According to this page, and other similar ones, the rules for detecting overflow when adding signed binary numbers in 2's complement form are the following:

  • If the sum of two positive numbers yields a negative result, the sum has overflowed.
  • If the sum of two negative numbers yields a positive result, the sum has overflowed.
  • Otherwise, the sum has not overflowed.

This, however, appears not to work for 3-bit signed numbers. For example consider adding -2 and -2, +2 is 010 and it's 2's complement is then 110. Then -2+-2 = 110+110 = 1100. When we discard the carry, the sign of the number is negative, because -4 can not be represented as a 3-bit signed number. I am trying to optimize a design for a school assignment and currently the best I can do is just handle the case where it gets to 100 (I just turn the overflow flag on if this happens). Any help is much appreciated.

  • \$\begingroup\$ -4 certainly can be represented as a 3-bit 2's comp number. However, +4 can't. \$\endgroup\$ – Brian Drummond Dec 15 '16 at 14:08
  • \$\begingroup\$ Where do you expect to encounter 3 bit wide 2's complement numbers anyway? \$\endgroup\$ – Lundin Dec 15 '16 at 14:16
  • \$\begingroup\$ It's for a school assignment. I don't know of any practical applications to it. \$\endgroup\$ – rka Dec 15 '16 at 14:19

I did not go to the website to read the text of the rules you listed, but I think there is a semantic error somewhere. First, your assertion that -4 cannot be represented in 2-s complement is incorrect. The range of 3 bit 2's complement numbers is indeed -4 (100)... 3 (011). Second, the detection of overflow is based on extending the carry bit. For adding numbers with the same sign, as long as the extra/ MSB carry bit is the same as the original MSB, an overflow has not occurred....For example

-2 (110) -2 (110) -4 (1100) No overflow

-2 (110) -3 (101) -5 (1011) overflow


I think your premise is wrong, just do the addition like so and throw away the carry out (which by itself is actually the overflow detection in logic terms).

  | 1 1 0 = -2
+ | 1 1 0 = -2
1 | 1 0 0 = -4
  • \$\begingroup\$ If we discard the carry, we are left with 100, Isn't that meaningless as a 3-bit signed number? \$\endgroup\$ – rka Dec 15 '16 at 14:00
  • 1
    \$\begingroup\$ @Enrev, No, 100 is (-4)*1 + (2)*0 + (1)*0 = -4 \$\endgroup\$ – Jotorious Dec 15 '16 at 14:12

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