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1 ECE369 ECE369 Chapter 2. 2 ECE369 Instruction Set Architecture A very important abstraction –interface between hardware and low-level software –standardizes.

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Presentation on theme: "1 ECE369 ECE369 Chapter 2. 2 ECE369 Instruction Set Architecture A very important abstraction –interface between hardware and low-level software –standardizes."— Presentation transcript:

1 1 ECE369 ECE369 Chapter 2

2 2 ECE369 Instruction Set Architecture A very important abstraction –interface between hardware and low-level software –standardizes instructions, machine language bit patterns, etc. –advantage: different implementations of the same architecture –disadvantage: sometimes prevents using new innovations Modern instruction set architectures: –IA-32, PowerPC, MIPS, SPARC, ARM, and others

3 3 ECE369 The MIPS Instruction Set Used as the example throughout the book Stanford MIPS commercialized by MIPS Technologies (www.mips.com)www.mips.com Large share of embedded core market –Applications in consumer electronics, network/storage equipment, cameras, printers, … Typical of many modern ISAs –See MIPS Reference Data tear-out card, and Appendixes B and E

4 4 ECE369 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 (we’ll talk about registers in a bit) “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”

5 5 ECE369 MIPS arithmetic Design Principle: simplicity favors regularity. Of course this complicates some things... C code: a = b + c + d; MIPS code: add a, b, c add a, a, d Operands must be registers, only 32 registers provided Each register contains 32 bits Design Principle: smaller is faster. Why?

6 6 ECE369 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

7 7 ECE369 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 8 ECE369 Memory Organization 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? bits of data Registers hold 32 bits of data

9 9 ECE369 Instructions Load and store instructions Example: C code: A[12] = h + A[8]; # $s3 stores base address of A and $s2 stores h MIPS code: lw $t0, 32($s3) add $t0, $s2, $t0 sw $t0, 48($s3) Can refer to registers by name (e.g., $s2, $t2) instead of number Store word has destination last Remember arithmetic operands are registers, not memory! Can’t write: add 48($s3), $s2, 32($s3)

10 10 ECE369 Instructions Example: C code:g = h + A[i]; # $s3 stores base address of A and # g,h and i in $s1,$s2 and $s4 Add $t1,$s4,$s4 Add $t1,$t1,$t1 Add $t1,$t1,$s3 Lw $t0,0($t1) Add $s1,$s2,$t0 t1 = 2*i t1 = 4*i t1 = 4*i + s3 t0 = A[i] g = h + A[i]

11 11 ECE369 So far we’ve learned: 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 lw $s1, 100($s2)$s1 = Memory[$s2+100] sw $s1, 100($s2)Memory[$s2+100] = $s1

12 12 ECE369 Policy of Use Conventions Register 1 ($at) reserved for assembler, for operating system

13 13 ECE369 MIPS Format Instructions, like registers and words –are also 32 bits long –add $t1, $s1, $s2 –Registers: $t1=9, $s1=17, $s2=18 Instruction Format: op rs rt rdshamtfunct

14 14 ECE369 Consider the load-word and store-word instructions, –What would the regularity principle have us do? –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 Where's the compromise? Machine Language

15 15 ECE369 Summary A[300]=h+A[300] # $t1 = base address of A, $s2 stores h # use $t0 for temporary register Lw $t0,1200($t1) Add $t0, $s2, $t0 Sw $t0, 1200($t1) Op rs,rt,address 35,9,8,1200 Op,rs,rt,rd,shamt,funct 0,18,8,8,0,32 Op,rs,rt,address 43,9,8,1200

16 16 ECE369 Summary of Instructions We Have Seen So Far

17 17 ECE369 Summary of New Instructions

18 18 ECE369 Example swap(int* v, int k); { int temp; temp = v[k] v[k] = v[k+1]; v[k+1] = temp; } swap: sll $t0, $a1, 4 add $t0, $t0, $a0 lw $t1, 0($t0) lw $t2, 4($t0) sw $t2, 0($t0) sw $t1, 4($t0) jr $31

19 19 ECE369 Control Instructions

20 20 ECE369 Using If-Else $s0 = f $s1 = g $s2 = h $s3 = i $s4 = j $s5 = k Where is 0,1,2,3 stored?

