Based on slides from D. Patterson and www-inst.eecs.berkeley.edu/~cs152/

1. COM 249 – Computer Organization andAssembly LanguageChapter 3Arithmetic for Computers Based on slides from D. Patterson and www-inst.eecs.berkeley.edu/~cs152/ Modified by S. J. Fritz Spring 2009 (1)

2. operation a ALU 32 result 32 b 32 Arithmetic • Where we've been: • Performance (seconds, cycles, instructions) • Abstractions: Instruction Set Architecture Assembly Language and Machine Language • What's up ahead: • Implementing the Architecture Modified by S. J. Fritz Spring 2009 (2)

3. Arithmetic for Computers §3.1 Introduction • Operations on integers • Addition and subtraction • Multiplication and division • Dealing with overflow • Floating-point real numbers • Representation and operations Modified by S. J. Fritz Spring 2009 (3)

4. Numbers • Bits are just bits (no inherent meaning)- conventions define relationship between bits and numbers • Binary numbers (base 2)0000 0001 0010 0011 0100 0101 0110 0111 1000 1001... decimal: 0...2n-1 • Of course it gets more complicated:-numbers are finite (overflow)-fractions and real numbers-negative numbers-no MIPS subi instruction; addi can add a negative number Modified by S. J. Fritz Spring 2009 (4)

5. Numbers • How do we represent negative numbers?-which bit patterns will represent which numbers? • What is the largest number that can be represented in a computer? • How does hardware REALLY multiply or divide numbers? Modified by S. J. Fritz Spring 2009 (5)

6. Possible Representations • Sign Magnitude: One's Complement Two's Complement • 000 = +0 000 = +0 000 = +0 001 = +1 001 = +1 001 = +1 010 = +2 010 = +2 010 = +2 011 = +3 011 = +3 011 = +3 100 = -0 100 = -3 100 = -4 101 = -1 101 = -2 101 = -3 110 = -2 110 = -1 110 = -2 111 = -3 111 = -0 111 = -1 • Issues: balance, number of zeros, ease of operations • Which one is best? Why? Modified by S. J. Fritz Spring 2009 (6)

7. maxint minint MIPS • 32 bit signed numbers:0000 0000 0000 0000 0000 0000 0000 0000two = 0ten0000 0000 0000 0000 0000 0000 0000 0001two = + 1ten0000 0000 0000 0000 0000 0000 0000 0010two = + 2ten...0111 1111 1111 1111 1111 1111 1111 1110two = + 2,147,483,646ten0111 1111 1111 1111 1111 1111 1111 1111two = + 2,147,483,647ten1000 0000 0000 0000 0000 0000 0000 0000two = – 2,147,483,648ten1000 0000 0000 0000 0000 0000 0000 0001two = – 2,147,483,647ten1000 0000 0000 0000 0000 0000 0000 0010two = – 2,147,483,646ten...1111 1111 1111 1111 1111 1111 1111 1101two = – 3ten1111 1111 1111 1111 1111 1111 1111 1110two = – 2ten1111 1111 1111 1111 1111 1111 1111 1111two = – 1ten Modified by S. J. Fritz Spring 2009 (7)

8. Two's Complement Operations • Negating a two's complement number: invert all bits and add 1 • remember: “negate” and “invert” are quite different! • Example: 8 bit representation of 5 • Binary 0000 0101 ( 5 two) • Invert 1111 1010 • Add 1 + 1 • 1111 1011 ( -5 two) • Interpreting a two’s complement number • First bit is sign bit • 1 x 2-7 + 1 x26 + 1x 25+ 1x24 + 1x 23 + 0x 22 + 1x21 + 1x 20 • -128 + 64 + 32 + 16 + 8 + 0 + 2 + 1 = -5 • Adding a number and it’s twos complement equals zero! Modified by S. J. Fritz Spring 2009 (8)

9. Integer Addition §3.2 Addition and Subtraction • Example: 7 + 6 • Overflow if result out of range • Adding + and –operands, no overflow • Adding two + operands • Overflow if result sign is 1 • Adding two – operands • Overflow if result sign is 0 Modified by S. J. Fritz Spring 2009 (9)

10. Integer Subtraction • Add negation of second operand • Example: 7 – 6 = 7 + (–6) +7: 0000 0000 … 0000 0111–6: 1111 1111 … 1111 1010+1: 0000 0000 … 0000 0001 • Overflow if result out of range • Subtracting two + or two – operands, no overflow • Subtracting + from – operand • Overflow if result sign is 0 • Subtracting – from + operand • Overflow if result sign is 1 Modified by S. J. Fritz Spring 2009 (10)

11. Addition & Subtraction • Just like in grade school (carry / borrow 1s) 0111 0111 0110+ 0110 - 0110 - 0101 • Two's complement operations easy • subtraction using addition of negative numbers 0111 + 1010 • Overflow (result too large for finite computer word): • e.g., adding two n-bit numbers does not always yield an n-bit number 1111 + 0001note that overflow term is somewhat misleading, 10000 it does not mean a carry “overflowed” Modified by S. J. Fritz Spring 2009 (11)

