in clojure language what <'a> really is

Question

actually i am trying to perfectly understand clojure and particularly symbols

(def a 1)
(type a)
;;=>java.lang.Long
(type 'a)
;;=>clojure.lang.Symbol

I know that type is a function so its arguments get evaluated first so i perfectly understand why the code above work this way .In the flowing code i decided to delay the evaluation using macro

 (defmacro m-type [x] (type x))
 (m-type a)
 ;;==>clojure.lang.Symbol

and i am fine with that but what i fail to uderstand is this:

 (m-type 'a)
 ;;=>clojure.lang.Cons

why the type of 'a is a cons

User who voted to close as "Too Broad" can you justify your decision? This has a concrete and simple answer — daniel gratzer, Jan 23 '14 at 18:07

score 6 · Accepted Answer · edited Jan 23 '14 at 18:04

6

the character ' is interpreted by the clojure reader as a reader-macro which expands to a list containing the symbol quote followed by whatever follows the ', so in your call to (m-type 'a) the 'a is expanding to:

user> (macroexpand-1 ''a)
(quote a)

then calling type on the list (quote a) which is a Cons.

This may be a bit more clear if we make the m-type macro print the arguments as it sees them while it is evaluating:

user> (defmacro m-type [x] (println "x is " x) (type x))
#'user/m-type
user> (m-type 'a)
x is  (quote a)
clojure.lang.Cons

edited Jan 23 '14 at 18:04

Vladimir Matveev

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answered Jan 23 '14 at 17:35

Arthur Ulfeldt

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1

I think the OP will still be confused by the result of (type 'a) - the call without Macro - – Chiron Jan 23 '14 at 17:37
1

Thats a good point, there is both macro expansion and quoting at play in that one. – Arthur Ulfeldt Jan 23 '14 at 17:41
1

Most likely he will ask again why (type 'a) is returning clojure.lang.Symbol . But I'm not illusionist! – Chiron Jan 23 '14 at 17:42
2

i am not an OP and i get it – user3228423 Jan 23 '14 at 17:49

in clojure language what <'a> really is

1 Answers1