Hmm, it does look neater with destructuring i have to admit, but it
won't handle nested parens will it? basically we 're doing the same
thing - mine is just more verbose and self-explanatory. what to do next
though? how do you expose all nested parens to the reader and feed the
result value back to the list in order to get the effect of 'parenthesis
first'?
Jim
ps: btw thanks for your time :)
On 02/05/12 23:14, Sam Ritchie wrote:
Hey Jim, what do you think of something like this?
(defmacro infix->prefix [form]
(loop [[a op b & more :as form] form]
(if (not op)
a
(recur (cons (list op a b) more)))))
This transforms the list by plucking off three items at a time,
rearranging them into prefix notation then popping them back onto the
list in the recursion. When only one form's left inside the list, the
macro returns that form, passing it into the compiler. Here's an
example expansion:
user> (infix->prefix (1 * 2 + 3 - 6))
(1 * 2 + 3 - 6)
((* 1 2) + 3 - 6)
((+ (* 1 2) 3) - 6)
((- (+ (* 1 2) 3) 6))
-1
On Wed, May 2, 2012 at 1:24 PM, Jim - FooBar(); <jimpil1...@gmail.com
<mailto:jimpil1...@gmail.com>> wrote:
Hey everyone,
I've been trying all morning (more than 3 hours) to improve my
macro but with little success...basically what I had before i
started fiddling with it was a macro which would take an
expression of the form (1 + 4 * 5 - 1 / 2) and would return the
expression in prefix form: (/ (- (* (+ 1 4) 5) 1) 2)...it would
not handle nested parens and no operator precedence...This is it:
---------------------------------------------------------------------------------------------------------------------------------------
(defmacro functionize [macro]
`(fn [& args#] (eval (cons '~macro args#))))
(defmacro infix-to-prefix [expr]
(if-not (empty? (filter #(list? %)) '~expr))
(infix-to-prefix
(map #(if (list? %)
(apply (functionize infix-to-prefix) %) %) 'expr))
`(loop [
ops# (filter #(not (number? %)) '~expr)
args# (filter #(number? %) '~expr)
]
(if (empty? ops#) (first args#)
(let [[a# b# & more#] args#]
(recur
(rest ops#)
(conj more#
(list (first ops#) a# b#))))))))
;;usage : (infix-to-prefix (1 + 4 * 6))
;;=> (* (1 + 4) 6)
--------------------------------------------------------------------------------------------------------------------------------------
I know for a fact that the bit after the back-quote works exactly
as explained above...all the new code is above the back-quoted
loop - recur form. all i'm trying to do is replace all the parens
at each level with values calculated recursively using eval on the
returned expression....However i can't seem to go any further...I
am literally about to give up...I know this may not be necessarily
the best approach but i 've worked hard on this and i want to
see it succeed!
Any help is greatly appreciated...
Jim
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