12. Arithmetic Practice • Two’s Complement http://scholar.hw.ac.uk/site/computing/topic18.asp?outline= (explained) http://scholar.hw.ac.uk/site/computing/activity12.asp?outline (simulator) • Data Representation and Number Systems http://scholar.hw.ac.uk/site/computing/topic1.html • Subtraction http://scholar.hw.ac.uk/site/computing/activity14.html • Logical Operations and Operators http://scholar.hw.ac.uk/site/computing/topic35.html Modified by S. J. Fritz Spring 2009 (12)

13. Overflow • Overflow occurs when the result of an operation on n bits > n bits • Example: adding two 8 bit values 1001 1101 157 +1111 1011 +251 11001 1000 408 overflow! Modified by S. J. Fritz Spring 2009 (13)

14. Detecting Overflow • Overflow occurs when the result cannot be represented in the available hardware • No overflow when adding a positive and a negative number • No overflow when signs are the same for subtraction • Overflow occurs when the value affects the sign: • overflow when adding two positives yields a negative • or, adding two negatives gives a positive • or, subtract a negative from a positive and get a negative • or, subtract a positive from a negative and get a positive • Consider the operations A + B, and A – B (See table p. 226) • Can overflow occur if B is 0 ? • Can overflow occur if A is 0 ? Modified by S. J. Fritz Spring 2009 (14)

15. Effects of Overflow • An exception (interrupt) occurs • Control jumps to predefined address for exception • Interrupted address is saved for possible resumption • add, addi, sub cause exceptions on overflow but • addu, addiu, subu DO NOT cause exceptions • Details based on software system / language • example: flight control vs. homework assignment • Don't always want to detect overflow — new MIPS instructions: addu, addiu, subu note: addiu still sign-extends! note: sltu, sltiu for unsigned comparisons Modified by S. J. Fritz Spring 2009 (15)

16. Dealing with Overflow in MIPS • Some languages (e.g., C) ignore overflow • Use MIPS addu, addui, subu instructions • Other languages (e.g., Ada, Fortran) require raising an exception • Use MIPS add, addi, sub instructions • MIPS detects overflow with an exception or interrupt • On overflow, invoke exception handler • Address of instruction that overflowed is stored in a register • Saves address in exception program counter (EPC) register • Jumps to predefined exception handler address • mfc0 (move from coprocessor reg) instruction can retrieve EPC value, to return after corrective action Modified by S. J. Fritz Spring 2009 (16)

17. Signed and Unsigned • MIPS – two versions of set on less than comparison • for signed integers slt (set on less than) and slti (set on less than immediate) • for unsigned integers sltu(set on less than unsigned) and sltiu (set on less than immediate unsigned) slt $s1,$s2,$s3 means if ($s2 < $s3) $s1 = 1//compare two’s complement else $s1 =0 slti $s1,$s2, 100 means if ( $s2 < 100) $s1 = 1 //compare to constant else $s1 =0 sltu $s1,$s2,$s3 means if ($s2 < $s3) $s1 = 1 //compares unsigned else $s1 =0 sltiu $s1,$s2, 100 means if ($s2 < 100) $s1 =1 //compare to constant else $s1 =0 Modified by S. J. Fritz Spring 2009 (17)

18. Shortcuts • To negate two’s complement binary number • Invert each bit and add one to result • Based on based on observation that the sum of a number and the representation of its inverse = -1 • To convert a number with n bits to a number represented with more than n bits • Take the most significant bit (sign bit) from the shorter number, and replicate it to fill the new bits of the longer number. ( Copy the original bits into the right portion of the new word – called sign extension) Modified by S. J. Fritz Spring 2009 (18)

19. Arithmetic for Multimedia • Graphics systems originally used 8 bits for each of the three colors (RGB), plus 8 bits for the location of the pixel (picture element) • Sound required 16 bits for audio samples. • Microprocessors have special support so that bytes and half-words take up less space, when stored in memory • Graphics and audio perform simultaneous operations on vectors of data, using a partitioned adder. These are called SIMD ( Single Instruction, Multiple Data) Modified by S. J. Fritz Spring 2009 (19)

20. Arithmetic for Multimedia • Graphics and media processing operates on vectors of 8-bit and 16-bit data • Use 64-bit adder, with partitioned carry chain • Operate on 8×8-bit, 4×16-bit, or 2×32-bit vectors • SIMD (single-instruction, multiple-data) • Saturating operations – when a calculation overflows the result is set to the largest positive number, or most negative number • On overflow, result is largest representable value • Instead of 2s-complement modulo arithmetic • Like turning a volume knob • E.g., clipping in audio, saturation in video (See table page 228) Modified by S. J. Fritz Spring 2009 (20)

21. 1000 × 1001 1000 0000 0000 1000 1001000 Multiplication §3.3 Multiplication • Start with long-multiplication approach multiplicand multiplier product Modified by S. J. Fritz Spring 2009 (21)

22. Shift Registers If we analyze long multiplication in terms of how we do it in the decimal number system, we will see that it is a combination of SHIFTS and ADDS.

