Chapter 3

Arithmetic for Computers

CprE 381 Computer Organization and Assembly Level Programming, Fall 2013

Zhao ZhangIowa State UniversityRevised from original slides provided by MKP

Chapter 2 — Instructions: Language of the Computer — 2

Week 7 Overview Lecture:

Integer division IEEE floating point standard Floating-point arithmetic

Lab Mini-Project A: Arithmetic Unit 1-bit ALU 32-bit ALU Barrel shifter

Quiz resumes this week

§2.8 Supporting P

rocedures in Com

puter Hardw

are

Chapter 3 — Arithmetic for Computers — 3

Division Check for 0 divisor Long division approach

If divisor ≤ dividend bits 1 bit in quotient, subtract

Otherwise 0 bit in quotient, bring down next

dividend bit Restoring division

Do the subtract, and if remainder goes < 0, add divisor back

Signed division Divide using absolute values Adjust sign of quotient and remainder

as required

10011000 1001010 -1000 10 101 1010 -1000 10

n-bit operands yield n-bitquotient and remainder

quotient

dividend

remainder

divisor

§3.4 Division

Chapter 3 — Arithmetic for Computers — 4

Division Hardware

Initially dividend

Initially divisor in left half

Chapter 3 — Arithmetic for Computers — 5

Optimized Divider

One cycle per partial-remainder subtraction Looks a lot like a multiplier!

Same hardware can be used for both

Chapter 3 — Arithmetic for Computers — 6

Faster Division Can’t use parallel hardware as in multiplier

Subtraction is conditional on sign of remainder Faster dividers (e.g. SRT devision)

generate multiple quotient bits per step Still require multiple steps

Chapter 3 — Arithmetic for Computers — 7

MIPS Division Use HI/LO registers for result

HI: 32-bit remainder LO: 32-bit quotient

Instructions div rs, rt / divu rs, rt No overflow or divide-by-0 checking

Software must perform checks if required Use mfhi, mflo to access result

Chapter 3 — Arithmetic for Computers — 8

Floating Point Representation for non-integral numbers

Including very small and very large numbers Like scientific notation

–2.34 × 1056

+0.002 × 10–4

+987.02 × 109

In binary ±1.xxxxxxx2 × 2yyyy

Types float and double in C

normalized

not normalized

§3.5 Floating P

oint

Chapter 3 — Arithmetic for Computers — 9

Floating Point Standard Defined by IEEE Std 754-1985 Developed in response to divergence of

representations Portability issues for scientific code

Now almost universally adopted Two representations

Single precision (32-bit) Double precision (64-bit)

Chapter 3 — Arithmetic for Computers — 10

IEEE Floating-Point Format

S: sign bit (0 non-negative, 1 negative) Normalize significand: 1.0 ≤ |significand| < 2.0

Always has a leading pre-binary-point 1 bit, so no need to represent it explicitly (hidden bit)

Significand is Fraction with the “1.” restored Exponent: excess representation: actual exponent + Bias

Ensures exponent is unsigned Single: Bias = 127; Double: Bias = 1203

S Exponent Fraction

single: 8 bitsdouble: 11 bits

single: 23 bitsdouble: 52 bits

Bias)(ExponentS 2Fraction)(11)(x

Chapter 3 — Arithmetic for Computers — 11

Single-Precision Range Exponents 00000000 and 11111111 reserved Smallest value

Exponent: 00000001 actual exponent = 1 – 127 = –126

Fraction: 000…00 significand = 1.0 ±1.0 × 2–126 ≈ ±1.2 × 10–38

Largest value exponent: 11111110

actual exponent = 254 – 127 = +127 Fraction: 111…11 significand ≈ 2.0 ±2.0 × 2+127 ≈ ±3.4 × 10+38

Chapter 3 — Arithmetic for Computers — 12

Double-Precision Range Exponents 0000…00 and 1111…11 reserved Smallest value

Exponent: 00000000001 actual exponent = 1 – 1023 = –1022

Fraction: 000…00 significand = 1.0 ±1.0 × 2–1022 ≈ ±2.2 × 10–308

Largest value Exponent: 11111111110

actual exponent = 2046 – 1023 = +1023 Fraction: 111…11 significand ≈ 2.0 ±2.0 × 2+1023 ≈ ±1.8 × 10+308

