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I have a question about the binary of negative integer number in Mathematica.

In most implementations, negative signed integers are stored in what is called two's complement. The other major way of storing negative signed numbers is called one's complement.

The two's complement of an N-bit number x is defined as 2^N - x. For example, the two's complement of 8-bit 1 is 2^8 - 1, or 1111 1111. The two's complement of 8-bit 8 is 2^8 - 8, which in binary is 1111 1000. This can also be calculated by flipping the bits of x and adding one. For example:

 1 = 0000 0001
~1 = 1111 1110
-1 = ~1 + 1 = 1111 1111

 21 = 0001 0101
~21 = 1110 1010
-21 = 1110 1011

 128 = 1000 0000
~128 = 0111 1111
-128 = 1000 0000

They can also be calculated as:

-1   = 2^8 - 1   = 255 = 1111 1111
-21  = 2^8 - 21  = 234 = 1110 1010
-128 = 2^8 - 128 = 128 = 1000 0000

These results are exactly identical to the results calculated in C language. However, when using Mathematica (I use Mathematica 11.1.0), the results are quite different:

BaseForm[-1, 2]   = -1
BaseForm[-21, 2]  = -10101
BaseForm[-128, 2] = -1000 0000

My question is: How to get the right results using Mathematica? Am I use the WRONG command?

Thanks a lot!

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  • $\begingroup$ Mathematica assumes an infinite two's complement, since there is otherwise no notion of finite bits; see e.g. the docs for BitNot[]. You can use Mod[] if you wish to simulate a finite amount of bits that spill over. $\endgroup$ – J. M. is away Oct 2 '17 at 23:33
  • $\begingroup$ Another note: when you post the same question on another place like Community, it is common courtesy to include a link to the other post, so that you don't waste other folks' time. $\endgroup$ – J. M. is away Oct 3 '17 at 0:04
  • $\begingroup$ @J.M. Thank you for your explanation and nice reminder. I read the document and found there is no implementation of finite two's complement. I have to write it by myself. This is my code: $\endgroup$ – zchenkui Oct 8 '17 at 1:49

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