Transcript Document
William Stallings Computer Organization and Architecture 6th Edition Chapter 9 Computer Arithmetic Arithmetic & Logic Unit • Performs arithmetic and logical operations on data. • Everything else in the computer is there to service this unit • Handles integers • May handle floating point (real) numbers • May be separate FPU (maths co-processor) • May be on chip separate FPU (486DX +) ALU Inputs and Outputs • Data are presented to the ALU in registers, and the results of an operation are stored in registers. • The ALU may also set flags as the result of an operation. • The control unit provides signals that control the operation of the ALU and the movement of the data into and out of the ALU. Integer Representation • Only have 0 & 1 to represent everything • Positive numbers stored in binary —e.g. 41=00101001 • No minus sign • No period • Negative number representation —Sign-Magnitude —Two’s complement Sign-Magnitude • Left most bit is sign bit —0 means positive —1 means negative • +18 = 00010010 • -18 = 10010010 • Problems —Need to consider both sign and magnitude in arithmetic —Two representations of zero (+0 and -0) Two’s Complement • Two’s complement — Take the Boolean complement of each bit of the integer. i.e., set each 1 to 0 and each 0 to 1. — Treating the result as an unsigned binary integer, add 1. — Example: – 3 = 00000011 – Boolean complement gives – Add 1 to LSB — +3 = 00000011 — +2 = 00000010 — +1 = 00000001 — +0 = 00000000 — -1 = 11111111 — -2 = 11111110 — -3 = 11111101 11111100 11111101 Benefits • One representation of zero • Arithmetic works easily (see later) • Negating is fairly easy Geometric Depiction of Twos Complement Integers 若加上k，則往順時針方 若減去k，則往逆時針方 向移動k個位置 向移動k個位置 Negation Special Case 1 • • • • • • • 0= 00000000 Bitwise not 11111111 Add 1 to LSB +1 Result 1 00000000 Overflow is ignored, so: -0=0 Overflow —When the result is larger than can be held in the word size being used. Negation Special Case 2 • • • • • • -128 = -10000000 bitwise not 01111111 Add 1 to LSB +1 Result 10000000 So: -(-128) = -128 X Range of Numbers • 8 bit 2s complement —+127 = 01111111 = 27 -1 — -128 = 10000000 = -27 • 16 bit 2s complement —+32767 = 011111111 11111111 = 215 - 1 — -32768 = 100000000 00000000 = -215 • N bits 2s complement —+(2n-1-1) —-(2n-1) Conversion Between Lengths • Positive number pack with leading zeros —+18 = 00010010 —+18 = 00000000 00010010 • Negative numbers pack with leading ones —-18 = 10010010 —-18 = 11111111 10010010 • i.e. pack with MSB (sign bit) —Sign extension Addition and Subtraction • Normal binary addition • Monitor sign bit for overflow • Take twos compliment of subtrahend and add to minuend —i.e. a - b = a + (-b) • So we only need addition and complement circuits Addition of Numbers in Twos Complement Representation Subtraction of Numbers in Twos Complement Representation (M-S) Subtraction of Numbers in Twos Complement Representation (M-S) Overflow • Overflow — The result of addition is larger than can be held in the word size. • 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 • Effects of Overflow — An exception (interrupt) occurs – Control jumps to predefined address for exception Hardware for Addition and Subtraction Multiplication • • • • Complex Work out partial product for each digit Take care with place value (column) Add partial products Multiplication Example • 1011 Multiplicand (11 dec) • x 1101 Multiplier (13 dec) • 1011 Partial products • 0000 Note: if multiplier bit is 1 copy • 1011 multiplicand (place value) • 1011 otherwise zero • 10001111 Product (143 dec) • Note: need double length result Unsigned Binary Multiplication Execution of Example Flowchart for Unsigned Binary Multiplication Multiplying Negative Numbers • Solution 1 —Convert to positive if required —Multiply as above —If signs were different, negate answer • Solution 2 —Booth’s algorithm Comparison of Multiplication of Unsigned and Twos Complement Integer 部分積以2n bit表示 可解決被乘數為負值的 問題，但若乘數為負值 時，仍無解 Booth’s Algorithm Example of Booth’s Algorithm Examples Using Booth’s Algorithm Examples Using Booth’s Algorithm (cont.) Division • More complex than multiplication • Negative numbers are really bad! • Based on long division Division of Unsigned Binary Integers 00001101 Divisor 1011 10010011 1011 001110 Partial 1011 Remainders 001111 1011 100 Quotient Dividend Remainder Flowchart for Unsigned Binary Division Real Numbers • Numbers with fractions • Could be done