1. COMPUTER ARCHITECTURE & OPERATIONS I Instructor: Hao Ji

2. Review Last Class Handling Character Data, Starting and Loading a Program, Linking, Dynamic Linking, and Multiplication. This Class Division Floating Point Numbers Floating Point Operations Quiz Final Exam June 24, 2015

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 subtraction, and if remainder goes < 0, add divisor back Signed division Divide using absolute values Adjust sign of quotient and remainder as required 1001 1000 1001010 -1000 1 10 101 1010 -1000 10 n-bit operands yield n-bit quotient and remainder quotient dividend remainder divisor

4. Division Hardware Initially dividend Initially divisor in left half First version of Division Hardware 32-bit quotient 64-bit divisor 64-bit ALU 64-bit remainder

5. First Version Division Hardware Initially dividend Initially divisor in left half

6. Example 7/2

7. Optimized Divider ALU and Divisor registers are halved The remainder is shifted left. Quotient is in the right half of the reminder register.

8. Faster Division Can’t use parallel hardware as in multiplier Subtraction is conditional on sign of remainder Faster dividers (e.g. SRT devision) Predict several quotient bits per step Use a table lookup based on the upper bits of the dividend and remainder.

9. 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

10. Example Revisit A better way to do 7/2? 7>>2 SRL Most compiler will replace divide by power of 2 using right shift operations

11. Time for a Break (10 mins)

12. Scientific Notation

13. Floating Point Normalized Number Scientific notation without leading 0s Like scientific notation –2.34 × 10 56 +0.002 × 10 –4 +987.02 × 10 9 normalized not normalized §3.5 Floating Point

14. Floating Point Binary numbers in scientific notation ±1.xxxxxxx 2 × 2 yyyy (binary point) Floating Point Numbers Is used to support such number in computer. Computer arithmetic that represents numbers in which the binary point is not fixed Types float and double in C §3.5 Floating Point

15. Floating-Point Representation Two components Fraction (mantissa) between 0 and 1 precision of the floating point number Exponent numerical value range of the floating point number Representation of floating point Determine the sizes of fraction and exponent Tradeoff between range and precision SExponentFraction

16. Terms in Floating Point Representation Overflow A positive exponent becomes too large to fit in the exponent field Underflow A negative exponent becomes too large to fit in the exponent field Double precision A floating-point value represented in 64 bits Single precision A floating-point value represented in 32 bits

17. Floating Point Standard Defined by IEEE Std 754-1985, now 2008 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)

18. IEEE 754 Encoding

19. 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 = 1023 SExponentFraction single: 8 bits double: 11 bits single: 23 bits double: 52 bits

20. Floating-Point Example Represent –0.375 –0.375 = -3/8=-3/2 3 =(–1) 1 × 11 2 × 2 –3 Normalization=(–1) 1 × 1.1 2 × 2 –2 S = 1 Fraction = 1000…00 2 Exponent = –2 + Bias Single: –2 + 127 = 125 = 01111101 2 Double: –2 + 1023 = 1021 = 01111111101 2

21. Floating Point Example Single Precision Double Precision 3131 3030 2929 2828 2727 2626 2525 2424 23232 2121 2020 1919 1818 1717 1616 1515 1414 1313 12121 1010 9876543210 10111110110000000000000000000000 SExponent (8-bit)Fraction (23 bit) 6363 6262 6161 6060 5959 5858 5757 56565 5454 5353 5252 5151 5050 4949 4848 4747 4646 45454 4343 4242 4141 4040 3939 3838 3737 3636 3535 34343 3232 10111111110110000000000000000000 SExponent (11-bit)Fraction (52 bit) 3131 3030 2929 2828 2727 2626 2525 2424 23232 2121 2020 1919 1818 1717 1616 1515 1414 1313 12121 1010 9876543210 00000000000000000000000000000000 Fraction (52 bits)

22. Floating-Point Example What number is represented by the single- precision float S = 1 Fraction = 01000…00 2 Exponent = 10000001 2 = 129 x = (–1) 1 × (1 + 0.01 2 ) × 2 (129 – 127) = (–1) × 1.25 × 2 2 = –5.0 3131 3030 2929 2828 2727 2626 2525 2424 23232 2121 2020 1919 1818 1717 1616 1515 1414 1313 12121 1010 9876543210 11000000101000000000000000000000 SExponent (8-bit)Fraction (23 bit)

23. 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

24. 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

25. Floating-Point Precision Relative precision all fraction bits are significant Single: approx 2 –23 Equivalent to 23 × log 10 2 ≈ 23 × 0.3 ≈ 6 decimal digits of precision Double: approx 2 –52 Equivalent to 52 × log 10 2 ≈ 52 × 0.3 ≈ 16 decimal digits of precision

26. 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!

27. 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

28. Time for a Break (10 mins)

29. Review Last Session Exponent Fraction Std 754-1985 Single Precision Double Precision This Session Floating Point Operations Quiz

30. Floating-Point Addition Consider a 4-digit decimal example 9.999 × 10 1 + 1.610 × 10 –1 1. Align decimal points Shift number with smaller exponent 9.999 × 10 1 + 0.016 × 10 1 2. Add significands 9.999 × 10 1 + 0.016 × 10 1 = 10.015 × 10 1 3. Normalize result & check for over/underflow 1.0015 × 10 2 4. Round and renormalize if necessary 1.002 × 10 2

31. Floating-Point Addition Now consider a 4-digit binary example 1.000 2 × 2 –1 + –1.110 2 × 2 –2 (0.5 + –0.4375) 1. Align binary points Shift number with smaller exponent 1.000 2 × 2 –1 + –0.111 2 × 2 –1 2. Add significands 1.000 2 × 2 –1 + –0.111 2 × 2 –1 = 0.001 2 × 2 –1 3. Normalize result & check for over/underflow 1.000 2 × 2 –4, with no over/underflow 4. Round and renormalize if necessary 1.000 2 × 2 –4 (no change) = 0.0625

32. Floating-Point Multiplication Consider a 4-digit decimal example 1.110 × 10 10 × 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 × 10 5 3. Normalize result & check for over/underflow 1.0212 × 10 6 4. Round and renormalize if necessary 1.021 × 10 6 5. Determine sign of result from signs of operands +1.021 × 10 6

33. Floating-Point Multiplication Now consider a 4-digit binary example 1.000 2 × 2 –1 × –1.110 2 × 2 –2 (0.5 × –0.4375) 1. Add exponents –1 + –2 = –3 Biased: –3 + 127 2. Multiply significands 1.000 2 × 1.110 2 = 1.110 2  1.110 2 × 2 –3 3. Normalize result & check for over/underflow 1.110 2 × 2 –3 (no change) with no over/underflow 4. Round and renormalize if necessary 1.110 2 × 2 –3 (no change) 5. Determine sign: +value × –value  –ve –1.110 2 × 2 –3 = –0.21875

34. 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

35. 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) 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

36. Summary Floating Pointer Operations Addition Multiplication Floating point processor MIPS floating point operations Next Class Final Review

37. What I want you to do Review Chapters 2 and 3 Work on your assignment 3 and 4 Prepare for your final