21 21 ECE369 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: op rs rt 16 bit address I Addresses in Branches What if the “Label” is too far away (16 bit address is not enough)

22 22 ECE369 Instructions: bne $t4,$t5,Label if $t4 != $t5 beq $t4,$t5,Label if $t4 = $t5 j Label j Label Next instruction is at Label Formats: op rs rt 16 bit address op 26 bit address IJIJ Addresses in Branches and Jumps

23 23 ECE369 We have: beq, bne, what about Branch-if-less-than? If (a

24 24 ECE369 While Loop While (save[i] == k) # i, j and k correspond to registers i = i+j; # $s3, $s4 and $s5 # array base address at $s6 Loop: add $t1, $s3, $s3 add $t1, $t1, $t1 add $t1, $t1, $s6 lw $t0, 0($t1) bne $t0, $s5, Exit add $s3, $s3, $s4 j loop Exit:

25 25 ECE369 What does this code do?

26 26 ECE369 simple instructions all 32 bits wide very structured, no unnecessary baggage only three instruction formats op rs rt rdshamtfunct op rs rt 16 bit address op 26 bit address RIJRIJ Overview of MIPS

27 27 ECE369 Arrays vs. Pointers clear1( int array[ ], int size) { int i; for (i=0; i

28 28 ECE369 Clear1 clear1( int array[ ], int size) { int i; for (i=0; i

29 29 ECE369 Clear2, Version 2 clear2(int* array, int size) { int* p; for( p=&array[0]; p<&array[size]; p++) *p=0; } Array and size to registers $a0 and $a1 loop2: sw $zero,0($t0) # memory[p]=0 add $t0,$a0,$zero# p = address of array[0] addi $t0,$t0,4# p = p+4 add $t1,$a1,$a1# $t1 = size*2 add $t1,$t1,$t1# $t1 = size*4 Distance of last element add $t2,$a0,$t1# $t2 = address of array[size] slt $t3,$t0,$t2# $t3=(p<&array[size]) bne $t3,zero,loop2# if (p<&array[size]) go to loop2

30 30 ECE369 Array vs. Pointer loop2: sw $zero,0($t0) # memory[p]=0 add $t0,$a0,$zero# p = address of array[0] addi $t0,$t0,$4# p = p+4 add $t1,$a1,$a1# $t1 = size*2 add $t1,$t1,$t1# $t1 = size*4 add $t2,$a0,$t1# $t2 = address of array[size] slt $t3,$t0,$t2# $t3=(p<&array[size]) bne $t3,zero,loop2# if (p<&array[size]) go to loop2 add $t0,$zero,$zero # i=0, register $t0=0 loop1: add$t1,$t0,$t0 # $t1=i*2 add$t1,$t1,$t1# $t1=i*4 add$t2,$a0,$t1# $t2=address of array[i] sw$zero, 0($t2) # array[i]=0 addi$t0,$t0,1# i=i+1 slt$t3,$t0,$a1# $t3=(i

31 31 ECE369 Summary

32 32 ECE369 More reading: support for procedures linkers, loaders, memory layout stacks, frames, recursion manipulating strings and pointers interrupts and exceptions system calls and conventions Some of these we'll talk more about later We have already talked about compiler optimizations Other Issues

33 33 ECE369 Elaboration What if there are more than 4 parameters for a function call? Addressable via frame pointer References to variables in the stack have the same offset

34 34 ECE369 What is the Use of Frame Pointer? Variables local to procedure do not fit in registers !!!

35 35 ECE369 Nested Procedures, function_main(){ function_a(var_x); /* passes argument using $a0 */ : /* function is called with “jal” instruction */ return; } function_a(int size){ function_b(var_y); /* passes argument using $a0 */ : /* function is called with “jal” instruction */ return; } function_b(int count){ : return; } Resource Conflicts ???

