For demonstration purposes, this section gives a few usage examples of some hash table procedures, together with some explanation what they do.
First we start by creating a new hash table with 31 slots, and populate it with two key/value pairs.
(define h (make-hash-table 31)) ;; This is an opaque object h ⇒ #<hash-table 0/31> ;; Inserting into a hash table can be done with hashq-set! (hashq-set! h 'foo "bar") ⇒ "bar" (hashq-set! h 'braz "zonk") ⇒ "zonk" ;; Or with hash-create-handle! (hashq-create-handle! h 'frob #f) ⇒ (frob . #f)
You can get the value for a given key with the procedure
hashq-ref, but the problem with this procedure is that you
cannot reliably determine whether a key does exists in the table. The
reason is that the procedure returns
#f if the key is not in
the table, but it will return the same value if the key is in the
table and just happens to have the value
#f, as you can see in
the following examples.
(hashq-ref h 'foo) ⇒ "bar" (hashq-ref h 'frob) ⇒ #f (hashq-ref h 'not-there) ⇒ #f
Better is to use the procedure
hashq-get-handle, which makes a
distinction between the two cases. Just like
procedure returns a key/value-pair on success, and
#f if the
key is not found.
(hashq-get-handle h 'foo) ⇒ (foo . "bar") (hashq-get-handle h 'not-there) ⇒ #f
Interesting results can be computed by using
hash-fold to work
through each element. This example will count the total number of
(hash-fold (lambda (key value seed) (+ 1 seed)) 0 h) ⇒ 3
The same thing can be done with the procedure
can also count the number of elements matching a particular predicate.
For example, count the number of elements with string values:
(hash-count (lambda (key value) (string? value)) h) ⇒ 2
Counting all the elements is a simple task using
(hash-count (const #t) h) ⇒ 3