MIPS指令集架构

1 . ECE4680 Computer Organization & Architecture MIPS Instruction Set Architecture Why is MIPS a good example? Learn ISA further by this example. How does IS fill up the gap between HLL and machine? ECE4680 Lec4 MIPS.1 February 6, 2002 RISC Vs. CISC RISC CISC °Determined by VLSI technology. °Software cost goes up constantly. To be convenient for programmers. °To shorten the semantic gap between HLL and architecture without advanced compilers. °To reduce the program length because memory was expensive. °VAX 11/780 reached the climax with >300 instructions and >20 addressing modes. CISC RISC °Things changed: HLL, Advanced Compiler, Memory size, … °Finding: 25% instructions used in 95% time. °Size: usually <100 instructions and <5 addressing modes. °Other properties: fixed instruction format, register based, hardware control… °Gains: CPI is smaller, Clock cycle shorter, Hardware simpler, Pipeline easier °Loss: Program becomes longer, but memory becomes larger and larger, cheaper and cheaper. Programmability becomes poor, but people use HLL instead of IS. °Result: although program is prolonged, the total gain is still a plus. ECE4680 Lec4 MIPS.2 February 6, 2002 

2 . MIPS R2000 / R3000 Registers °32-bit machine --> Programmable storage 232 x bytes °31 x 32-bit GPRs (R0 = 0) r0 0 r1 °32 x 32-bit FP regs (f0 - f31, paired DP) ° °HI, LO, PC: SPRegisters ° ° °Big Endian r31 °2 nomenclatures(next slide) PC lo °See Fig. A.18 at P. A-50 for more details hi °Addressing modes: • immediate • register • displacement °All instructions are 32-bit wide and must be aligned. ECE4680 Lec4 MIPS.3 February 6, 2002 2 Nomenclatures of MIPS Registers (p.140, A-23) Name number Usage Reserved on call? zero 0 constant value =0 n.a. at 1 reserved for assembler (p.147,157) n.a. v0 – v1 2–3 values for results and expression evaluation no a0 – a3 4–7 arguments no t0 – t7 8 – 15 temporaries no s0 – s7 16 – 23 saved yes t8 – t9 24 – 25 more temporaries no k0 – k1 26 – 27 Reserved for kernel n.a. gp 28 global pointer yes sp 29 stack pointer yes fp 30 frame pointer yes ra 31 return address yes zero at v0-v1 a0 - a3 t0 - t7 s0 - s7 t8 - t9 k0 - k1 gp sp fp ra 0 1 2 - 3 4-7 8 --- 15 16 --- 23 24 - 25 26 - 27 28 29 30 31 ECE4680 Lec4 MIPS.4 February 6, 2002 

3 . MIPS arithmetic and logic instructions Instruction Example Meaning Comments add add $1,$2,$3 $1 = $2 + $3 3 operands; exception possible subtract sub $1,$2,$3 $1 = $2 – $3 3 operands; exception possible add immediate addi $1,$2,100 $1 = $2 + 100 + constant; exception possible multiply mult $2,$3 Hi, Lo = $2 x $3 64-bit signed product divide div $2,$3 Lo = $2 ÷ $3, Lo = quotient, Hi = remainder Hi = $2 mod $3 Move from Hi mfhi $1 $1=Hi get a copy of Hi Move from Lo mflo $1 $1=lo Instruction Example Meaning Comment and and $1,$2,$3 $1 = $2 & $3 Logical AND or or $1,$2,$3 $1 = $2 | $3 Logical OR xor xor $1,$2,$3 $1 = $2 ⊕ $3 Logical XOR nor nor $1,$2,$3 $1 = ~($2 |$3) Logical NOR ECE4680 Lec4 MIPS.5 February 6, 2002 Example (p110) E.g. f= (g+h) - (i+j), assuming f, g, h, i, j be assigned to $1, $2, $3, $4, $5 add $7, $2, $3 add $8, $4, $5 sub $1, $7, $8 ECE4680 Lec4 MIPS.6 February 6, 2002 

