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Chapter 2 Instructions: Language of the Computer CprE 381 Computer Organization and Assembly Level Programming, Fall 2013 Zhao Zhang Iowa State University Revised from original slides provided by MKP

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Chapter 2 — Instructions: Language of the Computer — 2 Review of Week 4 MIPS procedure/function call convention Leaf and non-leaf examples Clearing array example String copy example Other issues: Load 32-bit immediate Assembler, loader, and compiler effects §2.8 Supporting Procedures in Computer Hardware

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Announcements Exam 1 on Friday Oct. 4 Course review on Wednesday Oct. 2 HW4 is due on Sep. 27 HW5 will be due on Oct. 11 Do HW5 as exercise before Exam 1 No HW and quizzes next week Lab 2 demo is due this week and Lab 3 demo due next week Lab 4 starts next week, due in one week Chapter 1 — Computer Abstractions and Technology — 3

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Exam 1 Open book, open notes, calculator are allowed E-book reader is allowed Must be put in airplane mode Coverage Chapter 1, Computer Abstraction and Technology Chapter 2, Instructions: Language of the Computer Some contents from Appendix B MIPS floating-point instructions Chapter 1 — Computer Abstractions and Technology — 4

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Exam Question Types Short conceptual questions Calculation: speedup, power saving, CPI, etc. MIPS assembly programming Translate C statements to MIPS (arithmetic, load/store, branch and jump, others) Translate C functions to MIPS (call convention) Among others Suggestions: Review slides and textbook Review homework and quizzes Chapter 1 — Computer Abstractions and Technology — 5

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Overview for Week 5 Overview for Week 5, Sep. 23 - 27 Bubble sorting example It will be used in Mini-Projects Floating point instructions ARM and x86 instruction set overview Chapter 1 — Computer Abstractions and Technology — 6

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Classic Bubble Sorting Bubble sort: Swap two adjacent elements if they are out of order Pass the array n times, each time a largest element will float to the top Look at the first pass of five elements 1 st try: 5 3 8 2 7 => 3 5 8 2 7 2 nd try: 3 5 8 2 7 => 3 5 8 2 7 3 rd try: 3 5 8 2 7 => 3 5 2 8 7 4 th try: 3 5 2 7 8 => 3 5 2 7 8 Chapter 1 — Computer Abstractions and Technology — 7

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Classic Bubble Sorting Pass i only has to check for (n-i) swaps In each pass, an element may float up until it meets a larger element The sorted sub-array increments by one 1 st pass: 5 3 8 2 7 => 3 5 2 7 8 2 nd pass: 3 5 2 7 8 => 3 2 5 7 8 3 nd pass: 3 2 5 7 8 => 2 3 5 7 8 4 nd pass: 2 3 5 7 8 => 2 3 5 7 8 Chapter 1 — Computer Abstractions and Technology — 8

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Revised Bubble Sorting The textbook bubble-sort is optimized to reduce comparisons void sort (int v[], int n) { int i, j; for (i = 0; i < n; i++) { for (j = i – 1; j >= 0 && v[j] > v[j+1]; j--) swap(v, j); } Chapter 1 — Computer Abstractions and Technology — 9

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Revised Bubble Sorting The classic one let a largest element float to the top of the unsorted sub-array The revised one let an element float to its right place in the sorted sub-array 1 st pass: 5 3 8 2 7 => 3 5 8 2 7 2 nd pass: 3 5 8 2 7 => 3 5 8 2 7 3 nd pass: 3 5 8 2 7 => 2 3 5 8 7 4 nd pass: 2 3 5 8 7 => 2 3 5 7 8 Chapter 1 — Computer Abstractions and Technology — 10

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Chapter 2 — Instructions: Language of the Computer — 11 The Swap Function The swap function is a leaf function void swap(int v[], int k) { int temp; temp = v[k]; v[k] = v[k+1]; v[k+1] = temp; } v in $a0, k in $a1, temp in $t0 §2.13 A C Sort Example to Put It All Together

