deftype failing in lein repl what am I doing wrong?

Question

Per http://www.assembla.com/spaces/clojure/wiki/Datatypes

I should be able to type the following into a lein reply:

(deftype Bar [a b c d e])

(def b (Bar 1 2 3 4 5))

However when I do I get the following output:

java.lang.Exception: Expecting var, but Bar is mapped to class user.Bar (NO_SOURCE_FILE:31)

I'm confused and am a complete newb to clojure all help is appreciated!

NOTE: Tried same code in standard clojure repl and get same problem.

ANSWER: Well I answered my own question with a little additional searching. Turns out the sample was bad. The correct way to instantiate Bar would be:

(def b (Bar. 1 2 3 4 5))

The . at the end of Bar in that usage is important. Still don't quite understand why (so you clojure experts please elaborate if you have time since I would like to know the details ;) ).

Thanks everyone!

Answer 1

I'm not that familiar with deftype, but from what I see you need a point to instantiate a type, try this:

(deftype Bar [a b c d e])

(def b (Bar. 1 2 3 4 5))

Note it's not Bar, but Bar.. See examples e.g. here:

• http://pastebin.com/9vqPpebT
• http://onclojure.com/2010/08/26/reusable-method-implementations-for-deftypes/ (which uses (new foo ...) instead of (foo. ...))

This thread seems to indicate this was a change in deftype:

• http://osdir.com/ml/clojure/2010-04/msg00526.html

or, to put it the other way around, the docs on the page you link to seem outdated.

Hope this helps.

Answer 2

There are two ways of achieving what you are attempting to do. First, let's go over the deftype example.

user=> (deftype Bar [a b c d e])
user.Bar
user=> (def b (Bar. 1 2 3 4 5))
#'user/b
user=> (:a b)
nil
user=> (.a b)
1

You'll notice a few things, here. As was mentioned, you need the "." to instantiate your type. Furthermore, you don't get keyword access (":" notation), only field access (again, with a ".").

The other method is by defining a record with defrecord:

user=> (defrecord Bar [a b c d e])
user.Bar
user=> (def b (Bar. 1 3 5 7 9))
#'user/b
user=> (:a b)
1
user=> (.a b)
1

This gives you both field and keyword access. This access is easily nested, as well, should you decide to have one of your fields contain another record.