4 . MIPS data transfer instructions Instruction Comment SW 500($4), $3 Store word SH 502($2), $3 Store half SB 41($3), $2 Store byte LW $1, 30($2) Load word LH $1, 40($3) Load half a word LB $1, 40($3) Load byte ECE4680 Lec4 MIPS.7 February 6, 2002 Example (pp112-114) Assume A is an array of 100 words, and compiler has associated the varialbes g and h with the register $1 and $2. Assume the base address of the array is in $3. Translate g = h + A[8] lw $4, 8($3); $4 <-- A[8] add $1, $2, $4; offset or displacement base register lw $4, 32($3); add $1, $2, $4 A[12] = h+A[8] SW $1, 48($3) ECE4680 Lec4 MIPS.8 February 6, 2002 

5 . Example (p114) Assume A is an array of 100 words, and compiler has associated the varialbes g, h, and i with the register $1, $2, $5. Assume the base address of the array is in $3. Translate g = h + A[i] addi $6, $0, 4; $6 = 4 mult $5, $6; Hi,Lo = i*4 mflo $6; $6 = i*4, assuming i is small add $4, $3, $6; $4 address of A[i] add $1, $2, $4 ECE4680 Lec4 MIPS.9 February 6, 2002 MIPS jump, branch, compare instructions Instruction Example Meaning branch on equal beq $1,$2,100 if ($1 == $2) go to PC+4+100 Equal test; PC relative branch branch on not eq. bne $1,$2,100 if ($1!= $2) go to PC+4+100 Not equal test; PC relative Pseudoinstruction blt, ble, bgt, bge not implemented by hardware, but synthesized by assembler set on less than slt $1,$2,$3 if ($2 < $3) $1=1; else $1=0 Compare less than; 2’s comp. set less than imm. slti $1,$2,100 if ($2 < 100) $1=1; else $1=0 Compare < constant; 2’s comp. jump j 10000 go to 10000 Jump to target address jump register jr $31 go to $31 For switch, procedure return jump and link jal 10000 $31 = PC + 4; go to 10000 For procedure call ECE4680 Lec4 MIPS.10 February 6, 2002 

6 . Example (p123) if (i==j) go to L1; if (i >=j) go to L1; f = g+ h; L1: f = f - i; Assuming f, g, h, i, j ~ $1, $2, $3, $4, $5 pseudoinstruction beq $4, $5, L1 bqe $4, $5, L1 add $1, $2, $3 assembler L1: sub $1, $1, $4 slt $1, $4, $5 beq $0, $1, L1 ECE4680 Lec4 MIPS.11 February 6, 2002 Example (p126) Loop: g = g +A[i]; i = i+ j; if (i != h) go to Loop: Assuming variables g, h, i, j ~ $1, $2, $3, $4 and base address of array is in $5 Loop: add $7, $3, $3; i*2 add $7, $7, $7; i*4 add $7, $7, $5 lw $6, 0($7); $6=A[i] add $1, $1, $6; g= g+A[i] add $3, $3, $4 bne $3, $2, Loop; ECE4680 Lec4 MIPS.12 February 6, 2002 

7 . Example (p127) while (A[i]==k) i = i+j; Assume i, j, and k ~ $17, $18, $19 and base of A is in $3 Loop: add $20, $17, $17 add $20, $20, $20 add $20, $20, $3 lw $21,0($20) bne $21, $19, Exit add $17, $17, $18 j Loop Exit: ECE4680 Lec4 MIPS.13 February 6, 2002 MIPS Addressing Modes/Instruction Formats (p118,148,152) R-format: 6 5 5 5 5 6 Register (direct) op rs rt rd smt func register I-format: Immediate op rs rt immed Base+index op rs rt immed Memory register + PC-relative op rs rt immed Memory PC + 4 + J-format: Memory op addr. ECE4680 Lec4 MIPS.14 February 6, 2002 