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Chapter 2 — Instructions: Language of the Computer — 12 The Swap Function swap: sll $t1, $a1, 2 # $t1 = k * 4 add $t1, $a0, $t1 # $t1 = v+(k*4) # (address of v[k]) lw $t0, 0($t1) # $t0 (temp) = v[k] lw $t2, 4($t1) # $t2 = v[k+1] sw $t2, 0($t1) # v[k] = $t2 (v[k+1]) sw $t0, 4($t1) # v[k+1] = $t0 (temp) jr $ra # return to calling routine

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The Sort Function for (i = 0; i < n; i++) { for (j = i – 1; j >= 0 && v[j] > v[j+1]; j--) swap(v, j); } Save $ra to stack, as it’s a non-leaf function Assign i and j to $s0 and $s1 They must be preserved when calling swap() Move v, n from $a0 and $a1 to $s2 and $s2 They must be preserved, too $a0 and $a1 are used when calling swap() We need a stack frame of 5 words or 20 bytes Chapter 1 — Computer Abstractions and Technology — 13

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Chapter 2 — Instructions: Language of the Computer — 14 sort: addi $sp,$sp, –20 # make room on stack for 5 registers sw $ra, 16($sp) # save $ra on stack sw $s3,12($sp) # save $s3 on stack sw $s2, 8($sp) # save $s2 on stack sw $s1, 4($sp) # save $s1 on stack sw $s0, 0($sp) # save $s0 on stack … # procedure body … exit1: lw $s0, 0($sp) # restore $s0 from stack lw $s1, 4($sp) # restore $s1 from stack lw $s2, 8($sp) # restore $s2 from stack lw $s3,12($sp) # restore $s3 from stack lw $ra,16($sp) # restore $ra from stack addi $sp,$sp, 20 # restore stack pointer jr $ra # return to calling routine Sort Prologue and Epilogue Entry: Get a frame, save $ra and $s3-$s0 Exit: Restore $s0-$s3 and $ra, free the frame

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Sort Function Body A new pseudo instruction move rd, rs is equivalent to add rd, rs, $zero Example move $s2, $a0 # $s2 = $zero move $s3, $a1 # $s3 = $a1 No use of pseudo assembly instructions in Exam 1 Chapter 1 — Computer Abstractions and Technology — 15

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Chapter 2 — Instructions: Language of the Computer — 16 Sort Function Body move $s2, $a0 # save $a0 into $s2 move $s3, $a1 # save $a1 into $s3 move $s0, $zero # i = 0 for1tst: slt $t0, $s0, $s3 # $t0 = 0 if $s0 ≥ $s3 (i ≥ n) beq $t0, $zero, exit1 # go to exit1 if $s0 ≥ $s3 (i ≥ n) addi $s1, $s0, –1 # j = i – 1 for2tst: slti $t0, $s1, 0 # $t0 = 1 if $s1 < 0 (j < 0) bne $t0, $zero, exit2 # go to exit2 if $s1 < 0 (j < 0) sll $t1, $s1, 2 # $t1 = j * 4 add $t2, $s2, $t1 # $t2 = v + (j * 4) lw $t3, 0($t2) # $t3 = v[j] lw $t4, 4($t2) # $t4 = v[j + 1] slt $t0, $t4, $t3 # $t0 = 0 if $t4 ≥ $t3 beq $t0, $zero, exit2 # go to exit2 if $t4 ≥ $t3 move $a0, $s2 # 1st param of swap is v (old $a0) move $a1, $s1 # 2nd param of swap is j jal swap # call swap procedure addi $s1, $s1, –1 # j –= 1 j for2tst # jump to test of inner loop exit2: addi $s0, $s0, 1 # i += 1 j for1tst # jump to test of outer loop Pass params & call Move params Inner loop Outer loop Inner loop Outer loop

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Sort Function Optimized Old version: void sort(int v[], int n) int i, j; for (i = 0; i < n; i++) { for (j = i – 1; j >= 0 && v[j] > v[j+1]; j--) swap(v, j); } New version: void sort(int v[], int n) { int *pi, *pj; for (pi = v; pi < &v[n]; pi++) for (pj = pj - 1; pj >= v && swap(pj); pj--) {} } Chapter 1 — Computer Abstractions and Technology — 17

