Editing Two's complement (section)

===Addition===
Adding two's complement numbers requires no special processing even if the operands have opposite signs; the sign of the result is determined automatically. For example, adding 15 and −5:
<pre style="width:25em">
   0000 1111  (15)
 + 1111 1011  (−5)
 ===========
   0000 1010  (10)
</pre>

Or the computation of 5 − 15 = 5 + (−15):

<pre style="width:25em">
   0000 0101  (  5)
 + 1111 0001  (−15)
 ===========
   1111 0110  (−10)
</pre>

This process depends upon restricting to 8 bits of precision; a carry to the (nonexistent) 9th most significant bit is ignored, resulting in the arithmetically correct result of 10<sub>10</sub>.

The last two bits of the [[Carry flag|carry]] row (reading right-to-left) contain vital information: whether the calculation resulted in an [[arithmetic overflow]], a number too large for the binary system to represent (in this case greater than 8 bits). An overflow condition exists when these last two bits are different from one another. As mentioned above, the sign of the number is encoded in the MSB of the result.

In other terms, if the left two carry bits (the ones on the far left of the top row in these examples) are both 1s or both 0s, the result is valid; if the left two carry bits are "1 0" or "0 1", a sign overflow has occurred. Conveniently, an [[XOR]] operation on these two bits can quickly determine if an overflow condition exists. As an example, consider the signed 4-bit addition of 7 and 3:
<pre style="width:25em">
  0111   (carry)
   0111  (7)
 + 0011  (3)
 ======
   1010  (−6)  invalid!
</pre>
In this case, the far left two (MSB) carry bits are "01", which means there was a two's-complement addition overflow. That is, 1010<sub>2</sub> = 10<sub>10</sub> is outside the permitted range of −8 to 7. The result would be correct if treated as unsigned integer.

In general, any two {{mvar|N}}-bit numbers may be added ''without'' overflow, by first sign-extending both of them to {{math|''N'' + 1}} bits, and then adding as above. The {{math|''N'' + 1}} bits result is large enough to represent any possible sum ({{math|1=''N'' = 5}} two's complement can represent values in the range −16 to 15) so overflow will never occur. It is then possible, if desired, to 'truncate' the result back to {{mvar|N}} bits while preserving the value if and only if the discarded bit is a proper sign extension of the retained result bits. This provides another method of detecting overflow – which is equivalent to the method of comparing the carry bits – but which may be easier to implement in some situations, because it does not require access to the internals of the addition.