With the following scheme, I built myself a 3-bit full adder. What I'm wondering is, and I couldn't figure this out myself, does the adder calculate with signed or unsigned values?

  • Is the range for the 3-bit adder between (-4) and 3, or is it between 0 and 7?

  • Alternatively, is it whatever. As in, can I choose whether I want the binary number 111 to represent (-1) or if I want it to represent 7?

The adder of course can make both addition and subtraction. The scheme I took came from here. The site has a scheme with overflow and subtraction/addition added to the scheme.

The site claims: "When is the result negative? When its most-significant bit is 1. " But I'm still not sure if that must be the ABSOLUTE truth. I'm thinking the range (-4) and 3 is necessary to calculate with subtraction. Am I correct?

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Answer to my problem. The main reason we use 2s complement is in fact the advantage that we can display unsigned and signed integers on the same circuitry.

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Yes, it can be either. It will work with unsigned 3-bit numbers. It will also work with 2s complement signed numbers. You'll need to do a bit of study, but the thing to remember with 2s complement is that a 0 sign bit means positive, and a 1 means negative.

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  • \$\begingroup\$ Sorry to bother you. Found a good source that explained what I was wondering. I attached an explanation to future viewers of this thread. Thank you for your help :) \$\endgroup\$ – B. Lee Feb 17 '15 at 2:30

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