Racket Binary Addition

Question

I have a question concerning Racket. Given 2 string binary numbers, I would like to obtain their sum in a string binary number. The function should take into account carry bits and a possible overflow. So the function should look like this:

(define (binary-sum a b) ...)

For example:

(binary-sum "010" "001")

"011"

Or:

(binary-sum "0010" "1011")

"1101"

Or in case of overflow:

(binary-sum "1111" "0001")

"overflow"

Please identify any packages I need.

Thank you!

Answer 1

Binary values in Racket are specified using the prefix #b . Using that, we can create the following function:

> (define (binary-sum a b)
  (+ (string->number (string-append "#b" a))
     (string->number (string-append "#b" b))))

> (binary-sum "010" "001")
3

To format the output as binary as well, we can use ~r , like so:

> (define (binary-sum a b)
  (~r (+ (string->number (string-append "#b" a))
      (string->number (string-append "#b" b))) #:base 2))

> (binary-sum "010" "001")
011

And finally, to add your overflow feature, we can simply compare the length of our result as a string to the length of one of the parameters.

(define (binary-sum a b)
  (let ((sum (~r (+ (string->number (string-append "#b" a))
                    (string->number (string-append "#b" b)))
                 #:base 2)))
    (if (> (string-length sum)
           (string-length a))
        "overflow"
        sum)))

Which behaves exactly as your question specifies. Let me know if you have any remaining doubts or confusion. Good luck!

Edit: Just wanted to note that if you ever need to merely add two binary values and don't mind getting an integer back, you can always just do it like so:

> (+ #b010 #b001)
3