in pure binary — 1001.1010 = 23 + 20 +2-1 + 2-3 =9.625 • Where is the binary point? — Fixed-point notation – Very limited + Very large numbers cannot be represented, nor can very small fractions. — Floating-point notation – Scientific notation E SB – – – – Sign: plus or minus Significand S Exponent E Base B Sign bit Floating Point Biased Exponent Significand or Mantissa • +/- .significand x 2exponent • Point is actually fixed between sign bit and body of mantissa — There is one bit to the left of the radix point. • Exponent value is biased representation — A fixed value is subtracted from the field to get the true exponent value — The bias equals (2k-1-1) Floating Point Examples Signs for Floating Point • Exponent is in excess or biased notation —e.g. Excess (bias) 127 means —8 bit exponent field —Pure value range 0-255 —Subtract 127 to get correct value —Range -127 to +128 • Significand also called the mantissa Normalization • FP numbers are usually normalized — i.e. exponent is adjusted so that leading bit (MSB) of mantissa is 1 — Since it is always 1 there is no need to store it — e.g. 3.123 x 103 — The normalized nonzero number is one in the form 1.bbb...b 2 E — See Fig. 9.18b – The sign is stored in the first bit of the word – The first bit of the true significand is always 1 and need not be stored in the significand field. – The value 127 is added to the true exponent to be stored in the exponent field. – The base is 2. FP Ranges • For a 32 bit number — Using 2s complement integer representation 231 ~ 231 1 — Using 8 bit exponent, 23 bit significand – Negative number –(2-2-23)*2128~-2-127 – Positive number: 2-127~(2-2-23)*2128 — Five regions on the number line are not included in these ranges: – – – – – Negative overflow Negative underflow Zero Positive underflow Positive overflow • Accuracy — The effect of changing lsb of mantissa — 23 bit mantissa 2-23 1.2 x 10-7 — About 6 decimal places Expressible Numbers IEEE 754 • Standard for floating point storage —32 bit single precision —64 bit double precision • 8 and 11 bit exponent respectively • 23 and 52 bit significand respectively • The exponent is biased, —the rang of exponents is -126~+127，-1022~+1023 —exponent 與significand均為零時， 由 sign決定為 positive or negative zero. —Exponent均為1，significand為零時,由 sign決定為正 負無窮大 IEEE 754 Formats IEEE 754 format parameter Interpretation of IEEE 754 FP numbers Example 1 • (a) Convert the decimal number -1.5 to a 32-bit floating-point number with IEEE 754 format. (b) Convert a 32-bits IEEE 754 format floatingpoint number 43A0C000 to the decimal number. • Solution: • (a) 1.5=1.12=1.1*20 S=1, C=E+127=127=01111111, F=1000 0000 0000 0000 0000 000 (b)43A0C000=0100 0011 1010 0000 1100 0000 0000 00002 S=0, C=10000111=135=E+127, =>E=8, F= 0100 0001 1000 0000 0000 1.01000001102*28=101000001.12= 321.5 Example 2 • Convert the decimal number -1.5 to a 32-bit floating-point number with IBM S/390 format. • Solution: —Base of 16, 7 bit exponent, 24 bit significand —-1.5=-1.8(16)=-0.18*161 —S=1, C=E+64=1+64=65=100 0001 F=0001 1000 0000 0000 0000 0000 —The IBM S/390 format is 1 100 0001 0001 1000 0000 0000 0000 0000 Floating-point Arithmetic • A FP operation may produce one of these conditions: —Exponent overflow —Exponent underflow —Significand underflow —Significand overfolw FP numbers and arithmetic operations FP Arithmetic +/• • • • Check for zeros Align significands (adjusting exponents) Add or subtract significands Normalize result FP Addition & Subtraction Flowchart FP Arithmetic x/ • • • • • • Check for zero Add/subtract exponents Multiply/divide significands (watch sign) Normalize Round All intermediate results should be in double length storage Floating Point Multiplication Floating Point Division Precision Considerations • Guard bit —Used to pad out the right end of the significand with 0s. • Rounding —When the result is put back into the floating-point format, the extra bits must be disposed of. —IEEE standard lists four alternative approaches: – Round to nearest + 使用最接近結果的表示值(四捨五入) – Round toward +∞ – Round toward –∞ – Round toward 0 + 單純的將多餘的位元截斷。 IEEE standard for binary FP arithmetic • Infinity —The limiting case of real arithmetic. • Quiet and Signaling NaNs —Signaling NaN afford values for uninitialized variables and arithmetic-like enhancement. —Quiet NaN propagates through almost every arithmetic operation without signaling an exception. • Denormalized Numbers —To handle cases of exponent underflow.