 Example: C assignments C: x = (a + b) - c; Assembler: ADR r4,a; get address for a LDR r0,[r4]; get value of a ADR r4,b; get address for b, reusing r4 LDR r1,[r4]; get value of b ADD r3,r0,r1; compute a+b ADR r4,c; get address for c LDR r2[r4]; get value of c SUB r3,r3,r2; complete computation of x ADR r4,x; get address for x STR r3[r4]; store value of x

29 29 of 35 Example: C assignment C: y = a*(b+c); Assembler: ADR r4,b ; get address for b LDR r0,[r4] ; get value of b ADR r4,c ; get address for c LDR r1,[r4] ; get value of c ADD r2,r0,r1 ; compute partial result ADRr4,a ; get address for a LDR r0,[r4] ; get value of a MUL r2,r2,r0 ; compute final value for y ADR r4,y ; get address for y STR r2,[r4] ; store y

30 30 of 35 Example: C assignment C: z = (a << 2) | (b & 15); Assembler: ADR r4,a ; get address for a LDR r0,[r4] ; get value of a MOV r0,r0,LSL 2 ; perform shift ADR r4,b ; get address for b LDR r1,[r4] ; get value of b AND r1,r1,#15 ; perform AND ORR r1,r0,r1 ; perform OR ADR r4,z ; get address for z STR r1,[r4] ; store value for z

31 31 of 35 Example: FIR filter C: for (i=0, f=0; i

32 32 of 35 FIR filter, cont’.d ADR r3,c ; load r3 with base of c ADR r5,x ; load r5 with base of x ; loop body loop LDRr4,[r3,r8] ; get c[i] LDR r6,[r5,r8] ; get x[i] MUL r4,r4,r6 ; compute c[i]*x[i] ADD r2,r2,r4 ; add into running sum ADD r8,r8,#4 ; add 1 word offset to array index ADD r0,r0,#1 ; add 1 to i CMP r0,r1 ; exit? BLT loop ; if i < N, continue

33 33 of 35 Barrel Shifter executing Fixed Point Arithmetic

34 34 of 35 Summary All data processing operations work only on internal registers, immediate operands and NOT on memory. Data processing instructions Arithmetic – ADD, ADC, SUB, SBC, RSB, RSC Comparison – CMP, CMN, TST, TEQ Logical – ORR, AND, EOR, BIC Data Movement – MOV, MVN Shifter Operations – LSL, LSR, ASR, ROR, RRX Multiplication Instructions – MUL, MLA, MULL, MLAL

35 35 of 35 Thank You, Any Questions ?


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