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1 Systems Architecture Lecture 11: Arithmetic for Computers Jeremy R. Johnson Anatole D. Ruslanov William M. Mongan Some or all figures from Computer Organization and Design: The Hardware/Software Approach, Third Edition, by David Patterson and John Hennessy, are copyrighted material (COPYRIGHT 2004 MORGAN KAUFMANN PUBLISHERS, INC. ALL RIGHTS RESERVED). Lec 11 Systems Architecture 1
2 Introduction Objective: To learn about computer arithmetic and how number are represented in computers Review binary and hexadecimal numbers Negative numbers signed-magnitude two's complement Two's complement numbers Addition and subtraction Representation: a number A = n 1 i 0 i d Example for base 2 (binary): 1011 = 1 x x x x 2 0 i Lec 11 Systems Architecture 2
3 Numbers Bits have no inherent meaning conventions define relationship between bits and numbers Binary numbers (base 2) decimal: n -1 Complications: numbers are finite (overflow) fractions and real numbers negative numbers e.g., no MIPS subi instruction addi can add a negative number How do we represent negative numbers? Which bit patterns will represent which numbers? Lec 11 Systems Architecture 3
4 Possible Representations Sign Magnitude: One's Complement: Two's Complement: 000 = = = = = = = = = = = = = = = = = = = = = = = = -1 Issues: balance, number of zeros, ease of operations Which one is best? Why? Lec 11 Systems Architecture 4
5 The Complement Theory Radix-minus-one complement Radix-complement (the complement) Radix-complement is the inverse with respect to addition A 4-digit decimal example: The following equation holds when subtracting one number from another in FIXED decimal width: Y = B A + ( ) = B + ( A + 1) = B + ([9999 A] + 1) Observe: 9999 A is 9-complement (radix-minus-one complement) Observe: [9999 A] + 1 is 10-complement (THE radix complement) Lec 11 Systems Architecture 5
6 Why the Complement? Example No borrowing is necessary when subtracting Due to fixed width of the registers, the leading 1 is lost automatically due to carry overflow. Lec 11 Systems Architecture 6
7 Two's Complement Why does it work? Modular arithmetic (mod n): a b (mod n), i.e., a = b + qn [a] = { a + qn for all q in Z }, i.e., all number equivalent to a mod n. This forms a residue class ring Z n. What is a ring? It is a group where addition is as expected and addition has well-defined inverse. (This is not precise.) That is [a] + [b] = [a + b] and [a] x [b] = [a x b] Lec 11 Systems Architecture 7
8 Two's Complement Why it works? Example: Z 8 [0], [1], [2],, [7] is a residue class ring of 8 elements. For this example, we can use other representations: 0, 1, 2, 3, -4, -3, -2, -1 because these numbers are equivalent, e.g. 4-4 (mod 8), etc Binary addition in k bits is equivalent to addition in Z k 2 ring. In k-bits, the largest number will be = = 2 k-1-1 In k-bits, the smallest number will be = = -2 k-1 Representation: b k-1 b 0 = -b k-1 2 k-1 +b k-2 2 k-2 + b b Lec 11 Systems Architecture 8
9 Two's Complement Inversion Let a k-bit number X have a two s complement inverse 2 X with respect to Z k 2 ring, i.e., X + 2 X 0 (mod 2 k ). For instance, mod 8 0! Let 1 X be a one s complement of X (i.e., radix-minus-one complement, which is equivalent to flipping all the bits in binary). By definition, 2 X = 1 X + 1 and X + 1 X = 2 k 1 Therefore, X + 2 X = 2 k 0 mod 2 k Lec 11 Systems Architecture 9
10 MIPS signed numbers 32 bit signed numbers: two = 0 ten two = + 1 ten two = + 2 ten two = + 2,147,483,646 ten two = + 2,147,483,647 ten two = 2,147,483,648 ten two = 2,147,483,647 ten two = 2,147,483,646 ten two = 3 ten two = 2 ten two = 1 ten maxint minint Lec 11 Systems Architecture 10
11 Two's Complement Operations Negating a two's complement number: invert all bits and add 1 remember: negate and invert are quite different! Converting n bit numbers into numbers with more than n bits: MIPS 16 bit immediate gets converted to 32 bits for arithmetic Copy the most significant bit (the sign bit) into the other bits > > "sign extension" (lbu vs. lb) Lec 11 Systems Architecture 11
12 Two's Complement Sign Extension Sign extension involves propagating the sign bit to increase the bit-size of the number. Why this works? Number before sign extension: b k-1 b 0 = -b k-1 2 k-1 +b k-2 2 k-2 + b b Number after sign extension by one bit: b k b 0 = -b k 2 k +b k-1 2 k-1 +b k-2 2 k-2 + b b This is equivalent to adding 2 k-1 and subtracting 2 k, i.e., subtracting 2 k - 2 k-1 = 2 k-1 *(2-1) It is obvious why this works for positive numbers Lec 11 Systems Architecture 12
13 Addition & Subtraction Just like in grade school (carry/borrow 1s) Two's complement operations easy subtraction using addition of negative numbers Overflow (result too large for finite computer word): e.g., adding two n-bit numbers does not yield an n-bit number note that overflow term is somewhat misleading, 1000 it does not mean a carry overflowed Lec 11 Systems Architecture 13
14 Detecting Overflow 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 Can overflow occur if B is 0? Can overflow occur if A is 0? Lec 11 Systems Architecture 14
15 Effects of Overflow An exception (interrupt) occurs Control jumps to predefined address for exception Interrupted address is saved for possible resumption 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 Lec 11 Systems Architecture 15
Oct: 50 8 = 6 (r = 2) 6 8 = 0 (r = 6) Writing the remainders in reverse order we get: (50) 10 = (62) 8
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