计算机组成原理.docx

1、计算机组成原理第三章3.1 已知x补和y 补的值，用补.码加减法计算x+y和x-y（写出结果的二进制表示和十进制表示），并指出结果是否溢出以及溢出的类型(建议采用变形补码什算)。 (1) x 补=0.11011 , y补=0.00011 (2) x 补=0.10111 y补=1.00101 (3) x 补=1.01010 y补=1.10001 (4) x 补=1.10011 y补=0.11001解: (1) x补=0.11011 , y补=0.00011y补=1.11101X+Y补=x补+y补= 00.11011+ 00.00011_ 00.11110X+Y= +15/16 =+0.1111B

2、 XY补=x补+y补=00.11011+11.11101_ 00.11000XY= +12/16 =3/4=+0.11B检验【 X= + 27/ 32Y= +3/32X+Y = +(27+3)/32 =+30/32=0.11110BXY = (273)/32=+24/32=+3/4=0.11B 】(2) x补=0.10111 , y补=1.00101 Y原=1.11011y补=0.11011X+Y补=x补+y补= 00.10111+ 11.00101_ 11.11100X+Y= 0.001B=1/8 XY补=x补+y补= 00.10111+00.11011_ 01.10010 (上溢)XY补=

3、1.10010 (上溢)【 修正：XY=0.01110 = 7 / 16+2 = 25 / 16 】检验【 X= +23/ 32Y= 27/32X+Y = (2327)/32 =-4/32=1/8XY = (23+27)/32=+50/32 =25/16 (上溢) 】(3) x补=1.01010 , y补=1.10001 y补=0.01111X+Y补=x补+y补= 11.01010+ 11.10001_ 10.11011X+Y=0.11011B=+27/32 修正 27/322=2764=37/32 XY补=x补+y补= 11.01010+ 00.01111_ 11.11001 XY=0.00

4、111B=7/32检验【X= 22/ 32=11/16Y=15/32X+Y = (1522) /32 =37/32 (下溢)XY = (22(15) / 32 = 7/32 】 (4) x补=1.10011 , y补=0.11001 X= 0.01101=13/ 32Y= +0.11001=25/32X+Y = (13+25)/32 =12/32=3/8XY = (1325)/32=38/32 = -19 / 16 (下溢) y补=1.00111X+Y补=x补+y补= 11.10011+ 00.11001_ 00.01100X+Y=0.01100B=3/8 XY补=x补+y补= 11.1001

5、1+ 11.00111_ 10.11010 (下溢)XY补=0.11010 (下溢)【 修正： XY = 13 / 16 2 = 19 / 16 】检验【 X= 0.01101=13/ 32Y= +0.11001=25/32X+Y = (13+25)/32 =12/32=3/8XY = (1325)/32=38/32 = -19 / 16 (下溢) 】3.2 给出x和y的二进制值，用补码加减法计算x+y和x-y（写出结果的二进制表示和十进制表示），并指出结果是否溢出以及溢出的类型(建议采用变形补码什算)、 (1) x=0.10111 y=0.11011 (2) x=0.11101 y=0.10

6、011 (3) x=0.11011 y=-0.1010 (4) x=-0.11111 y=0.11011 (5) x=-0.11011 y=0.1010 (6) x=-0.11010 y=-0.11001 (7) x=-1011101 y=+1101101 (8) x=+1110110 y=-1001101 (9) x=+1101110 y=+1010101 (10) x=-1111111 y=-1101101解：(1) X=0.10111 Y=0.11011 X补=0.10111 Y补=0.11011 Y补=1.00101 X+Y补=X补+Y补= 00.10111+ 00.11011 _01

7、.10010 (上溢) X+Y 上溢 XY补=X补Y补= 00.10111+ 11.00101_ 11.11100X-Y=0.00100 =1/8检验【X=23/32 Y=27/32X+Y=50/32 修正：X+Y=-0.0111 = 7 / 16 +2 =25 / 16 XY=1/8】(2) X=0.11101 Y=0.10011 X补=0.11101 Y补=0.10011 Y补=1.01101 X+Y补=X补+Y补= 00.11101+ 00.10011 _01.10000 (上溢) X+Y 上溢 XY补=X补Y补= 00.11101+ 11.01101_ 00.01010X-Y=0.01

