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Chapter 2 Instructions: Language of the Computer.

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2 Chapter 2 Instructions: Language of the Computer

3 Instructions: Language of the Machine Instruction set: the vocabulary of commands understood by a given architecture. More primitive than higher level languages e.g., no sophisticated control flow Very restrictive e.g., MIPS Arithmetic Instructions We’ll be working with the MIPS instruction set architecture similar to other architectures developed since the 1980's used by NEC, Nintendo, Silicon Graphics, Sony Design goals: maximize performance and minimize cost, reduce design time

4 MIPS arithmetic Operations and Operands All instructions have 3 operands Need to break a C statement into several assembly instructions ( f = (g + h) – ( I + j)) Operand order is fixed (destination first) Example: C code: A = B + C MIPS code: add $s0, $s1, $s2 (associated with variables by compiler) add t0,g,h add t1,i,j sub f,t0,t1

5 MIPS arithmetic Design Principle 1: simplicity favors regularity. Why? Of course this complicates some things... C code: A = B + C + D; E = F - A; MIPS code: add $t0, $s1, $s2 add $s0, $t0, $s3 sub $s4, $s5, $s0 Operands must be registers, only 32 registers provided

6 Registers vs. Memory Design Principle 2: smaller is faster. Why? Arithmetic instructions operands must be registers, — only 32 registers provided Compiler associates variables with registers What about programs with lots of variables?

7 Memory Organization Viewed as a large, single-dimension array, with an address. A memory address is an index into the array "Byte addressing" means that the index points to a byte of memory bits of data

8 Bytes are nice, but most data items use larger "words" For MIPS, a word is 32 bits or 4 bytes bytes with byte addresses from 0 to words with byte addresses 0, 4, 8, Words are aligned i.e., what are the least 2 significant bits of a word address? Memory Organization bits of data Registers hold 32 bits of data

9 Instructions Load and store instructions Example: C code: A[8] = h + A[8]; MIPS code: lw $t0, 32($s3) add $t0, $s2, $t0 sw $t0, 32($s3) Store word has destination last Remember arithmetic operands are registers, not memory!

10 Another Example What is the MIPS assembly code for A[12] = h + A[8] assuming that h is associated with register $s2 and the base address of the array A is in $s3? lw $t0,32($s3) # temporary reg $t0 gets A[8] add $t0,$s2,$t0 # $t0 gets h + A[8] sw $t0,48($s3) # stores h + A[8] into A[12]

11 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 one. 3 Constants

12 Constants (cont’d) MIPS Instructions: addi $s3, $s3, 4 slti $s1, $t1, 10 andi $s3, $s3, 6 ori $s3, $s3, 4 Design Principle 3: Make the common case fast.

13 So far we’ve learned: (p.59) MIPS — loading words but addressing bytes — arithmetic on registers only InstructionMeaning add $s1, $s2, $s3$s1 = $s2 + $s3 sub $s1, $s2, $s3$s1 = $s2 – $s3 addi $s1, $s2,100$s1 = $s lw $s1, 100($s2)$s1 = Memory[$s2+100] sw $s1, 100($s2)Memory[$s2+100] = $s1

14 Instructions, like registers and words of data, are also 32 bits long Example: add $t0, $s1, $s2 Registers have numbers, $t0=8, $s1=17, $s2=18 Instruction Format: Can you guess what the field names stand for? Machine Language oprsrtrdshamtfunct

15 MIPS Fields R-type op: basic operation of the instruction, or opcode (6 bits) rs: the first register source operand (5 bits) rt: the second register source operand (5 bits) rd: the register destination operand (5 bits) shamt: shift amount (5 bits) funct: function code (6 bits)

16 Consider the load-word and store-word instructions, What would the regularity principle have us do? New principle: Good design demands good compromises. Machine Language

17 I-Type Instruction 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) Another example: addi (R-type or I-Type?) Where's the compromise? oprsrt 16 bit number

18 Translating MIPS Assembly Language into Machine Language(p 65-66) Example (Figure 2.6 MIPS Instruction Encoding) A[300] = h + A[300] lw $t0,1200($t1) add $t0,$s2,$t0 sw $t0,1200($t1) MIPS Instruction Encoding: MIPS Reference Data Card

19 Instructions are bits Programs are stored in memory — to be read or written just like data Fetch & Execute Cycle Instructions are fetched and put into a special register Bits in the register "control" the subsequent actions Fetch the “next” instruction and continue Stored Program Concept

