I am designing a block that performs addition and subtraction with the IC74283. In short, it receives three inputs: \$A\$, \$B\$ and \$subtract\$. When \$subtract\$ is 1, the block performs \$A - B\$ (in 2's complement), otherwise, it will just do \$A + B\$. The outputs are the result of the operation, \$R\$, and \$C_{out}\$.

But I'm still struggling to understand the meaning of \$C_{out}\$ in this context, I can't found a pattern to identify when \$C_{out}\$ is 0 or 1 and what information it gives about the operation/result. Is it just an overflow indication, or does it give me a hint about the sign of the result?

  • \$\begingroup\$ You are wrong about the function of the 74283. It's not a "subtract" input you have there but a carry input \$C_{in}\$. For subtraction, it has to be set in the lowest stage to produce a correct result. In higher stages, it has to be connected to \$C_{out}\$ of the previous stage. For the most significant stage, \$C_{out}\$ indeed denotes an overflow. \$\endgroup\$ – Janka Mar 17 '17 at 22:29
  • \$\begingroup\$ Yes, I just gave a name to it. There is a block that does the 1's complement of \$B\$ and the \$subtract\$ is connected to \$C_{in}\$, so I can get the 2's complement. \$\endgroup\$ – Vinícius Lopes Simões Mar 17 '17 at 22:44

In 2's-complement (signed) arithmetic, the MSB of each number is its sign (1 indicates negative values).

Therefore, you have four possibilities for the result of an addition (or subtraction) operation:

  • Cout = 0, MSB = 0 — normal positive result (no overflow)
  • Cout = 0, MSB = 1 — positive overflow
  • Cout = 1, MSB = 1 — normal negative result (no overflow)
  • Cout = 1, MSB = 0 — negative overflow

In other words, as long as Cout and the MSB match, you have a good result, no overflow. Whenever they don't match, you experienced an overflow.


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