将Clojure宏作为函数 [英] Treat Clojure macro as a function
问题描述
如何使一个Clojure宏作为一个函数,所以我可以传递它作为参数的例子?
我不会期望被封装的版本与原始宏的行为完全一样(名称的调用与调用的差异)值),但在一些不重要的情况下会很有用。
将平方函数的这个(愚蠢)实现视为一个宏:
(defmacro square [x]
(list * xx))
直接作为参数传递不会工作,因为你知道:
user = > (map square [1 2 3 4 5])
java.lang.Exception:不能取宏的值:#'user / square(NO_SOURCE_FILE:8)
但是将它包装在一个函数中将会做到:
user => (map#(square%)[1 2 3 4 5])
(1 4 9 16 25)
或者(和更多的邪恶),你可以使另一个宏做一个更通用的包装:
(defmacro make-fn [m]
`(fn [& args#]
(eval`(〜'〜m〜@ args#))))
user => (map(make-fn square)[1 2 3 4 5])
(1 4 9 16 25)
b $ b我会坚持正常的函数包装,并跳过这最后一次黑客! :)
How can I cause a Clojure macro to act as a function, so I can pass it as an argument for example? I would expect to have to wrap it somehow.
I would not expect the wrapped version to behave exactly the same as the original macro (differences of call by name vs call by value), but it would be useful in a few cases where this wasn't important.
解决方案If I'm understanding you correctly, you can just wrap it in a function.
Consider this (silly) implementation of a squaring function as a macro:
(defmacro square [x] (list * x x))
Passing it directly as an arg won't work, as you know:
user=> (map square [1 2 3 4 5]) java.lang.Exception: Can't take value of a macro: #'user/square (NO_SOURCE_FILE:8)
But wrapping it in a function will do the trick:
user=> (map #(square %) [1 2 3 4 5]) (1 4 9 16 25)
Alternatively (and quite a bit more evil-ly), you could make another macro to do a more generic wrapping:
(defmacro make-fn [m] `(fn [& args#] (eval `(~'~m ~@args#)))) user=> (map (make-fn square) [1 2 3 4 5]) (1 4 9 16 25)
I'd stick with the normal function wrapping, and skip this last hack! :)
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