This can get really complex quite fast. Consider (1 – 2 * 3). If your calculator only supports addition and negation, this is still not trivial. Consider (5 – 5 – 5). If you consider the input as left associative, you’ll get the correct -5, but if you read it as right associative (which is … Read more
Here’s pseudo-code: Initialize stack to empty list For each element e in the expression: if e is a number, push it onto the stack otherwise, call the function corresponding to the operator (you can use an association list or cond to find this), passing the stack as the parameter – The function pops the needed … Read more
It depends. You really should try to narrow you question further with an actual problem in scheme since the answer will depend on what you are trying to do. In idiomatic Scheme most should be done without side effects so you have (if predicate-expression consequent-expression alternative-expression) ;; alternative is optional but should be used anyway. … Read more
Try this: (define (rswap lst) ;; Create a helper function to do the recursive work. (define (helper in out) ;; If the input is not a list, simply return it. ;; There is nothing to be done to rswap it. (if (not (list? in)) in ;; If in is an empty list, simply return the … Read more
For the simplest solution, use apply for this: (define (multiply-all lst) (apply * lst)) If you need to build the procedure from scratch, just remember that the base case (an empty list) should return 1, and the recursive step should multiply the current value using the standard solution template, like this: (define (multiply-all lst) (if … Read more
EDITED: A complete solution has been posted, so I guess it’s ok for me to show you how to write an idiomatic procedure. Notice that number->list uses a technique called tail recursion for efficiently expressing a loop. Although (coming from a Python background) it might be tempting to use an explicit looping construct (as shown … Read more