20 Logical Operations Useful to operate on fields of bits within a word or on individual bits. Logical operations C operatorsJava operators MIPS Shift left<< sll Shift right>>>>>srl Bit-by-bit AND &&and,andi Bit-by-bit OR||or,ori Bit-by-bit NOT ~~nor

21 shamt sll $t2,$s0,4#reg $t2 = reg $s0 << 4 bits oprsrtrd shamt funct Shift amount = 4 (What is the max shift amount?) Shifting left by i bits gives the same result as multiplying by 2 i

22 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:.... Instructions for Making Decisions

23 MIPS unconditional branch instructions: j label Example: if (i==j) bne $s3, $s4, Else h=g+h;add $s0, $s1, $s2 else j Exit f=g-h;Else:sub $s0, $s1, $s2 Exit: If-then-else conditional branches

24 Loops while loop in C: while (save[i]==k) i+=1; i  $s3, k  $s5, base of the array save is in $s6. Loop: sll$t1,$s3,2# reg $t1=4*i add $t1,$t1,$s6# $t1= address of save[i] lw$t0,0($t1)# reg t0=save[i] bne$t0, $s5, Exit # go to Exit if save[i]!=k addi$s3,$s3,1# i=i+1 jLoop# go to Loop Exit:

25 We have: beq, bne, what about Branch-if-less-than? New instruction: if $s3 < $s4 then $t0 = 1 slt $t0, $s3, $s4 else $t0 = 0 slti $t0,$s2,10# $t0=1 if $s2 < 10 Can use the above instructions to build " blt $s1, $s2, Label " — can now build general control structures Note that the assembler needs a register to do this, there are policy of use conventions for registers ($zero) Control Flow

26 Case/Switch Statement Two possible approaches Convert the switch statement into a chain of if- then-else statements Build a jump address table MIPS instruction: jr (jump register) Unconditional jump to the address specified in the register.

27 So far (p.77): 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 and $s1,$s2,$s3 $s1 = $s2 & $s3 or $s1,$s2,$s3 $s1 = $s2 | $s3 nor $s1,$s2,$s3 $s1 = ~($s2 | $s3) andi $s1,$s2,100 $s1 = $s2 & 100 ori $s1,$s2,100 $s1 = $s2 ! $100 sll $s1,$s2,10 $s1=s2 >10 bne $s4,$s5,LNext instr. is at Label if $s4!= $s5 beq $s4,$s5,LNext instr. is at Label if $s4 = $s5 slt$s1,$s2,$s3 if ($s2<$s3) $s1=1, else $s1=0 slti $s1,$s2,100 if ($s2<100) $s1=1, else $s1=0 j LabelNext instr. is at Label

28 Formats op rs rt rdshamtfunct op rs rt 16 bit address/constant op 26 bit address RIJRIJ

29 Supporting Procedures in Computer Hardware Six steps the program must follow in an execution of a procedure 1. Place parameters in a place where the procedure can access them 2. Transfer control to the procedure 3. Acquire the storage resources needed for the procedure 4. Perform the desired task 5. Place the result value in a place where the calling program can access it 6. Return control to the point of origin.

30 MIPS Registers MIPS registers: $a0-a3: four argument registers in which to pass parameters $ v0-$v1: two value registers in which to return values $ra: one return address register to return to the point of origin MIPS instruction: jal ProcedureAddress (jump and link)

31 Using More Registers Use stack MIPS allocates a register for the stack: the stack pointer ($sp) Example: p81-82.

32 Policy of Use Conventions

33 Communicating with People Load and store bytes lb $t0,0($sp) sb $t0,0($gp) Example: String Copy Procedure Unicode: 16 bits to represent a character  load halfwords lh $t0,0($sp) sh $t0,0($gp)

34 String Copy Procedure C version void strcpy char x[], char y[] { int i; i=0; while ((x[i]=y[i])!=‘\0’ i+=1; }

35 String Copy Procedure (MIPS) strcpy: addi$sp,$sp,-4#adjust stack for 1 more item sw$s0,0($sp)#save $s0 add$s0,$zero,$zero #i=0+0 L1: add$t1,$s0,$a1# address of y[i] in $t1 lb$t2,0($t1)#$t2=y[i] add$t3,$s0,$a0 #address of x[i] in $t3 sb$t2,0($t3) #x[i]=y[i] beq$t2,$zero,L2 # if y[i]==0 goto L2 addi$s0,$s0,1#i=i+1 jL1# goto L1 L2:lw$s0,0($sp)#y[i]==0; end of string, restore #old $s0 addi$sp,$sp,4#pop 1 word off stack jr$ra#return

