I am suppose to design a carry look ahead adder with ...

Input flag bits

  • E (Enable)
  • S (Subtraction)


  • The usual sum & carry out
  • O (Overflow)
  • N (Negative)
  • Z (Zero)

But the question is vague ... It says if

  • E = 0, S = 1, do subtraction
  • E = 0, S = 0, do addition
  • then if E = 1? What do I do? What do usual circuits do? I suppose if 0 is enable, 1 is disable, then I put all outputs 0?
  • For overflow bit, I suppose the below, where \$C_{in}, C_{out}\$ are carry in/out of MSB/Sign Bit and \$S\$ is subtract bit. The 1st half checks if Cin is different from Cout (of MSB), when doing subtraction. And the 2nd half checks if theres a carry out when doing addition. Is my implementation correct?

    enter image description here

  • For negative bit I also check if doing subtraction using \$S \cdot MSB\$?


1 Answer 1


Ask your professor if the question is vague. But your assumption that E=1 means disable is probably correct.

  • \$\begingroup\$ I read on and discovered when E = 1, do negation ... my mistake \$\endgroup\$
    – Jiew Meng
    Sep 20, 2011 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.