1 .已知X 和Y ,试用它们的变形补码计算出X + Y ,并指出结果是否溢出。
(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= 0 .11011 ,[Y ]补= 0 .11111
00 .11011 [X]补
+ 00 .11111 [Y ]补
01 .11010 [X + Y ]补结果正溢
(2) [X]补= 0 .11011 ,[Y ]补= 1 .01011
00 .11011 [X]补
+ 11 .01011 [Y ]补
00 .00110 [X + Y ]补
X + Y = 0 .00110
(3) [X]补= 1 .01010 ,[Y ]补= 1 .11111
11 .01010 [X]补
+ 11 .11111 [Y ]补
11 .01001 [X + Y ]补
X + Y = - 0 .10111
(4) [X]补= 1 .00101 ,[Y ]补= 0 .11110
11 .00101 [X]补
+ 00 .11110 [Y ]补
00 .00011 [X + Y ]补
X + Y = 0 .00011
2 .已知X 和Y ,试用它们的变形补码计算出X - Y ,并指出结果是否溢出。
(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 [X]补
+ 00 .11111 [ - Y ]补
01 .11010 [X - Y ]补结果正溢
(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
X - Y = - 0 .00100
(3) [X]补= 0 .11011 ,[Y ]补= 1 .01101 ,[ - Y ]补= 0 .10011
00 .11011 [X]补
+ 00 .10011 [ - Y ]补
01 .01110 [X - Y ]补结果正溢
(4) [X]补= 1 .01010 ,[Y ]补= 1 .11111 ,[ - Y ]补= 0 .00001
11 .01010 [X]补
+ 00 .00001 [ - Y ]补
11 .01011 [X - Y ]补
X - Y = - 0 .10101
3 .已知:X = 0 .1011 ,Y = - 0 .0101
求: [1/2X]补, [1/4X]补,[ - X]补, [1/2Y]补, [1/4Y]补,[ - Y ]补。
解:[X]补= 0 .1011
1/2X补
= 0 .0101 , 1/4X补
= 0 .0010 ,[ - X]补= 1 .0101
[Y ]补= 1 .1011
1/2Y补
= 1 .1101 , 1/4Y补
= 1 .1110 ,[ - Y ]补= 0 .0101
4 .设下列数据长8 位,包括1 位符号位,采用补码表示,分别写出每个数据右移或左移2 位之后的结果。
(1) 0 .1100100
(2) 1 .0011001
(3) 1 .1100110
(4) 1 .0000111
解:(1) [X]补= 0 .1100100
1/4X补= 0 .0011001 ,[4 X]补= 0 .0010000
2) [X]补= 1 .0011001
1/4X补= 1 .1100110 ,[4 X]补= 1 .1100100
(3) 1 .1100110
1/4X补= 1 .1111001 ,[4 X]补= 1 .0011000
(4) 1 .0000111
1/4X补= 1 .1100001 ,[4 X]补= 1 .0011100
5.证明在全加器里,进位传递函数P = A i + Bi = Ai⊕ Bi 。
解:并行加法器中的每一个全加器都有一个从低位送来的进位和一个传送给较高位
的进位。进位表