Hi Laurent,
Thanks for your detailed explanation! It greatly helped me understand
the usage of quoting.
Stefan
On Sep 10, 5:27 pm, Laurent PETIT <laurent.pe...@gmail.com> wrote:
> 2010/9/10 Stefan Rohlfing <stefan.rohlf...@gmail.com>:
>
>
>
>
>
>
>
>
>
> > @ Nicolas and ajuc
>
> > Thank you very much for showing me where I went wrong! With so many
> > parentheses it sometimes happens that I misplace one
>
> > Now the output of the function is as expected:
>
> > (defn prefix->postfix [expr]
> > (if (coll? expr)
> > (let [ [op arg1 arg2] expr]
> > [ (prefix->postfix arg1) (prefix->postfix arg2) op]) ;; <--
> > removed on parenthesis here
> > expr))
>
> > (prefix->postfix '(+ 2 (* 3 2)))
> > ;; --> [2 [3 2 *] +]
>
> > There is just one point I still don't quite understand. That is,
> > during one of the recursive calls, the expression (* 3 2) is passed as
> > an argument to prefix->postfix:
>
> > (prefix->postfix (* 3 2))
>
> > As prefix->postfix is a normal function and not a macro, (* 3 2)
> > should get evaluated BEFORE being passed to the function. However,
> > this is not the case here.
>
> > Could it be that because (* 3 2) is quoted, because the initial
> > argument to the function, '(+ 2 (* 3 2)), was quoted?
>
> > Could it be that because the initial argument to prefix->postfix, '(+
> > 2 (* 3 2)), is quoted, (* 3 2) is also quoted and therefore does not
> > get evaluated?
>
> Yes. '(+ 2 (* 3 2)) being quoted means that the function receives a
> datastructure : a list composed of the symbol +, the number 2, and
> finally a list composed of the symbol *, the number 3 and the number
> 2.
> ' is a syntactic sugar for the special form named quote , whose
> purpose is to not evaluate everything which is quoted, and just keep
> as a datastructure what the reader has read.
>
> And now, food for thought :-) :
>
> user=> (quote (defn x))
> (defn x)
>
> ;; => See, no macroexpansion, just what the reader read
>
> user=> (read-string "(defn x)")
> (defn x)
>
> ;; => I prove the point of above: just what the reader read
>
> user=> (quote (defn ^{:foo :bar} x))
> (defn x)
>
> ;; => but in the above ^^^, where's :foo :bar ?
>
> user=> (meta (second (quote (defn ^{:foo :bar} x))))
> {:foo :bar}
>
> ;; => it's there, but the reader has already interpreted the ^
> (because it is a reader macro) and placed {:foo :bar} as the metadata
> map for the symbol x
>
> HTH,
>
> --
> Laurent
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