Lecture 16: Computer Arithmetic Today’s topic –Floating point numbers –IEEE 754 representations –FP arithmetic Reminder –HW 4 due Monday 1

Floating Point Representation for non-integral numbers –Including very small and very large numbers Like scientific notation ––2.34 × – × 10 –4 – × 10 9 In binary –±1.xxxxxxx 2 × 2 yyyy Types float and double in C normalized not normalized 2

Floating Point Standard Defined by IEEE Std Now almost universally adopted Two representations –Single precision (32-bit) –Double precision (64-bit) A standard notation enables easy exchange of data between machines and simplifies hardware algorithms – the IEEE 754 standard defines how floating point numbers are represented 3

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 SExponentFraction single: 8 bits double: 11 bits single: 23 bits double: 52 bits 4

5 Single Precision Sign Exponent Fraction 1 bit 8 bits 23 bits SEF More exponent bits  wider range of numbers (not necessarily more numbers – recall there are infinite real numbers) More fraction bits  higher precision

Single-Precision Range Exponents and reserved Smallest value –Exponent:  actual exponent = 1 – 127 = –126 –Fraction: 000…00  significand = 1.0 –±1.0 × 2 –126 ≈ ±1.2 × 10 –38 Largest value –exponent:  actual exponent = 254 – 127 = +127 –Fraction: 111…11  significand ≈ 2.0 –±2.0 × ≈ ±3.4 ×

Double-Precision Range Exponents 0000…00 and 1111…11 reserved Smallest value –Exponent:  actual exponent = 1 – 1023 = –1022 –Fraction: 000…00  significand = 1.0 –±1.0 × 2 –1022 ≈ ±2.2 × 10 –308 Largest value –Exponent:  actual exponent = 2046 – 1023 = –Fraction: 111…11  significand ≈ 2.0 –±2.0 × ≈ ±1.8 ×

Floating-Point Example Represent –0.75 ––0.75 = (–1) 1 × × 2 –1 –S = 1 –Fraction = 1000…00 2 –Exponent = –1 + Bias Single: – = 126 = Double: – = 1022 = Single: …00 Double: …00 8

Floating-Point Example What number is represented by the single-precision float …00 –S = 1 –Fraction = 01000…00 2 –Fxponent = = 129 x = (–1) 1 × ( ) × 2 (129 – 127) = (–1) × 1.25 × 2 2 = –5.0 9

10 Details The number “0” has a special code so that the implicit 1 does not get added: the code is all 0s (it may seem that this takes up the representation for 1.0, but given how the exponent is represented, we’ll soon see that that’s not the case) The largest exponent value (with zero fraction) represents +/- infinity The largest exponent value (with non-zero fraction) represents NaN (not a number) – for the result of 0/0 or (infinity minus infinity)

11 More Details To simplify sort, sign was placed as the first bit For a similar reason, the representation of the exponent is also modified: in order to use integer compares, it would be preferable to have the smallest exponent as 00…0 and the largest exponent as 11…1 This is the biased notation, where a bias is subtracted from the exponent field to yield the true exponent IEEE 754 single-precision uses a bias of 127 (since the exponent must have values between -127 and 128)…double precision uses a bias of 1023 Final representation: (-1) S x (1 + Fraction) x 2 (Exponent – Bias)

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

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

Floating-Point Multiplication Consider a 4-digit decimal example –1.110 × × × 10 –5 1. Add exponents –For biased exponents, subtract bias from sum –New exponent = 10 + –5 = 5 2. Multiply significands –1.110 × =  × Normalize result & check for over/underflow – × Round and renormalize if necessary –1.021 × Determine sign of result from signs of operands – ×

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

16 MIPS Instructions The usual add.s, add.d, sub, mul, div Comparison instructions: c.eq.s, c.neq.s, c.lt.s…. These comparisons set an internal bit in hardware that is then inspected by branch instructions: bc1t, bc1f Separate register file $f0 - $f31 : a double-precision value is stored in (say) $f4-$f5 and is referred to by $f4 Load/store instructions (lwc1, swc1) must still use integer registers for address computation

17 Code Example float f2c (float fahr) { return ((5.0/9.0) * (fahr – 32.0)); } (argument fahr is stored in $f12) 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