Chapter 3 — Arithmetic for Computers — 13

Floating-Point Precision Relative precision

all fraction bits are significant Single: approx 2–23

Equivalent to 23 × log102 ≈ 23 × 0.3 ≈ 6 decimal digits of precision

Double: approx 2–52

Equivalent to 52 × log102 ≈ 52 × 0.3 ≈ 16 decimal digits of precision

Chapter 3 — Arithmetic for Computers — 14

Floating-Point Example Represent –0.75

–0.75 = (–1)1 × 1.12 × 2–1

S = 1 Fraction = 1000…002

Exponent = –1 + Bias Single: –1 + 127 = 126 = 011111102

Double: –1 + 1023 = 1022 = 011111111102

Single: 1011111101000…00 Double: 1011111111101000…00

Chapter 3 — Arithmetic for Computers — 15

Floating-Point Example What number is represented by the single-

precision float11000000101000…00 S = 1 Fraction = 01000…002

Fxponent = 100000012 = 129 x = (–1)1 × (1 + 012) × 2(129 – 127)

= (–1) × 1.25 × 22

= –5.0

Chapter 3 — Arithmetic for Computers — 16

Denormal Numbers Exponent = 000...0 hidden bit is 0

Smaller than normal numbers allow for gradual underflow, with

diminishing precision

Denormal with fraction = 000...0

Two representations of 0.0!

BiasS 2Fraction)(01)(x

0.0 BiasS 20)(01)(x

Chapter 3 — Arithmetic for Computers — 17

Infinities and NaNs Exponent = 111...1, Fraction = 000...0

±Infinity Can be used in subsequent calculations,

avoiding need for overflow check Exponent = 111...1, Fraction ≠ 000...0

Not-a-Number (NaN) Indicates illegal or undefined result

e.g., 0.0 / 0.0 Can be used in subsequent calculations

Chapter 3 — Arithmetic for Computers — 18

Floating-Point Addition Consider a 4-digit decimal example

9.999 × 101 + 1.610 × 10–1

1. Align decimal points Shift number with smaller exponent 9.999 × 101 + 0.016 × 101

2. Add significands 9.999 × 101 + 0.016 × 101 = 10.015 × 101

3. Normalize result & check for over/underflow 1.0015 × 102

4. Round and renormalize if necessary 1.002 × 102

Chapter 3 — Arithmetic for Computers — 19

Floating-Point Addition Now consider a 4-digit binary example

1.0002 × 2–1 + –1.1102 × 2–2 (0.5 + –0.4375) 1. Align binary points

Shift number with smaller exponent 1.0002 × 2–1 + –0.1112 × 2–1

2. Add significands 1.0002 × 2–1 + –0.1112 × 2–1 = 0.0012 × 2–1

3. Normalize result & check for over/underflow 1.0002 × 2–4, with no over/underflow

4. Round and renormalize if necessary 1.0002 × 2–4 (no change) = 0.0625

Chapter 3 — Arithmetic for Computers — 20

FP Adder Hardware Much more complex than integer adder Doing it in one clock cycle would take too

long Much longer than integer operations Slower clock would penalize all instructions

FP adder usually takes several cycles Can be pipelined

Chapter 3 — Arithmetic for Computers — 21

FP Adder Hardware

Step 1

Step 2

Step 3

Step 4

A B C

D E

FP Adder HardwareAssume inputs are X and Y, X.exp ≥ Y.exp Steps 1 and 2

The Small ALU output the exponent differnce Mux A selects X.exp, the bigger one Mux B selects Y.frac, which will be shifted to right Mux C selects X.frac The Big ALU add X.frac and shifted Y.frac Muxs D and E selects X.exp and Bit ALU output (added frac)

Steps 3 and 4, if active Muxes D and E selects the exponent and fraction from the

round hardware output The control unit generates control signals to all muxes

and other components

Chapter 3 — Arithmetic for Computers — 22

Chapter 3 — Arithmetic for Computers — 23

Floating-Point Multiplication Consider a 4-digit decimal example

1.110 × 1010 × 9.200 × 10–5

1. Add exponents For biased exponents, subtract bias from sum New exponent = 10 + –5 = 5

2. Multiply significands 1.110 × 9.200 = 10.212 10.212 × 105

3. Normalize result & check for over/underflow 1.0212 × 106

4. Round and renormalize if necessary 1.021 × 106

5. Determine sign of result from signs of operands +1.021 × 106

Chapter 3 — Arithmetic for Computers — 24

Floating-Point Multiplication Now consider a 4-digit binary example

1.0002 × 2–1 × –1.1102 × 2–2 (0.5 × –0.4375) 1. Add exponents

Unbiased: –1 + –2 = –3 Biased: (–1 + 127) + (–2 + 127) = –3 + 254 – 127 = –3 + 127

