计算机组成原理第4章习题参考答案ppt课件

.,第4章数值的机器运算,位一组，写出进位信号,C6逻辑表,4-2某加法器采用组内并行，组间并行的进位链，达式。解组间并行的进位链，产生的最低一组的进位输出为：,C4=G1*+P1*C0,（其中：G1*=G4+P4G3+P4P3G2+P4P3P2G1；P1*=P4P3P2P1）,C6：,然后在组内产生进位信号C6=G6+P6C5=G6+P6G5+P6P5C4,X+Y，并指出结果是否溢出。,4-4已知X和Y，使用它们的变形补码计算出（1）X=0.11011，Y=0.11111（2）X=0.11011，Y=-0.10101（3）X=-0.10110，Y=-0.00001（4）X=-0.11011，Y=0.11110解,（1）X,补,补,=0.11011，Y00.11011+00.11111,=0.11111XY,补补,01.11010,X+Y,补,结果正溢出！,（2）X,补,补,=0.11011，Y00.11011+11.01011,=1.01011XY,补补,00.00110,X+Y,补,结果不溢出！,X+Y=0.00110,（3）X,补,补,=1.01010，Y11.01010+11.11111,=1.11111XY,补补,11.01001,X+Y,补,结果不溢出！,X+Y=-0.10111,（4）X,补,补,=1.00101，Y11.00101+00.11110,=0.11110XY,补补,00.00011,X+Y,补,结果不溢出！,X+Y=0.00011,X-Y，并指出结果是否溢出。,4-5已知X和Y，使用它们的变形补码计算出（1）X=0.11011，Y=-0.11111（2）X=0.10111，Y=0.11011（3）X=0.11011，Y=-0.10011（4）X=-0.10110，Y=-0.00001,.,解,（1）X,补,=0.11011，Y,补,=1.00001，-Y,补,=0.11111,00.11011+00.11111,X-Y,补补,01.11010,X-Y,补,结果正溢出！,（2）X,补,=0.10111，Y,补,=0.11011,-Y,补,=1.00101,00.10111,X,补,+11.00101,-Y,补,11.11100,X-Y,补,结果不溢出！,X-Y=11.11100,（3）X,补,=0.11011，Y,补,=1.01101,-Y,补,=0.10011,00.11011+00.10011,X-Y,补补,01.01110,X-Y,补,结果正溢出！,（4）X,补,=1.01010，Y,补,=1.11111，-Y,补,=0.00001,11.01010+00.00001,X-Y,补补,11.01011,X-Y,补,结果不溢出！,X-Y=-0.101014-6已知：X=0.1011，Y=-0.0101,求：X/2,补,X/4,补,-X,补,Y/2,补,Y/4,补,-Y,补,2Y,补,解由X,补,=0.1011,Y,补,=1.1011得：,X/2,补,=0.0101，X/4,补,=0.0010，-X,补,=1.0101,Y/2,补,=1.1101，Y/4,补,=1.1110，-Y,补,=0.0101，2Y,补,=1.0110,8位，包括一位符号位，采用补码表示，分别写出每个数右移或左移两,4-7.设下列数据长位之后的结果,（1）0.1100100,(2)1.0011001(3)1.1100110,（4）1.0000111,解（1）右移两位后为0.0011001（2）右移两位后为1.1100110,左移两位后为左移两位后为,1.0010000（溢出）或0.1100100（溢出）或,0.00100001.1100100,（3）右移两位后为,1.1111001,左移两位后为1.0011000,0.0011100（溢出）或,1.0011100,（4）右移两位后为1.110000014-8分别用原码乘法和补码乘法计算,左移两位后为XY.,(1)X=0.11011,Y=0.11111(2)X=0.11010,Y=0.01110解原码乘法：,（1）|X|=0.11011B,|Y|=0.11111,C，0A,.,=0，+0,00.00000+|X|00.11011,C,说明5=1，+|X|,00.01101101111,部分积右移,1位,C,5=1，+|X|,+|X|00.1101101.01000,00.10100010111,部分积右移,1位,C,5=1，+|X|,+|X|00.1101101.01111,00.10111101011,部分积右移,1位,C,5=1，+|X|,+|X|00.1101101.10010,00.11001010101,部分积右移,1位,C,5=1，+|X|,+|X|00.1101101.10100,部分积右移,1位,00.11010001010所以，|XY|=0.1101000101Y=-0.1101000101X,（2）|X|=0.11010,B，|Y|=0.01110,C，0