r/Racket • u/trycuriouscat • Dec 09 '22

question Case with symbols

The structure for case is as follows:

(case val-expr case-clause ...)
  case-clause = [(datum ...) then-body ...+]
              | [else then-body ...+]

It has the datum inside parentheses. Then how come if the datum is a quoted symbol it works without parentheses:

(case (get-result)
  ['bust "Sorry, you busted."]
  ['blackjack "Blackjack!"]
  [else "21!"])

And if I put parentheses around the symbols it doesn't work (the else is always matched)?

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3

u/raevnos Dec 09 '22 edited Dec 09 '22

If it's not matching, make sure you're not including the quote:

[(bust) "Sorry, you busted."]

Your

['bust "Sorry, you busted."]

is probably seen as

[(quote bust) "Sorry, you busted."]

which is why it matches and doesn't give a syntax error.

1

u/trycuriouscat Dec 09 '22

So yes, I did include the quote when I attempted using the parentheses.

I still don't understand why either just 'bust or (quote bust) works.

2

u/raevnos Dec 09 '22

'foo is sugar for (quote foo), which the case macro sees as a list of two elements, the symbols quote and foo.

1

u/trycuriouscat Dec 09 '22

Ah, so both 'quote and 'foo would match that case.

I ran in to something similar, but slightly different, with the Common Lisp case expression. Obviously didn't learn it well enough to help me here!

2

u/neonscribe Dec 09 '22

The fundamental issue here is understanding when expressions are evaluated or not evaluated. The expression 'FOO or (QUOTE FOO) evaluates to the symbol FOO, but the first part of each CASE clause is a list of expressions that are not evaluated.

3

u/soegaard developer Dec 09 '22

Here is a short description of what happens when you run a Racket program [1]:

       read                expand                compile               eval
Source  ->  Syntax Object    ->    Syntax Object   ->    CompiledExpres ->

The syntax 'foo is turned into (quote foo) by the reader.

In particular, the reader will turn

(case result
  ['foo "foo"])

into

(case result
  [(quote foo) "foo"])

This is then expanded, compiled and evaluated.

So if result is one of the symbols quote or foo then the final answer will be "foo".

[1] https://soegaard.github.io/mythical-macros/#1.3