Part 2 of 2 Method of complements作业
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Part 2 of 2 Method of complements
学号16340090
(1)证明:
设一个任意负二进制的a为
\(a=-(x_62^6+x_52^5+x_42^4+x_32^3+x_22^2+x_12^1+x_0)\)
\(x_n=1或0(n<=6)\)
即\(a=1x_6x_5x_4x_3x_2x_1x_0\)
a的相反数为正数\(-a=0x_6x_5x_4x_3x_2x_1x_0\)
设b为a的补码
若\(b=1(1-x_6)(1-x_5)(1-x_4)(1-x_3)(1-x_2)(1-x_1)(1-x_0)+00000001\)
则a-a=0=b+(-a)=000000000
造成溢出
则其补码b为
即\(1(1-x_6)(1-x_5)(1-x_4)(1-x_3)(1-x_2)(1-x_1)(1-x_0)+00000001\)
也就是补码=反码+1
证毕
2)Int8_t x = - 017=(10001111)2补码为11110001八进制为0361
0 0
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