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I have a problem of adding two numbers in base of two's complement (6 bits!!!)

1100(2's C) + 0101(2's C)

I notice that the first number is starting with 1 which means it's negative but since it's 6 bits, I have to change those two numbers into 6 bits and I have no clue how to find those numbers in 6 bits..

I need help please

Thank You

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  • 1
    \$\begingroup\$ Has your professor covered sign-extension yet? \$\endgroup\$
    – AaronD
    Commented Feb 23, 2015 at 20:53
  • \$\begingroup\$ No never heard of them \$\endgroup\$
    – Jack
    Commented Feb 23, 2015 at 20:54
  • \$\begingroup\$ Just have to use the two's complement \$\endgroup\$
    – Jack
    Commented Feb 23, 2015 at 20:54
  • 1
    \$\begingroup\$ Okay. Google it then. Now that you have the term, it should solve your problem. \$\endgroup\$
    – AaronD
    Commented Feb 23, 2015 at 20:55
  • \$\begingroup\$ Just extend it to the left as much as needed, replicating the leftmost bit (sign bit). \$\endgroup\$
    – Eugene Sh.
    Commented Feb 23, 2015 at 20:56

1 Answer 1

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1100 in four bits is -4

extending this to 6 bits, do sign extension to the left, adding 1's because the left bit above is 1, and get 111100

this is still -4 in two's complement

0101 is 5 in decimal

extending it to the left, adding 0's, because the left bit is 0, and get 000101

adding these two together:

 111100
 000101
 ======
1000001

the carry bit (first one on the left) is discarded.

The result then is just 1, which matches adding in decimal, -4 + 5 = 1.

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  • \$\begingroup\$ The truncated left most bit in your calculation result is not overflow. It's Carry (out). \$\endgroup\$
    – Paebbels
    Commented Feb 23, 2015 at 23:24

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