How Can 1111 1111 Be the Two-S Complement Representation of -1?
My Book Says That to Get the Two's-Complement Representation, to Just Flip the Bits and Add 1. Correct Me If I Am Wrong but the Binary Representation of -1...
My book says that to get the two's-complement representation, to just flip the bits and add 1. Correct me if I am wrong but the binary representation of -1 would be:
1000 0001
The MSB 1 denotes the sign (1 being negative number) and the 1 at the very end is where the 1 comes from.
So when I flip the bits:
0111 1110
So why does my book say that the two's complement representation of -1 is 1111 1111? I assume I am messing up somewhere.
3 Answers
The book has right. 1111 1111 is the representation of -1 in two's complement.
Try add one to 1111 1111. The result is:
1111 1111 +
0000 0001
---------
1 0000 0000
The "one" at the beginning of the result is the "carry bit". The result is your answer: 0. At first glance you would say that overflow has occurred, but not at this time, because the result (the zero) can be represented on 8 bits.
One more example:
If you add -1 to -1 then you should get -2:
1111 1111 +
1111 1111
---------
1 1111 1110
And so on...
So why does my book say that the two's complement representation of -1 is 1111 1111?
It's because the MSB is negative and the rest is positive:
1111 1111 = -128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 => -1
If you flip the bits of 1111 1111 you would get 0000 0000 if you add one then its 0000 0001 = 1 but that would be -1. Conversely if you want to represent -1 then think of it in reverse. 1 is represented 0000 0001 subtract 1 => 0000 0000, then flip it => 1111 1111. And -2 from above. 2 => 0000 0010. Subtract 1 => 0000 0001, flip it => 1111 1110.
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