Next: https://mitpress.mit.edu/sites/default/files/sicp/full-text/book/book-Z-H-11.html#%_sec_1.2
TODO:
Chapter 1.1
- Exercise 1.8
- Exercise 1.9
I'm using mit-scheme interpreter for exercises and will be referring to it as scm to save some keystrokes.
Improve reading experience:
$('body').style["width"] = "50%";
$('body').style["font-family"] = "Verdana";Bookmarklet:
javascript:(function()%7Bvar%20b%3Ddocument.getElementsByTagName(%22body%22)%5B0%5D%3Bb.style.width%3D%2250%25%22%2Cb.style%5B%22font-family%22%5D%3D%22Verdana%22%7D)()
brew install mit-scheme
If you are having problems with scheme interpreter history (e.g., arrow up doesn't show a previous command), install rlwrap:
brew install rlwrap
Run scheme interpreter like so:
rlwrap mit-scheme
Create an alias for convenience:
alias scm='rlwrap mit-scheme'
Exercise 1.5
A test to determine whether the interpreter he is faced with is using applicative-order evaluation or normal-order evaluation:
Define 2 procedures:
(define (p) (p))
(define (test x y)
(if (= x 0)
0
y))Evaluate the expression:
(test 0 (p))If the interpreter is using applicative-order evaluation, it will freeze as procedure p will recurse indefinitely. As we know from the textbook, scm uses applicative-order evaluation, so it is easy for us to test this in the interpreter. Interpreters using normal-order evalution will return 0.
Example in Haskell:
> p = p
> test x y = if x == 0 then 0 else y
> test 0 p
0In exercise 1.6 we are asked what will be if if is defined as a function instead of as a special form, and then previously defined sqrt is re-written using the above new-if.
scm uses applicative-order evaluation, so it has to evaluate all the arguments before passing them into a procedure. This means that it will try to evaluate a recursive branch even when the procedure has reached a base case and should stop the recursion.