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New Swap Function A more efficient swap function that reduces memory loads // swap two adjacent elements if they are // out of order. Return 1 if swapped, 0 // otherwise int swap(int *p) { if (p[0] > p[1]) { int tmp = p[0]; p[0] = p[1]; p[1] = tmp; return 1; } else return 0; } Chapter 1 — Computer Abstractions and Technology — 18

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New Swap Function A new swap function swap: lw $t0, 0($a0) # load p[0] lw $t1, 4($a0) # load p[1] slt $t2, $t1, $t0 # p[1] < p[0]? beq $t2, $zero, else sw $t1, 0($a0) # swap sw $t0, 4($a0) # swap addi $v0, $zero, 1 # $v0 = 1 jr $ra else: addi $v0, $zero, 0 # $v0 = 0 jr $ra Chapter 1 — Computer Abstractions and Technology — 19

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New Sort Function The sort() function optimized Register usage $s0: v $s1: &v[n] $s2: pi $s3: pj Need a frame of 5 words to save $ra and $s0-$s2 Chapter 1 — Computer Abstractions and Technology — 20

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Sort Prologue and Epilogue sort: addi $sp, $sp, -20 # frame of 5 words sw $ra, 16($sp) sw $s3, 12($sp) sw $s2, 8($sp) sw $s1, 4($sp) sw $s0, 0($sp) lw $s0, 0($sp) lw $s1, 4($sp) lw $s2, 8($sp) lw $s3, 12($sp) lw $ra, 16($sp) addi $sp, $sp, 20 # release frame jr $ra Chapter 1 — Computer Abstractions and Technology — 21 MIPS code for sort function body

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New Sort: Outer Loop for (pi = v; pi < &v[n]; pi++) for (pj = pj - 1; pj >= v && swap(pj); pj--) {} add $s0, $a0, $zero # $s0 = v sll $a1, $a1, 2 # $a1 = 4*n add $s1, $s0, $a1 # $s1 = &v[n] add $s2, $s0, $zero # pi = v j for1_tst for1_loop: addi $s2, $s2, 4 # pi++ for1_tst: slt $t0, $s2, $s1 # pi < &v[n]? bne $t0, $zero, for1_loop # yes? repeat Chapter 1 — Computer Abstractions and Technology — 22 C code for the inner loop MIPS code for the inner loop

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New Sort: Inner Loop for (pj = pi-1; pj >= v && swap(pj); pj--) {} addi $s3, $s2, -4 # pj = pi-1 j for2_tst for2_loop: addi $s3, $s3, -4 # pj-- for2_tst: slt $t0, $s3, $s0 # pj < v? bne $t0, $zero,for2_exit # yes? exit add $a0, $s3, $zero # $a0 = pj jal swap # swap(pj) bne $v0, $zero,for2_loop # ret 1? cont for2_exit: Chapter 1 — Computer Abstractions and Technology — 23

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Lab Mini-Projects You will use the sorting code to test your CPU design in the lab mini-projects Use the new sorting code The new code is more optimized It will simplify the debugging Chapter 1 — Computer Abstractions and Technology — 24

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FP Instructions in MIPS Reading: Textbook Ch. 3.5 and B-71 – B80 FP hardware is coprocessor 1 Adjunct processor that extends the ISA Separate FP registers 32 single-precision: $f0, $f1, … $f31 Paired for double-precision: $f0/$f1, $f2/$f3, … Release 2 of MIPS ISA supports 32 × 64-bit FP reg’s Chapter 3 — Arithmetic for Computers — 25

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FP Instructions in MIPS FP instructions operate only on FP registers Programs generally don’t do integer ops on FP data, or vice versa More registers with minimal code-size impact Chapter 1 — Computer Abstractions and Technology — 26

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FP Instructions in MIPS FP load and store instructions lwc1, ldc1, swc1, sdc1 e.g., ldc1 $f8, 32($sp) lwc1, swc1: Load/store single- precision ldc1, swc1: Load/store double- precision Chapter 1 — Computer Abstractions and Technology — 27