23. Shift Registers Example: 1910 x 510 1 0 0 1 1 = 19 x 1 0 1 = x5 1 0 0 1 1 0 0 0 0 0 0 1 0 0 1 1 0 0 1 0 1 1 1 1 1 = 9510 The multiplication is achieved by adding in the multiplicand every time there is a ‘1’ in the multiplier and adding in zero every time there is a ‘0’ in the multiplier, shifting up to next digit each time. Multiplicand Multiplier Product

24. Full Binary Multiplication • Multiplying two n-bit numbers together may possibly • generate a 2n-bit result. • So two registers are typically needed to hold the • result. • Consider the multiplication of 7 by 5 on a 4-bit computer

25. Full Binary Multiplication Add in here = 7 multiplicand 0111 Shift in here = 5 multiplier 0101 •Start at ALL zero’s... •Bit 0 is a 1 => Add in the Multiplier •Shift Across to the right. •Bit 1 is a 0 => Add in 0000 (zero) •Shift Across to the right •Bit 2 is a 1 => Add in the Multiplier •Shift Across to the right •Bit 3 is a 0 => Add in 0000 (zero) •Shift Across to the right. ANS=3510 0000 xxxx 0111 xxxx 0011 1xxx 0011 1xxx 0001 11xx 1000 11xx 0100 011x 0100 011x 0010 0011

26. Multiplication Hardware Initially 0 Modified by S. J. Fritz Spring 2009 (26)

27. Optimized Multiplier • Perform steps in parallel: add/shift • One cycle per partial-product addition • That’s ok, if frequency of multiplications is low Modified by S. J. Fritz Spring 2009 (27)

28. Faster Multiplier • Uses multiple adders • Cost/performance tradeoff • Can be pipelined • Several multiplication performed in parallel Modified by S. J. Fritz Spring 2009 (28)

29. MIPS Multiplication • Two 32-bit registers for product • HI: most-significant 32 bits • LO: least-significant 32-bits HI LO • Instructions • mult rs, rt / multu rs, rt • 64-bit product in HI/LO • mfhi rd / mflo rd • Move from HI/LO to rd • Can test HI value to see if product overflows 32 bits • mul rd, rs, rt • Least-significant 32 bits of product –> rd Modified by S. J. Fritz Spring 2009 (29)

30. Division §3.4 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 quotient dividend 1001 1000 1001010 -1000 10 101 1010 -1000 10 divisor remainder n-bit operands yield n-bitquotient and remainder Modified by S. J. Fritz Spring 2009 (30)

31. Division Hardware Initially divisor in left half Initially dividend Modified by S. J. Fritz Spring 2009 (31)

32. Optimized Divider • One cycle per partial-remainder subtraction • Looks a lot like a multiplier! • Same hardware can be used for both Modified by S. J. Fritz Spring 2009 (32)

33. Faster Division • Can’t use parallel hardware as in multiplier • Subtraction is conditional on sign of remainder • Faster dividers (e.g. SRT division) generate multiple quotient bits per step • Still require multiple steps Modified by S. J. Fritz Spring 2009 (33)

34. MIPS Division • Use HI/LO registers for result • HI: 32-bit remainder • LO: 32-bit quotient HI ( remainder) LO (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 Modified by S. J. Fritz Spring 2009 (34)

35. Floating Point §3.5 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 Modified by S. J. Fritz Spring 2009 (35)

36. 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) Modified by S. J. Fritz Spring 2009 (36)

37. IEEE Floating-Point Format single: 8 bitsdouble: 11 bits single: 23 bitsdouble: 52 bits • 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 Modified by S. J. Fritz Spring 2009 (37)

38. 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 Modified by S. J. Fritz Spring 2009 (38)

39. 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 Modified by S. J. Fritz Spring 2009 (39)

40. 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 Modified by S. J. Fritz Spring 2009 (40)

41. 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 Modified by S. J. Fritz Spring 2009 (41)

42. Floating-Point Example • What number is represented by the single-precision float 11000000101000…00 • S = 1 • Fraction = 01000…002 • Fxponent = 100000012 = 129 • x = (–1)1 × (1 + 012) × 2(129 – 127) = (–1) × 1.25 × 22 = –5.0 Modified by S. J. Fritz Spring 2009 (42)

43. 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 Modified by S. J. Fritz Spring 2009 (45)

44. 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 Modified by S. J. Fritz Spring 2009 (46)

45. 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 Modified by S. J. Fritz Spring 2009 (47)

46. FP Adder Hardware Step 1 Step 2 Step 3 Step 4 Modified by S. J. Fritz Spring 2009 (48)

47. 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 Modified by S. J. Fritz Spring 2009 (51)

48. 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) Modified by S. J. Fritz Spring 2009 (52)

49. 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 Modified by S. J. Fritz Spring 2009 (53)

50. 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, literals in global memory space • Compiled MIPS code: f2c: lwc1 $f16, const5($gp) lwc2 $f18, const9($gp) div.s $f16, $f16, $f18 lwc1 $f18, const32($gp) sub.s $f18, $f12, $f18 mul.s $f0, $f16, $f18 jr $ra Modified by S. J. Fritz Spring 2009 (54)