Editing Adder–subtractor (section)

==Construction==

[[Image:4-bit ripple carry adder-subtracter.svg|thumb|400px|A 4-bit ripple-carry adder–subtractor based on a 4-bit adder that performs [[two's complement]] on ''A'' when {{nowrap|1=''D'' = 1}} to yield {{nowrap|1=''S'' = ''B'' − ''A''}}.]]

Having an ''n''-bit adder for ''A'' and ''B'', then {{nowrap|1=''S'' = ''A'' + ''B''}}. Then, assume the numbers are in [[two's complement]]. Then to perform {{nowrap|''B'' − ''A''}}, two's complement theory says to invert each bit of ''A'' with a [[NOT gate]] then add one. This yields {{nowrap|1=''S'' = ''B'' + {{overline|''A''}} + 1}}, which is easy to do with a slightly modified adder.

By preceding each ''A'' input bit on the adder with a 2-to-1 [[multiplexer]] where:
* Input 0 (''I''<sub>0</sub>) is ''A''
* Input 1 (''I''<sub>1</sub>) is {{overline|''A''}}
that has control input ''D'' that is also connected to the initial carry, then the modified adder performs
* addition when {{nowrap|1=''D'' = 0}}, or
* subtraction when {{nowrap|1=''D'' = 1}}.
This works because when {{nowrap|1=''D'' = 1}} the ''A'' input to the adder is really {{overline|''A''}} and the carry in is 1.  Adding ''B'' to {{overline|''A''}} and 1 yields the desired subtraction of {{nowrap|''B'' − ''A''}}.

A way you can mark number ''A'' as positive or negative without using a multiplexer on each bit is to use an [[XOR gate]] to precede each bit instead. 
* The first input to the XOR gate is the actual input bit
* The second input for each XOR gate is the control input ''D''
This produces the same [[truth table]] for the bit arriving at the adder as the multiplexer solution does since the XOR gate output will be what the input bit is when {{nowrap|1=''D'' = 0}} and the inverted input bit when {{nowrap|1=''D'' = 1}}.