8 . Example: See machine code in memory (p149) while (A[i]==k) i = i+j; Assume i, j, and k ~ $17, $18, $19 and base of A is in $3 Assume the loop is placed starting at loc 8000 8000: 0 17 17 20 0 32 Loop: add $20, $17, $17 add $20, $20, $20 0 20 20 20 0 32 add $20, $20, $3 0 20 3 20 0 32 lw $21,0($20) 35 20 21 0 bne $21, $19, Exit 5 21 19 8 2 add $17, $17, $18 0 17 18 17 0 32 j Loop 2 8000 2000 Exit: Offset in branch is relative. Address in jump is absolute. Address in Branch or Jump instruction is word address so that they can go 4 times far as opposed to byte address. (p150) ECE4680 Lec4 MIPS.15 February 6, 2002 Procedure Call and Stack Stacking of Subroutine Calls & Returns and Environments: A: A CALL B B: A B CALL C C: A B C RET A B RET A Some machines provide a memory stack(special hardware) as part of the architecture (e.g. the VAX). Use special instructions, e.g. pop, push. Sometimes stacks are implemented via software convention (e.g. MIPS). Use same data transfer instructions, e.g., lw, sw. ECE4680 Lec4 MIPS.16 February 6, 2002 

9 . Example in C: swap (pp163-165) °Assume swap is called as a procedure °Assume temp is register $15; arguments v and k ~ $16 and $17; °Write MIPS code swap(int v[], int k) { int temp; temp = v[k]; v[k] = v[k+1]; sll $18, $17, 2 ; mulitply k by 4 v[k+1] = temp; addu $18, $18, $16 ; address of v[k] lw $15, 0($18) ; load v[k] } lw $19, 4($18) ; load v[k+1] sw $19, 0($18) ; store v[k+1] into v[k] sw $15, 4($18) ; store old v[k] into v[k+1] Registers $15, $16, $17, $18, $19 are occupied by caller ?? ECE4680 Lec4 MIPS.17 February 6, 2002 Example: Swap Given a procedure swap(v, j) Calling swap is as simple as jal swap jal --- jump and link $31 = PC+4; $31=$ra : always store return address goto swap ECE4680 Lec4 MIPS.18 February 6, 2002 

10 . swap: MIPS swap: addi $sp,$sp, –24 ; Make room on stack for 6 registers sw $31, 20($sp) ; Save return address sw $15, 16($sp) ; Save registers on stack sw $16, 12($sp) sw $17, 8($sp) sw $18, 4($sp) sw $19, 0(sp) .... lw $19, 0($sp) ; Restored registers from stack lw $18, 4($sp) lw $17, 8($sp) lw $16, 12($sp) lw $15, 16($sp) lw $31, 20($sp) ; Restore return address addi $sp,$sp, 24 ; restore top of stack jr $31 ; return to place that called swap ECE4680 Lec4 MIPS.19 February 6, 2002 Other ISAs °Intel 8086/88 => 80286 => 80386 => 80486 => Pentium => P6 • 8086 few transistors to implement 16-bit microprocessor • tried to be somewhat compatible with 8-bit microprocessor 8080 • successors added features which were missing from 8086 over next 15 years • product of several different Intel engineers over 10 to 15 years • Announced 1978 °VAX simple compilers & small code size => • efficient instruction encoding • powerful addressing modes • powerful instructions • few registers • product of a single talented architect • Announced 1977 ECE4680 Lec4 MIPS.20 February 6, 2002 

11 . Machine Examples: Address & Registers Intel 8086 2 20 x 8 bit bytes acc, index, count, quot AX, BX, CX, DX stack, string SP, BP, SI, DI code,stack,data segment CS, SS, DS IP, Flags 32 r15-- program counter VAX 11 2 x 8 bit bytes r14-- stack pointer 16 x 32 bit GPRs r13-- frame pointer r12-- argument ptr 24 MC 68000 2 x 8 bit bytes 8 x 32 bit GPRs 7 x 32 bit addr reg 1 x 32 bit SP 1 x 32 bit PC 32 MIPS 2 x 8 bit bytes 32 x 32 bit GPRs 32 x 32 bit FPRs HI, LO, PC ECE4680 Lec4 MIPS.21 February 6, 2002 Homework 2, due Feb. 20, 2002 °Questions 3.2, 3.3, 3.5, 3.6, 3.7, 3.9, 3.11 °To answer question 3.7, please refer to Figure 3.13 (page 140) for register convention °To answer 3.11, please refer to sort example in pages 166 for a skeleton of for loop ECE4680 Lec4 MIPS.22 February 6, 2002