36 We'd like to be able to load a 32 bit constant into a register Must use two instructions, new "load upper immediate" instruction lui $t0, Then must get the lower order bits right, i.e., ori $t0, $t0, ori filled with zeros MIPS Addressing for 32-bit Immediates and Addresses

37 Assembly provides convenient symbolic representation much easier than writing down numbers e.g., destination first Machine language is the underlying reality e.g., destination is no longer first Assembly can provide 'pseudoinstructions' e.g., “move $t0, $t1” exists only in Assembly would be implemented using “add $t0,$t1,$zero” When considering performance you should count real instructions Assembly Language vs. Machine Language

38 Simple instructions, all 32 bits wide Very structured, no unnecessary baggage Only three instruction formats Rely on compiler to achieve performance — what are the compiler's goals? Help compiler where we can op rs rt rdshamtfunct op rs rt 16 bit address op 26 bit address RIJRIJ Overview of MIPS

39 Instructions: bne $t4,$t5,Label Next instruction is at Label if $t4!= $t5 beq $t4,$t5,Label Next instruction is at Label if $t4 = $t5 j Label Next instruction is at Label Formats: Addresses are not 32 bits — How do we handle this with load and store instructions? op rs rt 16 bit address op 26 bit address IJIJ Addressing in Branches and Jumps

40 Instructions: bne $t4,$t5,Label Next instruction is at Label if $t4!=$t5 beq $t4,$t5,Label Next instruction is at Label if $t4=$t5 Formats: Could specify a register (like lw and sw) and add it to address use Instruction Address Register (PC = program counter) most branches are local (principle of locality) Jump instructions just use high order bits of PC address boundaries of 256 MB op rs rt 16 bit address I Addresses in Branches

41 MIPS Addressing Modes

42 Decoding Machine Code P102 Example: 00af8020 hex add $s0,$a1,$t7

43 To summarize:

44 Design alternative: provide more powerful operations goal is to reduce number of instructions executed danger is a slower cycle time and/or a higher CPI Sometimes referred to as “RISC vs. CISC” virtually all new instruction sets since 1982 have been RISC VAX: minimize code size, make assembly language easy instructions from 1 to 54 bytes long! We’ll look at PowerPC and 80x86 Alternative Architectures

45 PowerPC Indexed addressing example: lw $t1,$a0+$s3 #$t1=Memory[$a0+$s3] What do we have to do in MIPS? Update addressing update a register as part of load (for marching through arrays) example: lwu $t0,4($s3) #$t0=Memory[$s3+4];$s3=$s3+4 What do we have to do in MIPS? Others: load multiple/store multiple a special counter register “ bc Loop ” decrement counter, if not 0 goto loop

46 The Intel IA : The Intel 8086 is announced (16 bit architecture) 1980: The 8087 floating point coprocessor is added 1982: The increases address space to 24 bits, +instructions 1985: The extends to 32 bits, new addressing modes : The 80486, Pentium, Pentium Pro add a few instructions (mostly designed for higher performance) 1997: MMX is added 1999: Add 70 instructions labeled SSE (Streaming SIMD Extensions) 2001:Add 144 instructions labeled SSE2 2003: AMD : EM64T “ This history illustrates the impact of the “ golden handcuffs ” of compatibility “ adding new features as someone might add clothing to a packed bag ” “ an architecture that is difficult to explain and impossible to love ”

47 A dominant architecture: 80x86 See your textbook for a more detailed description Complexity: Instructions from 1 to 17 bytes long one operand must act as both a source and destination one operand can come from memory complex addressing modes e.g., “base or scaled index with 8 or 32 bit displacement” Saving grace: the most frequently used instructions are not too difficult to build compilers avoid the portions of the architecture that are slow “ what the 80x86 lacks in style is made up in quantity, making it beautiful from the right perspective ”

48 Instruction complexity is only one variable lower instruction count vs. higher CPI / lower clock rate Design Principles: simplicity favors regularity smaller is faster good design demands compromise make the common case fast Instruction set architecture a very important abstraction indeed! Summary


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