Adder hardware logic

The standard adder / subtract circuit is given by the figure below.For an explanation of how it works please see my previous answer here : Subtraction using adder circuit

Now notice that if you were to change the input that is tied to '0' to Cin, then the circuit would now perform the operation: A + ~B + (Cin ^ SUB), meaning that:

1) it would be a normal Adder/Subtract circuit when Cin is 0.

2) if Cin = 1 and Sub = 1 it would evaluate to:

A + ~B + (1 ^ 1) = A + ~B = A + ~B + 1 - 1 = A + (~B + 1) - 1 = A - B - 1 ,

This means that the normal Add/Sub circuit can achieve the operation you require, all you need to do is change the input tied to '0' your Cin.

You could also design the desired circuit by using the usual method of writing all your possible inputs into a truth table, then deriving the Boolean equations for them, then simplifying them to get your final circuit equation.

You are exactly right. And any resistor value less than about 10k will work fine, as long as it is more than about 330 ohms (Look at the data sheet for the maximum current into a low output, Iol(max). Assume 15 mA. Then a low output will nominally put 5 volts across the resistor, and R = V/I = 5/.015 = 333.)

2.2 k is a perfectly good value. If you're doing this a lot, you can consider using somewhat higher values to reduce the power wasted in the resistors.