I think you're missing an important part:
We're saying that you can take your third, private helper macro, and turn
it into a private helper function. If you invoke it at compile time from
the public macros, it will receive the unevaluated forms you want, and be
able to transform the unevaluated S-expression the way you'd like.
Instead of:
(defmacro helper
[s-expr]
(reverse s-expr))
(defmacro p1
[s-expr]
`(foo (helper s-expr) bar))
(defmacro p2
[s-expr]
`(baz ~(helper s-expr) quux))
You would have:
(defn ^:private helper
[s-expr]
(reverse s-expr))
(defmacro p1
[s-expr]
`(foo ~(helper s-expr) bar))
(defmacro p2
[s-expr]
`(baz ~(helper s-expr) quux))
(macroexpand (p1 (a b)) ;=> (foo (b a) bar)
(macroexpand (p2 (c d)) ;=> (baz (d c) quux)
Once you've entered a macro, you're evaluating normal expressions, and you
can call functions normally to manipulate the expressions.
On Tue, Mar 18, 2014 at 12:41 AM, Yoav Rubin <yoavru...@gmail.com> wrote:
> The case is that there are two public macro that call the same private
> helper macro, and that private macro is the one who does the heavy lifting,
> so I can either duplicate that private macro's code to be inside the public
> macros, or make it public and mark it somehow with a "do-not-touch" sign.
>
>
> On Tuesday, March 18, 2014 1:38:59 AM UTC+2, Michael Blume wrote:
>>
>> You don't have the macro generate a call to the private function, you
>> have the macro call the private function directly
>>
>> replace:
>>
>> (defmacro call-self* [x]
>> `(~x ~x))
>>
>> (defmacro call-self [x]
>> `(do
>> (println "calling form " ~(str x) " with itself")
>> (call-self ~x)))
>>
>> with:
>>
>> (defn- call-self* [x]
>> `(~x ~x))
>>
>> (defmacro call-self [x]
>> `(do
>> (println "calling form " ~(str x) " with itself")
>> ~(call-self x)))
>>
>> The function call-self* is still called at compile-time and is called *by
>> the call-self macro*, not the generated client code. Make sense?
>>
>>
>> On Monday, March 17, 2014 10:31:36 AM UTC-7, Yoav Rubin wrote:
>>>
>>> I need to do it, as I need the arguments to remain not evaluated until
>>> they get to that private macro. That private macro does some work on the
>>> arguments before they get evaluated (the arguments themselves are
>>> s-expressions).
>>>
>>> Still, even if it is a private function - how can I call it from a macro
>>> that is called from another namespace?
>>>
>>> On Monday, March 17, 2014 4:19:19 PM UTC+2, James Reeves wrote:
>>>>
>>>> Don't use a private macro: use a function that spits out an
>>>> s-expression.
>>>>
>>>> - James
>>>>
>>>>
>>>> On 17 March 2014 06:02, Yoav Rubin <yoav...@gmail.com> wrote:
>>>>
>>>>> Hi All,
>>>>>
>>>>> I have a namespace that has two macros as part of its public API, and
>>>>> another macro that act as helpers for the public macro
>>>>>
>>>>> (defmacro helper-mac [arg1 arg2 f]
>>>>> ;; do stuff with f , arg1 and arg2
>>>>> )
>>>>>
>>>>> (defmacro m0 [arg1 arg2]
>>>>> (priv-mac arg1 arg2 f1)
>>>>> )
>>>>>
>>>>> (defmacro m1 [arg1 arg2] (
>>>>> (priv-mac arg1 arg2 f2)
>>>>> )
>>>>>
>>>>> f1 and f2 are just two functions.
>>>>>
>>>>> I would like to make the helper macro private (using ^:private), but
>>>>> when I do it and call either m0 or m1 from another namespace, I get an
>>>>> exception saying that helper-mac is private.
>>>>>
>>>>> Is it possible to call from to a macro in another namespace when that
>>>>> macro is calling a private macro in its namespace?
>>>>>
>>>>> Thanks,
>>>>>
>>>>> Yoav
>>>>>
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