2. Multiply significands 1.0002 × 1.1102 = 1.1102 1.1102 × 2–3

3. Normalize result & check for over/underflow 1.1102 × 2–3 (no change) with no over/underflow

4. Round and renormalize if necessary 1.1102 × 2–3 (no change)

5. Determine sign: +ve × –ve –ve –1.1102 × 2–3 = –0.21875

Chapter 3 — Arithmetic for Computers — 25

FP Arithmetic Hardware FP multiplier is of similar complexity to FP

adder But uses a multiplier for significands instead of

an adder FP arithmetic hardware usually does

Addition, subtraction, multiplication, division, reciprocal, square-root

FP integer conversion Operations usually takes several cycles

Can be pipelined

Chapter 3 — Arithmetic for Computers — 26

FP Instructions in MIPS 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 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 FP load and store instructions

lwc1, ldc1, swc1, sdc1 e.g., ldc1 $f8, 32($sp)

Chapter 3 — Arithmetic for Computers — 27

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 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 3 — Arithmetic for Computers — 28

Accurate Arithmetic IEEE Std 754 specifies additional rounding

control Extra bits of precision (guard, round, sticky) Choice of rounding modes Allows programmer to fine-tune numerical behavior of

a computation Not all FP units implement all options

Most programming languages and FP libraries just use defaults

Trade-off between hardware complexity, performance, and market requirements

Chapter 3 — Arithmetic for Computers — 29

Associativity Parallel programs may interleave

operations in unexpected orders Assumptions of associativity may fail

§3.6 Parallelism

and Com

puter Arithm

etic: Associativity

(x+y)+z x+(y+z)x -1.50E+38 -1.50E+38y 1.50E+38z 1.0 1.0

1.00E+00 0.00E+00

0.00E+001.50E+38

Need to validate parallel programs under varying degrees of parallelism

Chapter 3 — Arithmetic for Computers — 30

x86 FP Architecture Originally based on 8087 FP coprocessor

8 × 80-bit extended-precision registers Used as a push-down stack Registers indexed from TOS: ST(0), ST(1), …

FP values are 32-bit or 64 in memory Converted on load/store of memory operand Integer operands can also be converted

on load/store Very difficult to generate and optimize code

Result: poor FP performance

§3.7 Real S

tuff: Floating P

oint in the x86

Chapter 3 — Arithmetic for Computers — 31

Streaming SIMD Extension 2 (SSE2)

Adds 4 × 128-bit registers Extended to 8 registers in AMD64/EM64T

Can be used for multiple FP operands 2 × 64-bit double precision 4 × 32-bit double precision Instructions operate on them simultaneously

Single-Instruction Multiple-Data

Chapter 3 — Arithmetic for Computers — 32

Right Shift and Division Left shift by i places multiplies an integer

by 2i

Right shift divides by 2i? Only for unsigned integers

For signed integers Arithmetic right shift: replicate the sign bit e.g., –5 / 4

111110112 >> 2 = 111111102 = –2 Rounds toward –∞

c.f. 111110112 >>> 2 = 001111102 = +62

§3.8 Fallacies and P

itfalls

Chapter 3 — Arithmetic for Computers — 33

Who Cares About FP Accuracy?

Important for scientific code But for everyday consumer use?

“My bank balance is out by 0.0002¢!”

The Intel Pentium FDIV bug The market expects accuracy See Colwell, The Pentium Chronicles

Chapter 3 — Arithmetic for Computers — 34

Concluding Remarks ISAs support arithmetic

Signed and unsigned integers Floating-point approximation to reals

Bounded range and precision Operations can overflow and underflow

MIPS ISA Core instructions: 54 most frequently used

100% of SPECINT, 97% of SPECFP Other instructions: less frequent

§3.9 Concluding R

emarks

Extra Slides

Chapter 1 — Computer Abstractions and Technology — 35

Chapter 3 — Arithmetic for Computers — 36

x86 FP Instructions

Optional variations I: integer operand P: pop operand from stack R: reverse operand order But not all combinations allowed

Data transfer Arithmetic Compare Transcendental

FILD mem/ST(i)FISTP mem/ST(i)FLDPIFLD1FLDZ

FIADDP mem/ST(i)FISUBRP mem/ST(i) FIMULP mem/ST(i) FIDIVRP mem/ST(i)FSQRTFABSFRNDINT

FICOMPFIUCOMPFSTSW AX/mem

FPATANF2XMIFCOSFPTANFPREMFPSINFYL2X

Exam 1 Review: Problem 1___ Performance is defined as 1/(Execution Time).