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Chapter 3 — Arithmetic for Computers — 28 FP Instructions in MIPS Single-precision arithmetic add.s, sub.s, mul.s, div.s e.g., add.s $f0, $f1, $f6 Double-precision arithmetic add.d, sub.d, mul.d, div.d e.g., mul.d $f4, $f4, $f6

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FP Instructions in MIPS Single- and double-precision comparison c.xx.s, c.xx.d (xx is eq, lt, le, …) Sets or clears FP condition-code bit e.g. c.lt.s $f3, $f4 Branch on FP condition code true or false bc1t, bc1f e.g., bc1t TargetLabel Chapter 1 — Computer Abstractions and Technology — 29

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MIPS Call Convention: FP The first two FP parameters in registers 1 st parameter in $f12 or $f12:$f13 A double-precision parameter takes two registers 2 nd FP parameter in $f14 or $f14:$f15 Extra parameters in stack $f0 stores single-precision FP return value $f0:$f1 stores double-precision FP return value $f0-$f19 are FP temporary registers $f20-$f31 are FP saved temporary registers Chapter 1 — Computer Abstractions and Technology — 30

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Chapter 3 — Arithmetic for Computers — 31 FP Example: °F to °C C code: float f2c (float fahr) { return ((5.0/9.0) * (fahr - 32.0)); } fahr in $f12, result in $f0 Assume literals in global memory space, e.g. const5 for 5.0 and const9 for 9.0 Can FP immediate be encoded in MIPS instructions?

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FP Example: °F to °C Compiled MIPS code: f2c: lwc1 $f16, const5($gp) lwc1 $f18, const9($gp) div.s $f16, $f16, $f18 lwc1 $f18, const32($gp) sub.s $f18, $f12, $f18 mul.s $f0, $f16, $f18 jr $ra Chapter 1 — Computer Abstractions and Technology — 32

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FP Example: Function Call extern float fahr, cel; cel = f2c(fahr); Assume fahr is at 100($gp), cel is at 104($gp) lwc1 $f12, 100($gp) # load 1 st para jal f2c swcl $f0, 104($gp); # save ret val Chapter 1 — Computer Abstractions and Technology — 33

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FP Example: Max double max(double x, double y) { return (x > y) ? x : y; } max: c.lt.d $f14, $f12 # y < x? bc1f else # if false, do else mov.d $f0, $f12 # $f0:$f1 = x jr $ra else: mov.d $f0, $f14 # $f0:$f1 = y jr $ra Chapter 1 — Computer Abstractions and Technology — 34

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FP Example: Max How to call max? Assume a, b, c at 100($gp), 108($gp), and 116($gp) extern double a, b, c; c = max(a, b); ldc1 $f12, 100($gp) # $f12:$f13 = a ldc1 $f14, 108($gp) # $f14:$f15 = b jal max sdc1 $f0, 116($gp) # c = $f0:$f1 Chapter 1 — Computer Abstractions and Technology — 35

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FP Example: Search Value int search(double X[], int size, double value) { for (int i = 0; i < size; i++) if (X[i] == value) return 1; return 0; } Note 1: There are integer and FP parameters, and the return value is integer Note 2: A real program may search a value in a range, e.g. [value - delta, value + delta] Chapter 1 — Computer Abstractions and Technology — 36

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FP Example: Search Value search: add $t0, $zero, $zero # i = 0 j for_cond for_loop: sll $t1, $t0, 3 # $t1 = 8*i add $t1, $a0, $t1 # $t1 = &X[i] lwc1 $f2, 0($t1) # $f2 = X[i] c.eq.d $f2, $f12 # X[i] == value? bc1f endif # if false, skip addi $v0, $zero, 1 # $v0 = 1 jr $ra # return endif: addi $t0, $t0, 1 # i++ for_cond: slt $t1, $t0, $a1 # i < size? bne $t1, $zero, for_loop # repeat if true add $v0, $zero, $zero # to return 0 jr $ra Chapter 1 — Computer Abstractions and Technology — 37

