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Chapter 2. Instructions: Language of the Machine We’ll be working with the MIPS instruction set architecture –MIPS (originally an acronym for Microprocessor.

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Presentation on theme: "Chapter 2. Instructions: Language of the Machine We’ll be working with the MIPS instruction set architecture –MIPS (originally an acronym for Microprocessor."— Presentation transcript:

1 Chapter 2

2 Instructions: Language of the Machine We’ll be working with the MIPS instruction set architecture –MIPS (originally an acronym for Microprocessor without Interlocked Pipeline Stages) is a RISC microprocessor architecture developed by MIPS Technologies. By the late 1990s it was estimated that one in three RISC chips produced were MIPS-based designsRISCMIPS TechnologiesBy the late 1990s –MIPS :It presently stands for million instructions per second –similar to other architectures developed since the 1980's –Almost 100 million MIPS processors manufactured in 2002 –used by NEC, Nintendo, Cisco, Silicon Graphics, Sony, …

3 MIPS arithmetic All instructions have 3 operands Operand order is fixed (destination first) Example: C code: a = b + c MIPS ‘code’: add a, b, c “The natural number of operands for an operation like addition is three…requiring every instruction to have exactly three operands, no more and no less, conforms to the philosophy of keeping the hardware simple”

4 4 Design Principles Design Principle 1: Simplicity favors regularity. Design Principle 2: Smaller is faster. Design Principle 3: Make the common case fast. Design Principle 4: Good design demands good compromises.

5 MIPS arithmetic Design Principle: Simplicity favors regularity. Of course this complicates some things... C code:a = b + c + d + e; MIPS code:add a, b, c add a, a, d add a, a, e Eg.1. a = b + c; d = a – e ; Show the MIPS code produced by the compiler Eg.2. f = (g + h) - ( i + j); Operands must be registers, only 32 registers provided Each register contains 32 bits. Design Principle: smaller is faster. Why?

6 Registers MIPS convention is to use two character names following a dollar sign to represent a register. Temporary registers –$t0 thru $t7 correspond to registers 8-15 Registers for holding variables –$s0 thru $s7 correspond to registers Constant 0 –$zero used for the constant 0 –Also corresponds to register 0 It is the compiler’s job to associate program variables with registers. Eg. F =(g + h) – (i +j ); the variables f,g,h,I,j are assigned to the registers $s0,$s1,$s2,$s3,$s4 respectively. What is the complied MIPS code?

7 Registers vs. Memory ProcessorI/O Control Datapath Memory Input Output Arithmetic instructions operands MUST be registers, — only 32 registers provided Compiler associates variables with registers What about programs with lots of variables

8 Memory Organization Viewed as a large, single-dimension array, with an address. A memory address is an index into the array Assume that A is an array of 100 words, the compiler has associated the variables g and h with the registers $s1 and $s2 as before. The base address of the array is in $s3. Compile the C assignment statement g = h + A[8]; The data transfer instruction that copies data from memory to a register is called “load”. "Byte addressing" means that the index points to a byte of memory bits of data

9 Instructions Load and store instructions Example: C code: A[12] = h + A[8]; MIPS code: lw $t0, 32($s3) add $t0, $s2, $t0 sw $t0, 48($s3) Remember arithmetic operands are registers, not memory! Can’t write: add 48($s3), $s2, 32($s3)

10 Small constants are used quite frequently (50% of operands) e.g., A = A + 5; B = B + 1; C = C - 18; Solutions? Why not? –put 'typical constants' in memory and load them. –create hard-wired registers (like $zero) for constants like zero. MIPS Instructions: addi $s3, $s3, 4# $s3 = $s3 + 4 Design Principle: Make the common case fast. Constant operands occur frequently, and by including constants inside arithmetic instructions, they are much faster than if constants were loaded from memory. Constants

11 Instructions, like registers, are also 32 bits long –Example: add $t1, $s1, $s2 –registers have numbers, $t1=9, $s1=17, $s2=18 Instruction Format: R-type (Register format) op rs rt rdshamtfunct Can you guess what the field names stand for? Machine Language

12 Consider the load-word and store-word instructions, –What would the regularity principle have us do? –A conflict between the desire to keep all instructions the same length and the desire to have a single instruction format. –New principle: Good design demands a compromise Introduce a new type of instruction format –I-type for data transfer instructions –other format was R-type for register Example: lw $t0, 32($s2) op rs rt 16 bit number Although multiple formats complicate the hardware, we can reduce the complexity by keeping the formats similar. Formats are distinguished by the values in the first field. Machine Language

13 Translating MIPS Assembly Level Language into machine language Eg. If $t1 has the base of the array A and $s2 corresponds to h, Compile into MIPS code. What is the machine language code for these instructions corresponding to the assignment statement A[300] = h + A[300];

14 Logical Operations Shift left logical( sll ) and shift right logical( srl ) Example Sll $t2, $s0, 4# $t2 = $s0 << 4 bits R-format Shifting left corresponds to what arithmetic operation? Shifting right corresponds to what arithmetic operation?

15 Logical Operations AND and $t0, $t2, $t1# $t0 = $t2 AND $t1 OR or $t0, $t2, $t1# $t0 = $t2 OR $t1 NOT –Implemented with NOR with one operand 0 (why?) nor $t0, $t2, 0# $t0 = NOT ($t2) Also have immediate versions andi $t0, $t1, 0 ori $t0, $t1, 0

16 Decision making instructions –alter the control flow, –i.e., change the "next" instruction to be executed MIPS conditional branch instructions: bne $t0, $t1, Label beq $t0, $t1, Label Example: if (i==j) h = i + j; bne $s0, $s1, Label add $s3, $s0, $s1 Label:.... Control

17 MIPS unconditional branch instructions: j label Example: if (i!=j) beq $s4, $s5, L1 h=i+j;add $s3, $s4, $s5 else j L2 h=i-j;L1:sub $s3, $s4, $s5 L2:... Control

18 So far: InstructionMeaning add $s1,$s2,$s3$s1 = $s2 + $s3 sub $s1,$s2,$s3$s1 = $s2 – $s3 lw $s1,100($s2)$s1 = Memory[$s2+100] sw $s1,100($s2)Memory[$s2+100] = $s1 bne $s4,$s5,LNext instr. is at Label if $s4 ≠ $s5 beq $s4,$s5,LNext instr. is at Label if $s4 = $s5 j LabelNext instr. is at Label Formats: op rs rt rdshamtfunct op rs rt 16 bit address op 26 bit address RIJRIJ

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