36 36 ECE369 Stack Last-in-first-out queue Register # 29 reserved as stack pointer Points to most recently allocated address Grows from higher to lower address Subtracting $sp Adding data – Push Removing data – Pop

37 37 ECE369 Function Call and Stack Pointer jr

38 38 ECE369 Recursive Procedures Invoke Clones !!! int fact (int n) { if (n < 1 ) return ( 1 ); else return ( n * fact ( n-1 ) ); } “n” corresponds to $a0 Program starts with the label of the procedure “fact” How many registers do we need to save on the stack? Registers $a0 and $ra

39 39 ECE369 Factorial Code 200 fact:addi$sp, $sp, -8 #adjust stack for 2 items 204sw$ra, 4($sp) #save return address sw$a0, 0($sp) #save argument n slti$t0, $a0, 1 # is n<1? beq$t0, $zero, L1 # if not  go to L1 addi$v0, $zero, 1#return result addi $sp, $sp, 8#pop items off stack jr $ra#return to calling proc. L1: addi$a0, $a0, -1 #decrement n 236jalfact # call fact(n-1) 240lw $a0, 0($sp) # restore “n” lw$ra, 4($sp) # restore address addi$sp, $sp,8 # pop 2 items mult$v0,$a0,$v0 # return n*fact(n-1) jr$ra # return to caller : 100 fact(3) 104 add …. ra = 104 a0= 3 sp= 40 vo= int fact (int n) { if (n < 1 ) return ( 1 ); else return ( n * fact ( n-1 ) ); }

40 40 ECE369 Assembly to Hardware Example int i; int k = 0; for (i=0; i<3; i++){ k = k + i + 1; } k = k/2; add $t1,$zero,$zero # i=0, register $t1=0 k is stored in $t0; i is stored in $t1 3 stored in $t2;$t3 used as temp add $t0,$zero,$zero # k=0, register $t0=0 loop: add $t0,$t0,$t1 # k = k + i addi$t0,$t0,1# k = k + 1 addi$t2,$zero,3# $t2=3 addi$t1,$t1,1# i=i+1 slt$t3,$t1,$t2# $t3= (i<3) bne$t3,$zero,loop# if (i<3) go to loop srl$t0,$t0,1# k = k / 2

41 41 ECE369 Assembly to Hardware Example add $t1,$zero,$zero add $t0,$zero,$zero loop: add $t0,$t0,$t1 addi$t0,$t0,1 addi$t2,$zero,3 addi$t1,$t1,1 slt$t3,$t1,$t2 bne$t3,$zero,loop srl$t0,$t0,1 Instruction Types? R-Type I-Type op rs rt rdshamtfunct op rs rt 16 bit address RIRI

42 42 ECE369 How do we represent in machine language? add $t1,$zero,$zero add $t0,$zero,$zero loop: add $t0,$t0,$t1 addi$t0,$t0,1 addi$t2,$zero,3 addi$t1,$t1,1 slt$t3,$t1,$t2 bne$t3,$zero,loop srl$t0,$t0,2 op rs rt rdshamtfunct op rs rt 16 bit address RIRI 6 bits 5 bits 5 bits 5 bits 5 bits 6 bits _00000_00000_01000_00000_ _00000_00000_01001_00000_ op rs rtrd shamt funct _00000_01010_ $t0 is reg 8 $t1 is reg 9 $t2 is reg 10 $t3 is reg _01000_01001_01000_00000_ _01000_01000_ _01001_01001_ _01001_01010_01011_00000_ _00000_01000_01000_00001_ _00000_01011_ : 4: 8: 12: 16: 20: 24: 28: 32: PC+4+BR Addr - 5

43 43 ECE369 How do we represent in machine language? add $t1,$zero,$zero add $t0,$zero,$zero loop: add $t0,$t0,$t1 addi$t0,$t0,1 addi$t2,$zero,3 addi$t1,$t1,1 slt$t3,$t1,$t2 bne$t3,$zero,loop srl$t0,$t0,2 op rs rt rd shamt funct

44 44 ECE369 Representation in MIPS Datapath op rs rt rd shamt funct

45 45 ECE369 Big Picture

46 46 ECE369 Compiler

47 47 ECE369 Addressing Modes

48 48 ECE369 Our Goal add $t1, $s1, $s2 ($t1=9, $s1=17, $s2=18) – op rs rt rdshamtfunct

49 49 ECE369 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

50 50 ECE369 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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