这是溢出算术计算吗? [英] Is this an overflow arithmetic calculation?

问题描述

所以当我读这本书时，它说添加不同的符号并减去相同的符号就不会发生溢出. 但是我在执行此操作时有疑问:185-122 我将122的二进制转换为2s的补码，并进行了加法运算，这是不同的符号: 185 +(-122)并将它们加在一起时，符号位溢出到100111111.但是，如果我切断左侧的MSB，则是正确的答案.它溢出了吗?

So when I read the book and it says that overflow can't occur when add different signs and subtraction of the same sign. But I have question when I do this: 185 - 122 I converted binary of 122 to 2s complement and did the addition, which is different signs: 185+(-122) and when I add them together, I got the sign bit overflow to 100111111. But if I cut off the MSB on the left, it is the correct answer. Is it an overflow?

推荐答案

不，它不是溢出-必须仅丢弃MSB中添加2 1's所导致的溢出.来自维基百科

No, it isn't overflow - the overflow resulting from the addition of 2 1's in the MSB must just be discarded. From Wikipedia

要获得二进制数的二进制补码，请使用按位NOT运算将这些位取反或翻转".然后将1的值添加到结果值中，而忽略了取2的补数为0时发生的溢出.

To get the two's complement of a binary number, the bits are inverted, or "flipped", by using the bitwise NOT operation; the value of 1 is then added to the resulting value, ignoring the overflow which occurs when taking the two's complement of 0.

所以在您的示例中

185 10111001
122 01111010 -

取122的2的补码(1的补码+1)

Taking the 2's complement of 122 (One's complement +1)

01111010 => 10000110

添加:

10111001 185
10000110 +(-122)
--------
00111111 (63)

= 63

overflow被忽略.

但是，在执行2的补码后，有一些检测溢出的规则:

There are however rules for detecting overflow after doing the 2's complement :

• 如果两个正数之和导致负数
• 如果两个负数的总和为正数

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