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我有一个程序可以添加 2 个二进制补码并输出输出。它们是从文件中读取的，因为第一个二进制数是二进制补码。第二个二进制数是有偏差的表示法。为了改变偏向二进制补码的形式，我们翻转最左边的位。然后我们将两个二进制数相加并将输出提供给文件。我遇到的问题是理解输出。这是输出，其中第一个二进制数是二进制补码，第二个是有偏表示法。

   01000000 <- 2's complement         64
+  11000000 <- biased notation        64
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   10000000 <- 2's complement   Overflow

   01111111 <- 2's complement        127
+  00000000 <- biased notation      -128
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   11111111 <- 2's complement         -1

   00000011 <- 2's complement          3
+  10000111 <- biased notation         7
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   00000110 <- 2's complement          6

   00001111 <- 2's complement         15
+  10000111 <- biased notation         7
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   00010010 <- 2's complement         18

   10000000 <- 2's complement       -128
+  11111111 <- biased notation       127
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   11111111 <- 2's complement         -1

   11110000 <- 2's complement        -16
+  10001000 <- biased notation         8
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   11111000 <- 2's complement         -8

   10000001 <- 2's complement       -127
+  00000001 <- biased notation      -127
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   00000010 <- 2's complement   Underflow

   01111111 <- 2's complement        127
+  00000000 <- biased notation      -128
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   11111111 <- 2's complement         -1

   01110101 <- 2's complement        117
+  11010001 <- biased notation        81
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   01000110 <- 2's complement         70

   00000000 <- 2's complement          0
+  10000000 <- biased notation         0
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   00000000 <- 2's complement          0

   00001111 <- 2's complement         15
+  11110000 <- biased notation       112
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   01111111 <- 2's complement        127

   00001111 <- 2's complement         15
+  10000001 <- biased notation         1
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   00010000 <- 2's complement         16

   00000111 <- 2's complement          7
+  11110000 <- biased notation       112
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   01110111 <- 2's complement        119

   11111111 <- 2's complement         -1
+  01111111 <- biased notation        -1
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   10101010 <- 2's complement        -86

   00000000 <- 2's complement          0
+  10000000 <- biased notation         0
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   00000000 <- 2's complement          0

   11111111 <- 2's complement         -1
+  11111111 <- biased notation       127
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   00101010 <- 2's complement         42

鉴于第三个示例说 3 + 7 = 6。这是否正确？这似乎不对，但其他例子是正确的。比如-16 + 8 = -8。

所以我的问题是......这个文件的输出是否正确？

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1 回答 1

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似乎每当您连续进行 3 个进位时，它就会掉 1。与下一个相同，15 + 7。它没有正确地进行进位。

  11  <- carry
 0011 = 3
+0111 = 7
 1010 = 10
于 2015-02-11T22:58:00.903 回答