 

___ CPU Time = Instruction Count × CPI × Clock Rate

 

___ Processor dynamic power consumption is proportional to the voltage (of power supply)

 

___ Improving the algorithm of a program may reduce the clock cycles of the execution

 

___ Compiler optimization may reduce the number of instructions being executed

Chapter 1 — Computer Abstractions and Technology — 37

Problem 2

For a given benchmark program, a processor spends 40% of time in integer instructions (including branches and jumps) and 60% of time in floating-point instructions. Design team A may improve the integer performance by 2.0 times, and design team B may improve the floating-point performance by 1.5 times, for the same cost. Which design team may improve the overall performance more than the other?

Chapter 1 — Computer Abstractions and Technology — 38

Problem 2

Use Amdahl’s Law

Overall speedup = 1/(1-f+f/s)

 

A: f = 40%, s = 2.0,

speedup = 1/(0.6+0.4/2) = 1.25 times.

B: f = 60%, s = 1.5,

speedup = 1/(0.6/1.5+0.4) = 1.25 times.

The speedups will be the same, or neither team is better than the other.

Chapter 1 — Computer Abstractions and Technology — 39

Problem 3

f = f & 0x00FFFF00;

lw $t0, 200($gp) # load flui $t1, 0x00FFori $t1, 0xFF00and $t0, $t0, $t1sw $t0, 200($gp)

Chapter 1 — Computer Abstractions and Technology — 40

Problem 3

g = (g >> 16) | (g << 16);

lw $t0, 204($gp)srl $t1, $t0, 16sll $t2, $t0, 16or $t0, $t1, $t2sw $t0, 204($gp)

Chapter 1 — Computer Abstractions and Technology — 41

Problem 4

void swap(int X[], int i, int j){ if (i != j) { char int tmp1 = X[i]; char int tmp2 = X[j]; X[i] = tmp2; X[j] = tmp1; }}

Chapter 1 — Computer Abstractions and Technology — 42

Problem 4

This question tests MIPS call convention, leaf function, if statement, and array access. Call convention: X in $a0, i in $a1, j in $a2 Leaf function: Prefer t-regs for new values If statement: branch out if condition is false Array access: &X[i] = X + i*sizeof(X[i])

Three instructions: SLL, ADD, LW

Chapter 1 — Computer Abstractions and Technology — 43

Problem 4swap: beq $a1, $a2, endif # i == j, skip sll $t0, $a1, 2 # $t0 = i*4 add $t0, $a0, $t0 # $t0 = &X[i] lw $t1, 0($t0) # $t1 = X[i] sll $t2, $a2, 2 # $t2 = i*4 add $t2, $a0, $t2 # $t2 = &Y[i] lw $t3, 0($t2) # $t2 = Y[i] sw $t3, 0($t0) # X[i] = $t2 sw $t1, 0($t2) # Y[i] = $t1endif: jr $ra

Chapter 1 — Computer Abstractions and Technology — 44

Problem 4 What about this C code? How would a

compiler generate the code?void swap(int X[], int i, int j){ if (i != j) { char tmp1 = X[i]; char tmp2 = X[j]; X[i] = tmp2; X[j] = tmp1; }}

Chapter 1 — Computer Abstractions and Technology — 45

Problem 4swap: beq $a1, $a2, endif # i == j, skip sll $t0, $a1, 2 # $t0 = i*4 add $t0, $a0, $t0 # $t0 = &X[i] lb $t1, 0($t0) # $t1 = X[i] sll $t2, $a2, 2 # $t2 = i*4 add $t2, $a0, $t2 # $t2 = &Y[i] lb $t3, 0($t2) # $t2 = Y[i] sw $t3, 0($t0) # X[i] = $t2 sw $t1, 0($t2) # Y[i] = $t1endif: jr $ra

Chapter 1 — Computer Abstractions and Technology — 46

Problem 5// Prototype of the floor function extern float floor(float x); // Floor all elements of array X. N is the size of the array.void array_floor(float X[], int N){ for (int i = 0; i < N; i++) X[i] = floor(X[i]);}

Chapter 1 — Computer Abstractions and Technology — 47

Problem 5This question tests MIPS call convention, non-leaf function, FP part of the call convention, for loop, FP load/store, and function call. Call convention with FP

void array_floor(float X[], int N): X in $a0, N in $a1 float floor(float x): x in $f12, return value in $f0

Non-leaf function: Prefer s-regs for new values, must have a stack frame

For loop: Follow the for-loop template FP load/store: lwc1/swc1 for float type Function call for “X[i] = floor(X[i])”