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Chapter 3 — Arithmetic for Computers — 38 FP Example: Array Multiplication X = X + Y × Z All 32 × 32 matrices, 64-bit double-precision elements C code: void mm (double x[][], double y[][], double z[][]) { int i, j, k; for (i = 0; i! = 32; i = i + 1) for (j = 0; j! = 32; j = j + 1) for (k = 0; k! = 32; k = k + 1) x[i][j] = x[i][j] + y[i][k] * z[k][j]; } Addresses of x, y, z in $a0, $a1, $a2, and i, j, k in $s0, $s1, $s2

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Chapter 3 — Arithmetic for Computers — 39 FP Example: Array Multiplication MIPS code: li $t1, 32 # $t1 = 32 (row size/loop end) li $s0, 0 # i = 0; initialize 1st for loop L1: li $s1, 0 # j = 0; restart 2nd for loop L2: li $s2, 0 # k = 0; restart 3rd for loop sll $t2, $s0, 5 # $t2 = i * 32 (size of row of x) addu $t2, $t2, $s1 # $t2 = i * size(row) + j sll $t2, $t2, 3 # $t2 = byte offset of [i][j] addu $t2, $a0, $t2 # $t2 = byte address of x[i][j] l.d $f4, 0($t2) # $f4 = 8 bytes of x[i][j] L3: sll $t0, $s2, 5 # $t0 = k * 32 (size of row of z) addu $t0, $t0, $s1 # $t0 = k * size(row) + j sll $t0, $t0, 3 # $t0 = byte offset of [k][j] addu $t0, $a2, $t0 # $t0 = byte address of z[k][j] l.d $f16, 0($t0) # $f16 = 8 bytes of z[k][j] …

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Chapter 3 — Arithmetic for Computers — 40 FP Example: Array Multiplication … sll $t0, $s0, 5 # $t0 = i*32 (size of row of y) addu $t0, $t0, $s2 # $t0 = i*size(row) + k sll $t0, $t0, 3 # $t0 = byte offset of [i][k] addu $t0, $a1, $t0 # $t0 = byte address of y[i][k] l.d $f18, 0($t0) # $f18 = 8 bytes of y[i][k] mul.d $f16, $f18, $f16 # $f16 = y[i][k] * z[k][j] add.d $f4, $f4, $f16 # f4=x[i][j] + y[i][k]*z[k][j] addiu $s2, $s2, 1 # $k k + 1 bne $s2, $t1, L3 # if (k != 32) go to L3 s.d $f4, 0($t2) # x[i][j] = $f4 addiu $s1, $s1, 1 # $j = j + 1 bne $s1, $t1, L2 # if (j != 32) go to L2 addiu $s0, $s0, 1 # $i = i + 1 bne $s0, $t1, L1 # if (i != 32) go to L1

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Chapter 2 — Instructions: Language of the Computer — 41 ARM & MIPS Similarities ARM: the most popular embedded core Similar basic set of instructions to MIPS §2.16 Real Stuff: ARM Instructions ARMMIPS Date announced1985 Instruction size32 bits Address space32-bit flat Data alignmentAligned Data addressing modes93 Registers15 × 32-bit31 × 32-bit Input/outputMemory mapped

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Chapter 2 — Instructions: Language of the Computer — 42 Compare and Branch in ARM Uses condition codes for result of an arithmetic/logical instruction Negative, zero, carry, overflow Compare instructions to set condition codes without keeping the result Each instruction can be conditional Top 4 bits of instruction word: condition value Can avoid branches over single instructions

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Chapter 2 — Instructions: Language of the Computer — 43 Instruction Encoding

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Chapter 2 — Instructions: Language of the Computer — 44 The Intel x86 ISA Evolution with backward compatibility 8080 (1974): 8-bit microprocessor Accumulator, plus 3 index-register pairs 8086 (1978): 16-bit extension to 8080 Complex instruction set (CISC) 8087 (1980): floating-point coprocessor Adds FP instructions and register stack 80286 (1982): 24-bit addresses, MMU Segmented memory mapping and protection 80386 (1985): 32-bit extension (now IA-32) Additional addressing modes and operations Paged memory mapping as well as segments §2.17 Real Stuff: x86 Instructions

