Getting started with Scheme, and SICP (Structure and Interpretation of Computer Programs) introduces some terms and ways of thinking about code that you don’t always run into while programming in a language like C#.
The first thing that you’re bound to notice are things like the use of parenthesis everywhere, and the use of prefix notation instead of infix notation for performing arithmetic and other operations. These are pretty easy to get used to, although reading code written in Scheme takes some work coming from a more “mainstream” imperative language.
One of the first things that SICP delves into in section 1.1.5 is the topic of the substition model for procedure application. In a nutshell, this is basically talking about how the expressions are evaluated and how the procedure is applied to these expressions. In Scheme, the interpreter follows the applicative-order evaluation model, which means that the arguments to the procedure are evaluated and then the procedure is applied.
SICP also talks about an alternative evaluation technique, known as normal-mode evaluation, in which the arguments are not evaluated before the procedure body is applied.
This early on in the book it doesn’t go too far into the differences between the two, besides mentioning that LISP uses applicative order evaluation partly because of the performance boost of not having to evaluate the expressions multiple times. Normal-mode evaluation can result in the expressions being evaluated multiple times, due to the fact that an expression passed in as an argument may be used further on in multiple expressions in the procedure.
That being said, you can also imagine how an expression passed as an argument may result in significant computation, and if it’s passed into a procedure which may not use the value of that expression, it would be beneficial to not spend the time evaluating it needlessly.
To demonstrate a method which shows that Scheme uses applicative mode evaluation, SICP gives us exersize 1.5, where one of the “characters” in the book, Ben Bitdiddle, writes a procedure to determine whether the interpreter uses applicative or normal-mode evaluation.
The procedure written is shown below:
If you type this into the REPL in DrRacket, and execute the ‘(test 0 (p))’ command, you’ll quickly see that it hangs the REPL. In effect, it loops infinitely due to (p) being evaluated over and over again. This effect is caused by the applicative evaluation mode that Scheme makes use of. Before (test) is applied the arguments have to be evaluated, and in this case evaluating (p) causes the infinite loop, since there is no base case which breaks the recursion in the evaluation
At this point you can probably figure out how this “test” works to see if the interpreter is using applicative or normal-mode evaluation. Had normal-mode evaluation been in use, the argument which evaluates to (p) would never had been evaluated, and the procedure would have returned 0 in this case.
Another example of evaluating normal-mode vs applicative-mode evaluation comes up in exercise 1.20, where the text asks you to evaluate how many times the remainder operation is performed for the given procedure which calculates the GCD of two numbers.
This example shows clearly how normal-mode evaluation can lead to an operation being evaluated many more times then it would be with applicative-mode evaluation.
Bill the Lizard? (name unknonwn) has a great write-up where he shows the process of normal-mode evaluation.
You can check it out here.