Load X[i] in $f12, make the call, save $f0 back to X[i]

Chapter 1 — Computer Abstractions and Technology — 48

Problem 5Function prologue: Save $ra and all s-regs in use

$s0 for X, $s1 for N, $s2 for i, $s3 for &X[i] May write the function body first before prologue

array_floor: addi $sp, $sp, -20 sw $ra, 16($sp) sw $s3, 12($sp) sw $s2, 8($sp) sw $s1, 4($sp) sw $s0, 0($sp)

Chapter 1 — Computer Abstractions and Technology — 49

Problem 5 # Move X, N to $s0, $s1; clear i add $s0, $a0, $zero # s0 = X add $s1, $a1, $zero # s1 = N add $s2, $zero, $zero # i = 0 j for_cond

Chapter 1 — Computer Abstractions and Technology — 50

Problem 5for_loop: sll $s3, $s2, 2 # $s3=4*i add $s3, $s0, $s3 # $s0=&X[i] lwc1 $f12, 0($s3) # $f12=X[i] jal floor # floor(X[i]) swc1 $f0, 0($s3) # save X[i] addi $s2, $s2, 1 # i++for_cond: slt $t0, $s2, $s1 # i<N

bne $t0, $zero, for_loop

Chapter 1 — Computer Abstractions and Technology — 51

Problem 5 # function epilogue, restore $s0-$s3 # and $ra and $sp then returnret: lw $s0, 0($sp) lw $s1, 4($sp) lw $s2, 8($sp) lw $s3, 12($sp) lw $ra, 16($sp) addi $sp, $sp, 20 jr $ra

Chapter 1 — Computer Abstractions and Technology — 52

In Class Exercise What single-precision FP number does 0xC0F00000 represent?

Procedure1. Covert it to 32-bit binary

2. Split to 1-bit sign, 8-bit exponent, 23-bit fraction

3. Get the actual exponent

4. Get the significand

5. Put everything together

Chapter 1 — Computer Abstractions and Technology — 53

Covert Hex to FP Binary: 1100 0000 1111 0000 0000 0000 0000 Split the bit fields

Sign bit: 1 Exponent: 10000001 => 12910

Fraction: 11100…0

Actual exponent = 129 – 127 = 2 Significand = 1.11100…02=1.1112

Putting together: -1.112×22 = 111.12

= -7.510

Chapter 1 — Computer Abstractions and Technology — 54

Convert FP to Hex What is the binary represent of single-

precision FP number of +4.2? Procedure

1. Convert it to binary floating point number: Sign, significand, exponent

2. Get the biased component, binary form

3. Get the fraction, binary form

4. Put every thing together

Chapter 1 — Computer Abstractions and Technology — 55

Convert FP to Hex Binary floating point: +4.210

4.0 = 1002

0.2 = ???

Chapter 1 — Computer Abstractions and Technology — 56

Convert FP to Hex Use a multiplication table

0.2 x 2 = 0.4 0

0.4 x 2 = 0.8 0

0.8 x 2 = 1.6 1

0.6 x 2 = 1.2 1

0.2 x 2 = 0.4 0

0.4 x 2 = 0.8 0

0.8 x 2 = 1.6 1

0.6 x 2 = 1.2 1

Repeat …

0.210 = 0.0011001100110011…2

Note 0.2 = 1/8+1/16+1/128+1/256+…Chapter 1 — Computer Abstractions and Technology — 57

Convert FP to Hex Binary floating point: +4.210=+100.00110011…2

Normalize: +4.210=+1.0000110011…2×22

Get the components Sign bit = 0 Actual exponent = 210

Significand = 1.0000110011…2

Biased exponent = 2+127 = 129 = 100_0000_12

Fraction = 0000110011…2 = 000_0110_0110_0110_0110_01102 (23-bit)

Putting together: 0100_0000_1000_0110_0110_0110_0110_0110= 0x40866666

Chapter 1 — Computer Abstractions and Technology — 58

Why Study It If it sounds tedious… The Pentium FDIV bug (1994)

Intel used look-up table to speed up FP division, a few entries in the table were missing

A professor of math noticed the bug, and proved it Intel initially said it was not serious (the probability

of error is one in every 9 billion FDIVs) But scientific programs execute billions and trillions

of FP ops! It costs Intel $475 million to replace those

processors

http://en.wikipedia.org/wiki/Pentium_FDIV_bugChapter 1 — Computer Abstractions and Technology — 59