3810-08.ppt
- 1. 1 Lecture 8: Binary Multiplication & Division • Today’s topics: Addition/Subtraction Multiplication Division • Reminder: get started early on assignment 3
- 2. 2 2’s Complement – Signed Numbers 0000 0000 0000 0000 0000 0000 0000 0000two = 0ten 0000 0000 0000 0000 0000 0000 0000 0001two = 1ten … 0111 1111 1111 1111 1111 1111 1111 1111two = 231-1 1000 0000 0000 0000 0000 0000 0000 0000two = -231 1000 0000 0000 0000 0000 0000 0000 0001two = -(231 – 1) 1000 0000 0000 0000 0000 0000 0000 0010two = -(231 – 2) … 1111 1111 1111 1111 1111 1111 1111 1110two = -2 1111 1111 1111 1111 1111 1111 1111 1111two = -1 Why is this representation favorable? Consider the sum of 1 and -2 …. we get -1 Consider the sum of 2 and -1 …. we get +1 This format can directly undergo addition without any conversions! Each number represents the quantity x31 -231 + x30 230 + x29 229 + … + x1 21 + x0 20
- 3. 3 Alternative Representations • The following two (intuitive) representations were discarded because they required additional conversion steps before arithmetic could be performed on the numbers sign-and-magnitude: the most significant bit represents +/- and the remaining bits express the magnitude one’s complement: -x is represented by inverting all the bits of x Both representations above suffer from two zeroes
- 4. 4 Addition and Subtraction • Addition is similar to decimal arithmetic • For subtraction, simply add the negative number – hence, subtract A-B involves negating B’s bits, adding 1 and A
- 5. 5 Overflows • For an unsigned number, overflow happens when the last carry (1) cannot be accommodated • For a signed number, overflow happens when the most significant bit is not the same as every bit to its left when the sum of two positive numbers is a negative result when the sum of two negative numbers is a positive result The sum of a positive and negative number will never overflow • MIPS allows addu and subu instructions that work with unsigned integers and never flag an overflow – to detect the overflow, other instructions will have to be executed
- 6. 6 Multiplication Example Multiplicand 1000ten Multiplier x 1001ten --------------- 1000 0000 0000 1000 ---------------- Product 1001000ten In every step • multiplicand is shifted • next bit of multiplier is examined (also a shifting step) • if this bit is 1, shifted multiplicand is added to the product
- 7. 7 HW Algorithm 1 In every step • multiplicand is shifted • next bit of multiplier is examined (also a shifting step) • if this bit is 1, shifted multiplicand is added to the product
- 8. 8 HW Algorithm 2 • 32-bit ALU and multiplicand is untouched • the sum keeps shifting right • at every step, number of bits in product + multiplier = 64, hence, they share a single 64-bit register
- 9. 9 Notes • The previous algorithm also works for signed numbers (negative numbers in 2’s complement form) • We can also convert negative numbers to positive, multiply the magnitudes, and convert to negative if signs disagree • The product of two 32-bit numbers can be a 64-bit number -- hence, in MIPS, the product is saved in two 32-bit registers
- 10. 10 MIPS Instructions mult $s2, $s3 computes the product and stores it in two “internal” registers that can be referred to as hi and lo mfhi $s0 moves the value in hi into $s0 mflo $s1 moves the value in lo into $s1 Similarly for multu
- 11. 11 Fast Algorithm • The previous algorithm requires a clock to ensure that the earlier addition has completed before shifting • This algorithm can quickly set up most inputs – it then has to wait for the result of each add to propagate down – faster because no clock is involved -- Note: high transistor cost
- 12. 12 Division 1001ten Quotient Divisor 1000ten | 1001010ten Dividend -1000 10 101 1010 -1000 10ten Remainder At every step, • shift divisor right and compare it with current dividend • if divisor is larger, shift 0 as the next bit of the quotient • if divisor is smaller, subtract to get new dividend and shift 1 as the next bit of the quotient
- 13. 13 Division 1001ten Quotient Divisor 1000ten | 1001010ten Dividend 0001001010 0001001010 0000001010 0000001010 100000000000 0001000000 00001000000000001000 Quo: 0 000001 0000010 000001001 At every step, • shift divisor right and compare it with current dividend • if divisor is larger, shift 0 as the next bit of the quotient • if divisor is smaller, subtract to get new dividend and shift 1 as the next bit of the quotient
- 14. 14 Divide Example • Divide 7ten (0000 0111two) by 2ten (0010two) Iter Step Quot Divisor Remainder 0 Initial values 1 2 3 4 5
- 15. 15 Divide Example • Divide 7ten (0000 0111two) by 2ten (0010two) Iter Step Quot Divisor Remainder 0 Initial values 0000 0010 0000 0000 0111 1 Rem = Rem – Div Rem < 0 +Div, shift 0 into Q Shift Div right 0000 0000 0000 0010 0000 0010 0000 0001 0000 1110 0111 0000 0111 0000 0111 2 Same steps as 1 0000 0000 0000 0001 0000 0001 0000 0000 1000 1111 0111 0000 0111 0000 0111 3 Same steps as 1 0000 0000 0100 0000 0111 4 Rem = Rem – Div Rem >= 0 shift 1 into Q Shift Div right 0000 0001 0001 0000 0100 0000 0100 0000 0010 0000 0011 0000 0011 0000 0011 5 Same steps as 4 0011 0000 0001 0000 0001
- 16. 16 Hardware for Division A comparison requires a subtract; the sign of the result is examined; if the result is negative, the divisor must be added back
- 18. 18 Divisions involving Negatives • Simplest solution: convert to positive and adjust sign later • Note that multiple solutions exist for the equation: Dividend = Quotient x Divisor + Remainder +7 div +2 Quo = Rem = -7 div +2 Quo = Rem = +7 div -2 Quo = Rem = -7 div -2 Quo = Rem =
- 19. 19 Divisions involving Negatives • Simplest solution: convert to positive and adjust sign later • Note that multiple solutions exist for the equation: Dividend = Quotient x Divisor + Remainder +7 div +2 Quo = +3 Rem = +1 -7 div +2 Quo = -3 Rem = -1 +7 div -2 Quo = -3 Rem = +1 -7 div -2 Quo = +3 Rem = -1 Convention: Dividend and remainder have the same sign Quotient is negative if signs disagree These rules fulfil the equation above