A,00.00000+000.00000,C,说明5,00.00000000111,部分积右移,1位,C,5=1，+|X|,+|X|00.1101000.11010,00.01101000011,部分积右移,1位,C,5=1，+|X|,+|X|00.1101001.00111,00.10011100001,部分积右移,1位,C,5=1，+|X|,+|X|00.1101001.01101,00.10110110000,部分积右移,1位,C,5=0，+0,+000.0000000.10110,部分积右移,1位,00.01011011000所以,|XY|=0.0101101100Y=+0.0101101100X,补码乘法：,（1）X,补,=0.11011B,Y,补,=1.00001C，0A,A0.1111100.11011,C,A寄存器0.0111000.00000,C寄存器,.,C6=00，+0,-X,补,=1.00101,附加位,说明,00.00000,+-X,补11.00101,C,6,5C=10，+-X,补,11.00101,11.100101100001,部分积右移,1位,+X,补,00.11011,C,5C6=01，+X,补,00.01101,00.001101110000+000.00000C00.0011000.000110111000+000.00000C00.0001100.000011011100+000.00000C00.0000100.000001101110,部分积右移5C6=00，+0部分积右移5C6=00，+0部分积右移5C6=00，+0部分积右移,1位1位1位1位,+-X,补,11.00101,C,5C6=10，+-X,补,11.00101,所以，XY,补,=1.0010111011,X,Y=-0.1101000101,（2）X,补,=1.00110B，Y补=1.10010C，0A,-X,补,=0.11010,附加位,说明,00.00000,1.100100,+000.00000,C,5,00.00000,00.000000110010,部分积右移,1位,+-X,补00.11010,C,5C6=10,+-X,补,00.11010,00.011010011001,部分积右移,1位,+X,补,11.00110,C,5C6=01，+X,补,11.10011,11.110011001100+000.00000C11.1100111.111001100110,部分积右移5C6=00，+0部分积右移,1位1位,+-X,补00.11010,C,5C6=10，+-X,补,00.10110,00.010110110011,部分积右移,1位,+0,00.00000,C,5C6=11，+0,00.01011所以，XY补=0.0101101100,X,Y=0.0101101100,X/Y。,4-10分别用原码和补码加减交替法计算（2）X=-0.10101，Y=0.11011（4）X=-0.10110，Y=-0.11011解原码除法：,（2）|X|=00.10101A，|Y|=00.11011B，0,变补,=11.00101,A寄存器1.000010,C寄存器,A寄存器,C寄存器,.,A,说明,00.101010.00000,+|Y|,变补,11.00101,-|Y|,11.110100.00000,余数为负，商,0,左移一位,+|Y|,11.1010000.11011,+|Y|,00.011110.00001,余数为正，商,1,00.11110,左移一位,+|Y|,变补,11.00101,-|Y|,00.000110.00011,余数为正，商,1,00.00110,左移一位,+|Y|,变补,11.00101,-|Y|,11.010110.00110,余数为负，商,0,左移一位,+|Y|,10.1011000.11011+|Y|,11.100010.01100,余数为负，商,0,左移一位,+|Y|,11.0001000.11011,+|Y|,11.111010.11000,余数为负，商,0,+|Y|,00.11011,+|Y|,恢复余数,00.11000,商符：Qs=XsYs=10=1,商=-0.11000,X/Y,=-0.11000+,(0.11000*2,-5,)/0.11011(,中括号内余数项的分子和分母都取正,),（4）X=-0.10110，Y=-0.11011,|X|=00.10110A，|Y|=00.11011B，0,变补,=11.00101,A,说明,00.101100.00000,+|Y|,变补,11.00101,-|Y|,11.110110.00000,余数为负，商,0,左移一位,+|Y|,11.1011000.11011+|Y|,00.100010.00001,余数为正，商,1,01.00010,左移一位,+|Y|,变补,11.00101,-|Y|,00.001110.00011,余数为正，商,1,00.01110,左移一位,+|Y|,变补