需要一些帮助，了解使用2＆＃x27的补充二进制加法/减法
如果a = 01110011，b = 10010100，我该如何添加？
我这样做了：
ie：01110011 + 10010100 = 100000111
，虽然不是本质上是115 +（-108）= 7，而我得到的-249
编辑：我看到那是删除的，最高订单位（溢出）我得到了7，这就是我要寻找的东西，但是我没有得到为什么您不会额外的。
编辑**：好的，我弄清楚了。我认为没有溢出，因为[-128，127]（8位）中有7个。取而代之的是，就像奥马尔（Omar）所暗示的那样，我应该从加法中删除“额外” 1。
If A = 01110011, B = 10010100, how would I add these?
I did this: 
i.e: 01110011 + 10010100 = 100000111
Though, isn't it essentially 115 + (-108) = 7, whereas, I'm getting -249
Edit: I see that removing the highest order bit (overflow) I get 7 which is what I'm looking for but I'm not getting why you wouldn't have the extra bit.
Edit**: Ok, I figured it out. There was no overflow as I had assumed there was because 7 is within [-128, 127] (8-bits). Instead, like Omar hinted at I was supposed to drop the "extra" 1 from addition.
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您的计算正确，结果是正确的。
您说第二个数字为-108，因此您的两个数字都被解释为签名的8位值。因此，您还应该将结果解释为8位签名值，这就是为什么必须删除第9位的原因，因此结果为7（00000111）。
在真实的硬件中，例如8位CPU，例如所有寄存器宽度为8位，您只能存储结果的最低8位，这是7（00000111）。
在某些情况下，第9位也可以放入随身携带/溢流标志中，以免完全“掉落”。
Your calculation is correct and the result is correct.
You stated that the second number is -108, so both your numbers are interpreted as signed 8-bit values. Thus, you should also interpret your result as an 8-bit signed value, this is why the 9th bit must be dropped, and so the result is 7 (00000111).
On a real hardware, like an 8-bit CPU for example, as all the registers are 8-bit wide, you are only be able to store the lowest 8-bit of the result, which here is 7 (00000111).
In some cases, the 9th bit may also be put inside a carry/overflow flag so it's not completely "dropped".