ELEN 468 Advanced Logic Design Lecture 18 MIPS Microprocessor ELEN 468 Lecture 18

Computer Architecture CISC Complex Instruction Set Computer Intel’s x86, Motorola’s 680x0 RISC Reduced Instruction Set Computer Any computer architecture defined after 1984 MIPS Microcomputer without Interlocked Pipeline Stages Millions of Instructions Per Second Strongly pipelined architecture DEC’s Alpha, HP’s Precision ELEN 468 Lecture 18

Registers 32 32-bit (word) registers $a0 - $a3: argument registers $v0 - $v1: return values $ra: return address register $sp: stack pointer $fp: frame pointer $gp: global pointer $zero: always equals 0 $s0 - $s7: preserved on a procedural call $t0 - $t9: not preserved by callee on a procedural call ELEN 468 Lecture 18

Arithmetic Operations add a, b, c # a = b + c add $t0, $s1, $s2 sub a, b, c # a = b – c sub $s0, $t0, $t1 Arithmetic operations occur only on registers ELEN 468 Lecture 18

Data Transfer lw $t0, 8($s3) # load $t0 with data from memory # base address in $s3, offset 8 sw $t0, 48($s3) # store word … … Byte – 8 bits Word – 32 bits Memory in words Address to byte level 12 101 110 10 1001 8 4 Address Memory ELEN 468 Lecture 18

MIPS Fields R-type op rs rt rd shamt funct 6 bits 5 bits 5 bits 5 bits I-type op rs rt address op: basic operation, also called opcode rs: the first register source operand rt: R-type: the second register source operand I-type: destination register rd: the register destination operand shamt: shift amount in shift instructions funct: selects the specific variant of the operation in op field, also called function code address: offset of memory address in data transfer instructions ELEN 468 Lecture 18

Examples of Machine Code op rs rt rd shamt funct 18 19 17 32 add $s1, $s2, $s3 18 19 17 34 sub $s1, $s2, $s3 op rs rt address 35 18 17 100 lw $s1, 100($s2) 43 18 17 100 sw $s1, 100($s2) ELEN 468 Lecture 18

Some Other Instructions beq $s3, $s4, L1 # go to branch L1, if equal bne $s3, $s4, L1 # go to branch L1, if not equal j L1 # jump to branch L1 jr $t0 # jump based on $t0 slt $t0, $s1, $s2 # set value of $t0 to 1, if less than sll $t2, $s0, 8 # reg $t2 = reg $s0 << 8 bits srl $t2, $s0, 8 # reg $t2 = reg $s0 >> 8 bits addi $sp, $sp, 4 # $sp = $sp + 4 nop # do nothing ELEN 468 Lecture 18

MIPS Addressing Mode Register addressing: the operand is a register Base or displacement addressing: the operand is at the memory whose address is the sum of a register and a constant Immediate addressing: the operand is a constant PC (Program Counter)-relative addressing: address is the sum of PC and a constant in the instruction ELEN 468 Lecture 18

Steps for MIPS Instructions Fetch instruction from memory Read registers while decoding the instruction Execute the operation or calculate an address Access an operand in data memory (for lw and sw) Write the result into a register ELEN 468 Lecture 18

Implement Instruction Fetch Add 4 Read address PC Instruction Instruction memory ELEN 468 Lecture 18

Datapath for R-type Instructions 5 Read register 1 Control 32 Read data 1 5 Read register 2 Instruction ALU Registers 32 5 Write register Result 32 Read data 2 Write data Reg_write ELEN 468 Lecture 18

Example of Instruction Execution Time Instruction fetch Register read ALU operation Memory access Register write Total time lw 2 1 8 sw 7 R-format (add, sub) 6 Branch 5 ELEN 468 Lecture 18

Unpipelined vs. Pipelined 2 4 6 8 10 12 14 16 18 lw $t1, 8($s1) lw $t2, 16($s2) lw $t3, 12($s3) IF ID ALU MEM WB IF: Instruction fetch ID: Instruction decode and read register ALU: Execution or address calculation IF ID ALU MEM WB MEM: Memory access WB: Write back to reg IF lw $t1, 8($s1) lw $t2, 16($s2) lw $t3, 12($s3) IF ID ALU MEM WB IF ID ALU MEM WB IF ID ALU MEM WB ELEN 468 Lecture 18

Instruction Sets for Pipelining All instructions have the same length There are only a few instruction formats Memory operands only appear in loads or stores Operands must be aligned in memory ELEN 468 Lecture 18

Pipeline Hazards Situations when the next instruction cannot execute in the following clock cycle Structural hazards Control hazards Data hazards ELEN 468 Lecture 18

Structural Hazards Hardware cannot support the combined instructions that we want to execute in the same clock cycle Example: if there is only one memory, then memory access and instruction fetch cannot be executed simultaneously Solution: add hardware ELEN 468 Lecture 18

Control Hazards Decision-making depends on the result of an instruction that has not been finished Example: PC following a branch instruction depends if branch is taken or not Solutions Predict: execute next instruction anyway, if branch is taken, retract the decision Dynamic prediction based on smart guess ELEN 468 Lecture 18

Data Hazards An instruction cannot be executed until a data is available from another instruction Example: add $s0, $t1, $t2 sub $t2, $s0, $t3 Solution: Bypassing: result can be fed to next instruction execution without loading to a register ELEN 468 Lecture 18