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Chapter 2 — Instructions: Language of the Computer — 45 The Intel x86 ISA Further evolution… i486 (1989): pipelined, on-chip caches and FPU Compatible competitors: AMD, Cyrix, … Pentium (1993): superscalar, 64-bit datapath Later versions added MMX (Multi-Media eXtension) instructions The infamous FDIV bug Pentium Pro (1995), Pentium II (1997) New microarchitecture (see Colwell, The Pentium Chronicles) Pentium III (1999) Added SSE (Streaming SIMD Extensions) and associated registers Pentium 4 (2001) New microarchitecture Added SSE2 instructions

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Chapter 2 — Instructions: Language of the Computer — 46 The Intel x86 ISA And further… AMD64 (2003): extended architecture to 64 bits EM64T – Extended Memory 64 Technology (2004) AMD64 adopted by Intel (with refinements) Added SSE3 instructions Intel Core (2006) Added SSE4 instructions, virtual machine support AMD64 (announced 2007): SSE5 instructions Intel declined to follow, instead… Advanced Vector Extension (announced 2008) Longer SSE registers, more instructions If Intel didn’t extend with compatibility, its competitors would! Technical elegance ≠ market success

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Chapter 2 — Instructions: Language of the Computer — 47 Basic x86 Registers

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Chapter 2 — Instructions: Language of the Computer — 48 Basic x86 Addressing Modes Two operands per instruction Source/dest operandSecond source operand Register Immediate RegisterMemory Register MemoryImmediate Memory addressing modes Address in register Address = R base + displacement Address = R base + 2 scale × R index (scale = 0, 1, 2, or 3) Address = R base + 2 scale × R index + displacement

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Chapter 2 — Instructions: Language of the Computer — 49 x86 Instruction Encoding Variable length encoding Postfix bytes specify addressing mode Prefix bytes modify operation Operand length, repetition, locking, …

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Chapter 2 — Instructions: Language of the Computer — 50 Implementing IA-32 Complex instruction set makes implementation difficult Hardware translates instructions to simpler microoperations Simple instructions: 1–1 Complex instructions: 1–many Microengine similar to RISC Market share makes this economically viable Comparable performance to RISC Compilers avoid complex instructions

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Chapter 2 — Instructions: Language of the Computer — 51 Fallacies Powerful instruction higher performance Fewer instructions required But complex instructions are hard to implement May slow down all instructions, including simple ones Compilers are good at making fast code from simple instructions Use assembly code for high performance But modern compilers are better at dealing with modern processors More lines of code more errors and less productivity §2.18 Fallacies and Pitfalls

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Chapter 2 — Instructions: Language of the Computer — 52 Fallacies Backward compatibility instruction set doesn’t change But they do accrete more instructions x86 instruction set

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Chapter 2 — Instructions: Language of the Computer — 53 Pitfalls Sequential words are not at sequential addresses Increment by 4, not by 1! Keeping a pointer to an automatic variable after procedure returns e.g., passing pointer back via an argument Pointer becomes invalid when stack popped

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Chapter 2 — Instructions: Language of the Computer — 54 Concluding Remarks Design principles 1.Simplicity favors regularity 2.Smaller is faster 3.Make the common case fast 4.Good design demands good compromises Layers of software/hardware Compiler, assembler, hardware MIPS: typical of RISC ISAs c.f. x86 §2.19 Concluding Remarks

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Chapter 2 — Instructions: Language of the Computer — 55 Concluding Remarks Measure MIPS instruction executions in benchmark programs Consider making the common case fast Consider compromises Instruction classMIPS examplesSPEC2006 IntSPEC2006 FP Arithmetic add, sub, addi 16%48% Data transfer lw, sw, lb, lbu, lh, lhu, sb, lui 35%36% Logical and, or, nor, andi, ori, sll, srl 12%4% Cond. Branch beq, bne, slt, slti, sltiu 34%8% Jump j, jr, jal 2%0%

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