blocks|key|3313475|text|二的补码的美妙之处在于，在二进制级别上，它是一个解释问题，而不是算法--用于将两个有符号数相加的硬件与用于无符号数的硬件相同(忽略标志位)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3313476|你的第一个例子--“只需添加它们”--就是正确的答案。您的示例数字|3313477|3313478|01001001+=+73|unordered-list-item|3313479|10101010+=+-86|3313480|3313481|因此，正确答案确实是-13。|3313482|减法也是一样的，因为不需要对二的补码数进行特殊处理:你只需要“减去它们”。|3313483|注意，有趣的是对溢出/下溢比特的处理。您不能将73+-+(-86)的结果表示为8位二进制补码数...|3313484|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|U|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|V|8|@]|9|@]|A|$]]|$1|H|3|I|5|G|7|W|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

The beauty of two's complement is that at the binary level it's a matter of interpretation rather than algorithm - the hardware for adding two signed numbers is the same as for unsigned numbers (ignoring flag bits). 

Your first example - "just add them" - is exactly the right answer. Your example numbers 

<ul>
<li>01001001 = 73</li>
<li>10101010 = -86</li>
</ul>

So, the correct answer is indeed -13.

Subtracting is just the same, in that no special processing is required for two's complement numbers: you "just subtract them".

Note that where things get interesting is the handling of overflow/underflow bits. You can't represent the result of 73 - (-86) as an 8-bit two's complement number...

blocks|key|3313433|text|当两个参数的符号相反时，添加二的补码不需要任何特殊处理。您只需像通常的二进制代码一样将它们相加，结果的符号就是您保留的符号。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3313434|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Adding in two's complement doesn't require any special processing when the signs of the two arguments are opposite. You just add them as you normally would in binary, and the sign of the result is the sign you keep.

blocks|key|4160095|text|为了确保您理解2的补码，将正数转换为负数(或反之亦然)：反转每一位，然后将结果加1。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4160096|例如，正数X=+01001001变成了负数10110101%2B1=10110110；负数Y=+10101010变成了正数01010101%2B1=01010110。|4160097|若要从X中减去Y，请对Y求反，然后将其相加。即01001001+%2B+01010110。|4160098|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

And just to make sure you understand two's complement, to convert from a positive to a negative number (or vice versa): invert each bit, then add 1 to the result.

For example, your positive number X = 01001001 becomes 10110101+1=10110110 as a negative number; your negative number Y = 10101010 becomes 01010101+1=01010110 as a positive number.

To subtract Y from X, negate Y and add. I.E. 01001001 + 01010110.

blocks|key|93548|text|您的困惑可能是因为涉及的数字范围太大。为了更好地理解这一点，你可以尝试从你的无符号整数中创建一个有符号整数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|93549|如果您的无符号整数的MSB已经是0，那么您可以将其作为有符号来读取，并获得相同的结果。|93550|如果MSB是1，那么你可以在左边附加一个0来得到一个有符号的数字。你应该对所有有符号的数字进行符号扩展(即，如果MSB为0，则添加0，如果MSB为1，则添加1)，以获得相同宽度的数字，这样您就可以“正常”进行运算。|93551|例如，使用你的数字：|93552|X=+01001001:无符号，MSB为0，不做任何操作。|93553|Y=+10101010:签名，对X不做任何事情，仍然什么都不做。|93554|但如果我们将X的MSB更改为1：|93555|X=+11001001:无符号，MSB为1，加0+-->+011001001|93556|Y=+10101010:带符号的扩展X，所以符号-扩展Y+-->+110101010|93557|现在你有了两个有符号的数字，你可以按照你已经知道的方式进行加减。|93558|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|X|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|Y|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|Z|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|10|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|11|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|12|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|13|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|14|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|U|$]]

Your confusion might be because of the widths of the numbers involved. To get a better feel for this you could try creating a signed integer out of your unsigned integer.

If the MSB of your unsigned integer is already 0, then you can read it as signed and get the same result. 
If the MSB is 1 then you can append a 0 to the left to get a signed number. You should sign-extend (that is, add 0s if the MSB is 0, add 1s if the MSB is 1) all the signed numbers to get a number of the same width so you can do the arithmetic "normally".

For instance, using your numbers:

X = 01001001: Unsigned, MSB is 0, do nothing.

Y = 10101010: Signed, did nothing with X, still do nothing.

But if we change the MSB of X to 1:

X = 11001001: Unsigned, MSB is 1, Add a 0 --> 011001001

Y = 10101010: Signed, extended X, so sign-extend Y --> 110101010

Now you have two signed numbers that you can add or subtract the way you already know.

X = 01001001 and Y = 10101010 
 
If I want to add them together how do I do that? They are "Two's Complement"...
I have tried a lots of things but I am not quite sure I am getting the right answer since there seems to be different type of rules. 
 
Just want to make sure it is correct: 
 1. Add them as they are do not convert the negative 
 2. Convert the negative number you get and that's the sum. 
 
f.eks 
01001001+10101010 = 11110011 => 00001100 => 1101 => -13 
 
Or? 
 1. Convert the negative 
 2. Add them together and convert the negative 
 
f.eks 
01001001+10101010 => 01001001 + 01010110 => 10011111 => 01100001 => -97 
 
 
So basically what I want to do is to take: X-Y, and X+Y 
Can someone tell me how to do that? 
 
Some resource sites:
<a href="http://www.cs.grinnell.edu/~rebelsky/Courses/CS152/97F/Readings/student-binary.html" rel="nofollow noreferrer">student-binary</a>
<a href="http://celtickane.com/computers/cs_binary.php" rel="nofollow noreferrer">celtickane</a>
<a href="http://www.swarthmore.edu/NatSci/echeeve1/Ref/BinaryMath/BinaryMath.html" rel="nofollow noreferrer">swarthmore</a>

Adding negative and positive binary?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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X= 01001001和Y= 10101010如果我想把它们加在一起，我该怎么做呢？他们是“二的互补”..。我尝试了很多方法，但我不太确定我是否得到了正确的答案，因为似乎有不同类型的规则。我只想确保它是正确的：将它们按原样相加，不要将negativeConvert转换为负数，这就是。f.eks01001001+10101010 = 11110011 => 00001100 => 1101 => -

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