计算机组成与设计 第五版答案_CH03_Solution

1、Solutions 3 Chapter 3 Solutions S-3 3.1 5730 3.2 5730 3.3 0101111011010100 Th e attraction is that each hex digit contains one of 16 diff erent characters (09, AE). Since with 4 binary bits you can represent 16 diff erent patterns, in hex each digit requires exactly 4 binary bits. And bytes are by defi nition 8 bits long, so two hex digits are all that are required to represent the contents of 1 byte. 3.4 753 3.5 7777 (?3777) 3.6 Neither (63) 3.7 Neither (65) 3.8 Overfl ow (result ? ?179, which

2、does not fi t into an SM 8-bit format) 3.9 ?105 ? 42 ? ?128 (?147) 3.10 ?105 ? 42 ? ?63 3.11 151 ? 214 ? 255 (365) 3.12 62?12 StepActionMultiplierMultiplicandProduct 0Initial Vals001 010000 000 110 010000 000 000 000 lsb=0, no op001 010000 000 110 010000 000 000 000 1Lshift Mcand001 010000 001 100 100000 000 000 000 Rshift Mplier000 101000 001 100 100000 000 000 000 Prod=Prod+Mcand000 101000 001 100 100000 001 100 100 2Lshift Mcand000 101000 011 001 000000 001 100 100 Rshift Mplier000 010000 011

3、 001 000000 001 100 100 lsb=0, no op000 010000 011 001 000000 001 100 100 3Lshift Mcand000 010000 110 010 000000 001 100 100 Rshift Mplier000 001000 110 010 000000 001 100 100 Prod=Prod+Mcand000 001000 110 010 000000 111 110 100 4Lshift Mcand000 001001 100 100 000000 111 110 100 Rshift Mplier000 000001 100 100 000000 111 110 100 lsb=0, no op000 000001 100 100 000000 111 110 100 5Lshift Mcand000 000011 001 000 000000 111 110 100 Rshift Mplier000 000011 001 000 000000 111 110 100 lsb=0, no op000 0

4、00110 010 000 000000 111 110 100 6Lshift Mcand000 000110 010 000 000000 111 110 100 Rshift Mplier000 000110 010 000 000000 111 110 100 S-4 Chapter 3 Solutions 3.13 62?12 StepActionMultiplicandProduct/Multiplier 0Initial Vals110 010000 000 001 010 1 lsb=0, no op110 010000 000 001 010 Rshift Product110 010000 000 000 101 2 Prod=Prod+Mcand110 010110 010 000 101 Rshift Mplier110 010011 001 000 010 3 lsb=0, no op110 010011 001 000 010 Rshift Mplier110 010001 100 100 001 4 Prod=Prod+Mcand110 010111 11

5、0 100 001 Rshift Mplier110 010011 111 010 000 5 lsb=0, no op110 010011 111 010 000 Rshift Mplier110 010001 111 101 000 6 lsb=0, no op110 010001 111 101 000 Rshift Mplier110 010000 111 110 100 3.14 For hardware, it takes 1 cycle to do the add, 1 cycle to do the shift , and 1 cycle to decide if we are done. So the loop takes (3 ? A) cycles, with each cycle being B time units long. For a soft ware implementation, it takes 1 cycle to decide what to add, 1 cycle to do the add, 1 cycle to do each shif

6、t , and 1 cycle to decide if we are done. So the loop takes (5 ? A) cycles, with each cycle being B time units long. (3?8)?4tu ? 96 time units for hardware (5?8)?4tu ? 160 time units for soft ware 3.15 It takes B time units to get through an adder, and there will be A ? 1 adders. Word is 8 bits wide, requiring 7 adders. 7?4tu ? 28 time units. 3.16 It takes B time units to get through an adder, and the adders are arranged in a tree structure. It will require log2(A) levels. 8 bit wide word requir

7、es 7 adders in 3 levels. 3?4tu ? 12 time units. 3.17 0 x33 ? 0 x55 ? 0 x10EF. 0 x33 ? 51, and 51 ? 32?16?2?1. We can shift 0 x55 left 5 places (0 xAA0), then add 0 x55 shift ed left 4 places (0 x550), then add 0 x55 shift ed left once (0 xAA), then add 0 x55. 0 xAA0?0 x550?0 xAA?0 x55 ? 0 x10EF. 3 shift s, 3 adds. (Could also use 0 x55, which is 64?16?4?1, and shift 0 x33 left 6 times, add to it 0 x33 shift ed left 4 times, add to that 0 x33 shift ed left 2 times, and add to that 0 x33. Same num

8、ber of shift s and adds.) Chapter 3 Solutions S-5 3.18 74/21 ? 3 remainder 9 StepActionQuotientDivisorRemainder 0Initial Vals000 000010 001 000 000000 000 111 100 1 Rem=RemDiv000 000010 001 000 000101 111 111 100 Rem0,R+D,Q000 000010 001 000 000000 000 111 100 Rshift Div000 000001 000 100 000000 000 111 100 2 Rem=RemDiv000 000001 000 100 000111 000 011 100 Rem0,R+D,Q000 000001 000 100 000000 000 111 100 Rshift Div000 000000 100 010 000000 000 111 100 3 Rem=RemDiv000 000000 100 010 000111 100 101

9、 100 Rem0,R+D,Q000 000000 100 010 000000 000 111 100 Rshift Div000 000000 010 001 000000 000 111 100 4 Rem=RemDiv000 000000 010 001 000111 110 110 100 Rem0,R+D,Q000 000000 010 001 000000 000 111 100 Rshift Div000 000000 001 000 100000 000 111 100 5 Rem=RemDiv000 000000 001 000 100111 111 111 000 Rem0,R+D,Q0,Q0,Q1 000 011000 000 010 001000 000 001 001 Rshift Div000 011000 000 001 000000 000 001 001 3.19. In these solutions a 1 or a 0 was added to the Quotient if the remainder was greater than or equal to 0. However, an equally valid solution is to shift in a 1 or 0, but if you do this you must do a compensating right shift of the remainder (only the remainder, not the entire remainder/quotient combination) aft er the last step. 74/21 ? 3 remainder 11 StepActionDivisorRemainder/Quotient 0Initial Vals010 001000 000 111 100 1 R010 001000 001 111 000 Rem=RemDiv010 001111 000 111 000 Rem0,R+D010 001000 001 111 000 2 R010 001000 011 110 000 Rem=RemDiv010 001

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