Understanding One's Complement

Question

I just have a quick question regarding one's complement with binary numbers that maybe someone could help me understand.

I am studying for a computer security test and I am practicing converting decimal numbers to binary numbers and then using one's and two's complement. I am given the decimal number 237 and converted it to 11101101 in binary. I understand that in one's complement you reverse the bits and it is simple. When I reversed it I get 00010010 but the answer leaves out the first three 0's. Any chance someone could explain to me why that is?

Govind Parmar · Answer 1 · 2019-02-07T00:38:03.047

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The value 237 is too large to be stored in a single 8 bit signed byte with one's complement. If it were stored in a 16-bit int instead, it, and its one's complement negations would have the respective representations in base 2:

0000000011101100

1111111100010011

When doing arithmetic like this by hand, keep a few extra digits on the left of your number to account for the sign of the number if you plan to negate the bits later.

edited Feb 07 '19 at 00:38

answered Feb 07 '19 at 00:32

Govind Parmar

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Still a little confused, is it because 237 is larger than 2^7? Also why take out 3 zeros and not just 1? – Cory Freeman Feb 07 '19 at 00:49
Like for example if it was 200, I would only include 6 digit. Just trying to grasp where the cut off is. – Cory Freeman Feb 07 '19 at 00:53
Yes, if the data type holding it has N bits, then at most N-1 bits are available for the magnitude of the number – Govind Parmar Feb 07 '19 at 00:54

score 0 · Answer 2 · answered Jul 13 '19 at 15:31

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Those leading 0's don't hold any significance. Basically, in decimal one can say 07 == 7 == 00000000007, so in binary equally 00010010 == 10010 == 0000000000000010010 holds.

answered Jul 13 '19 at 15:31

Bart Van Loon

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Understanding One's Complement

2 Answers2