,11.00101,-|Y|,11.100110.00110,余数为负，商,0,左移一位,+|Y|,11.0011000.11011+|Y|,00.000010.01101,余数为正，商,1,00.00010,左移一位,+|Y|,变补,11.00101,-|Y|,11.001110.11010,余数为负，商,0,+|Y|,恢复余数,00.11011+|Y|00.00010,A寄C存器,C寄存器,A寄C存器,C寄存器,.,（2）X=2,商符：Qs=XsYs=11=0,商=0.11010,-5,)/0.11011,X/Y=+0.11010+(0.00010*2补码除法：（2）X=-0.10101，Y=0.11011,X,补,=11.01011A，Y,补,=00.11011B，0C；-Y,补,=11.00101,C,说明,A11.010110.00000,+Y补00.11011X,补,与Y补异号，+Y补,1,r,1,00.001100.000000.01100,i补与Y补同号，商左移一位,+-Y补11.00101+-Y,补,11.100010.000,10,r,i,补与Y补异号，商0,11.00010,左移一位,+Y补00.11011X,补与Y,补异号，+Y,补,11.111010.00,100,r,i补与Y,补异号，商,0,11.11010,左移一位,+Y补00.11011X,补与Y,补异号，+Y,补,00.101010.0,1001,r,i补与Y,补同号，商,1,01.01010,左移一位,+-Y补11.00101+-Y,补,00.011110.,10011,r,i,补与Y补同号，商1,00.11110,左移一位,+-Y补11.00101+-Y,补,00.00011,1.0011,1,末位恒置1,商补=1.00111（4）X=-0.10110，Y=-0.11011,X,补,=11.01010A，Y,补,=11.00101B，0C；-Y,补,=00.11011,C,说明,A11.010100.00000,+-Y补00.11011X,补,与Y补同号，+-Y,补,0,r,0,00.001010.000000.01010,i补与Y补异号，商左移一位,+Y补11.00101+Y,补,11.011110.000,01,r,i,补与Y补同号，商1,10.11110,左移一位,+-Y补00.11011X,补与Y,补同号，+-Y,补,11.110010.00,011,r,i补与Y,补同号，商,1,11.10010,左移一位,+-Y补00.11011X,补与Y,补同号，+-Y,补,00.011010.0,0110,r,i补与Y,补异号，商,0,00.11010,左移一位,+Y补11.00101+Y,补,11.111110.,01101,r,i,补与Y补同号，商1,11.11110,左移一位,+-Y补00.11011X,补与Y,补同号，+-Y,补,00.11001,0.1101,1,末位恒置1,商补=0.110114-12设浮点数的阶码和尾数部分均用补码表示，按照浮点数的运算规则，计算下列各题：101100（1）X=2（-0.100010），Y=2（-0.111110）,-101-011（3）X=2,0.101100，Y=20.101100，Y=2,-100-001,（-0.101000）（-0.111100）,.,2,X-Y=2,注：此题中阶码用二进制表示。求：X+Y，X-Y解,（1）浮点数：XY,浮浮,=0101；1.011110=0100；1.000010,对阶：小阶向大阶看齐，E=EA-EB=1,Y,浮,=0101；1.100001,对阶之后，尾数相加和相减。,相加：11.011110,相减：11.011110,+00.01111111.111101,+11.10000110.111111规格化：,相加结果需要右规相减结果需需左规,1次，得：X+Y4次，得：X-Y,浮浮,=0110；1.011111=0001；1.010000,X+Y=2X-Y=2,110001,(-0.100001)(-0.110000),（2）浮点数：XY,浮浮,=1011；0.101100=1100；1.011000,对阶：小阶向大阶看齐，E=EA-EB=-1,X,浮,=1100；0.010110,对阶之后，尾数相加和相减。,相加：00.010110,相减：00.010110,+11.011000,+00.101000,1.101110,00.111110,1规格化：,相加结果需要左规一次，得：X+Y,浮,=1011；1.011100,X+Y=2X-Y=2,-101-100,(-0.100100)0.111110,（3）浮点数：X,浮,=1101；0.101100Y,浮,=1111；1.000100,对阶后：X,浮,=1111;0.001011,尾数相加减后：X+Y,浮,=1111；11.001111,X-Y,浮,=1111；01.000111,尾数规格化后：X+Y,浮,=1111；1.001111,X-Y,