8、010 =5/16检验【X=29/32 Y=19/32X+Y=48/32 =3 / 2 修正：X+Y = 0.1 = 1 / 2+2 =1.5 XY=10/32=5 / 16】(3) X=0.11011 Y=0.1010 X补=0.11011 Y补=1.01100 Y补=0.10100 X+Y补=X补+Y= 00.11011+ 11.01100 _00.00111 X+Y =7/32 XY补=X补Y补= 00.11011+ 00.10100_ 01.01111XY=0.10001=17/32 (上溢) 修正 17/32+2=47/32 检验【X=27/32 Y=20/32X+Y=7/32 XY

9、=47/32】(4) X=0.11111 Y=0.11011 X补=1.00001 Y补=0.11011 Y补=1.00101 X+Y补=X补+Y补= 11.00001+ 00.11011 _11.11100 X+Y =0.00100=1/8 XY补=X补Y补= 11.00001+ 11.00101_ 10.00110 (下溢)XY=0.00110=3/16 (下溢) 修正 6/322=58/32 =29/16 检验【X=31/32 Y=27/32X+Y=4/32 =1/8 XY=58/32=29/16】(5) X=0.11011 Y=0.10100 X补=1.00101 Y补=0.10100

10、 Y补=1.01100 X+Y补=X补+Y补= 11.00101+ 00.10100 _11.11011 X+Y =0.00111=7/32 XY补=X补Y补= 11.00101+ 11.01100_ 10.10001 (下溢)XY=17/32 (下溢) 修正 17/322=47/32 检验【X=27/32 Y=20/32X+Y=7/32 XY=47/32】(6) X=0.11010 Y=0.11001 X补=1.00110 Y补=1.00111 Y补=0.11001 X+Y补=X补+Y补= 11.00110+ 11.00111 _10.01101 X+Y =13/32 (下溢) 修正 13/

11、322=51/32 XY补=X补Y补= 11.00110+ 00.11001_ 11.11111 XY=0.00001= 1/32检验【X=26/32 Y=25/32X+Y=51/32 XY=1/32】 (7) X=1011101 Y=+1101101 X补=10100011 Y补=01101101 Y补=1.01100 X+Y补=X补+Y补= 110100011+ 001101101 _000010000 X+Y =00010000=16 XY补=X补Y补= 110100011+ 110010011_ 100110110 (下溢)XY=54 (下溢) 修正 54256=202 检验【X=93

12、 Y=109X+Y=16 XY=202】 (8) X=+1110110 Y=1001101 X补=01110110 Y补=10110011 Y补=01001101 X+Y补=X补+Y补= 001110110+ 110110011 _000101001 X+Y =41 XY补=X补Y补= 0011101101+ 001001101_ 011000011 (上溢)XY=61 (上溢) 修正 61+256=195 检验【X=118 Y=77X+Y=41XY=195】 (9) X=+1101110 Y=+1010101 X补=01101110 Y补=01010101 Y补=10101011 X+Y补=

13、X补+Y补= 001101110+001010101 _011000011 (上溢) X+Y =0111101=61 修正 61+256=195 XY补=X补Y补= 001101110+ 110101011_ 000011001XY=25检验【X=110 Y=85X+Y=195XY=25】 (10) X=1111111 Y=1101101 X补=10000001 Y补=10010011 Y补=01101101 X+Y补=X补+Y补= 110000001+ 110010100 _100010100 (下溢) X+Y =00010100=20 修正 20256=236 XY补=X补Y补= 1100

14、00001+ 001101101_ 111101110检验【XY=0010010=18X=127 Y=109X+Y=236XY=18】3.1 已知x补和y 补的值，用补.码加减法计算x+y和x-y （写出结果的二进制表示和十进制表示），并指出结果 是否溢出以及溢出的类型(建议采用变形补码计算)。 (3) x 补=1.01010 y补=1.10001 解：x补= 11.01010 y补= 11.10001 y补= 00.01111 X+Y补 = x补+y补= 11.01010 + 11.10001 = 10.11011 (下溢) X+Y = 1.00101B = (1 + 5/32) = 37/32 ：x补= 11.01010 y补= 11.10001 y补= 00.01111 X+Y补 = x补+y补= 11.01010 + 11.10001 = 10.11011 (下溢) X+Y = 1.00101B = (1 + 5/32) = 37/32 检验:X= - 0.10110 B =22/ 32 Y= - 0.01111 B =15/32X+Y = (22+15) /32 =37/32 (下溢)XY = （22(15)）/ 32 = 7/32