I tried a second approach where instead of keeping a function around I keep
just the symbol. I have a `transform` function that applies the function to
the resulting values. However, this only moves the problem to a new place.
In my `transform` method I have:
eval(Expr(:call, action, values))
where `action` is the symbol that I am now carrying around. The `eval`
still works in the module's namespace, so I get the same error.
If Julia had a way to access the namespace from the calling function, this
could be made to work. Or for that matter, if I could get the calling
namespace from the macro, I could then just get the macro that way.
I'm guessing Julia doesn't currently have this functionality, but would it
be something possible to add?
On Friday, August 1, 2014 12:22:46 PM UTC-4, Abe Schneider wrote:
>
> I think the problem is that I'm not accessing it directly through the
> macro parameters. The call:
>
> @grammar foo begin ... end
>
> passes just the single expression block to `foo`. Thus, I'm getting the
> function from the resulting Expr tree. Even if I quote it, it still ends up
> as an expression that contains a symbol that must be evaluated (unless I'm
> missing something...)
>
>
> On Friday, August 1, 2014 11:42:52 AM UTC-4, Simon Kornblith wrote:
>>
>> That looks like the output of :(esc(fn)), but you don't want to quote
>> that, you want to evaluate it when generating the code, i.e. use
>> $(esc(fn)) wherever you were previously using $fn
>>
>> On Friday, August 1, 2014 11:38:44 AM UTC-4, Abe Schneider wrote:
>>>
>>> That's correct, I'm generating code that keeps a pointer to that
>>> function to call later on.
>>>
>>> If I do that, I have the type `esc(fn)` in the macro (this is what I get
>>> when printing it out), and taking a dump of that argument gives:
>>>
>>> Expr
>>> head: Symbol call
>>> args: Array(Any,(2,))
>>> 1: Symbol esc
>>> 2: Symbol fn
>>> typ: Any
>>>
>>>
>>> Does that mean instead of carrying around a function I will need to
>>> carry around the Expr?
>>>
>>> Thanks!
>>>
>>> On Friday, August 1, 2014 11:17:50 AM UTC-4, Simon Kornblith wrote:
>>>>
>>>> Assuming you're generating code that calls fn, as opposed to trying to
>>>> call it when generating code in the macro (usually a bad idea), you should
>>>> use esc(fn) in place of fn
>>>>
>>>> On Friday, August 1, 2014 10:50:15 AM UTC-4, Abe Schneider wrote:
>>>>>
>>>>> I've come across a problem where I have macro to define a grammar
>>>>> (similar to how PEGParser currently works):
>>>>>
>>>>> @grammar foo begin
>>>>> rule[fn] = "some" + ("rule" | "test")
>>>>> end
>>>>>
>>>>> where the `[fn]` next to the rule defines a function to call on the
>>>>> results (in this case it's an expansion). The issue is that parsing the
>>>>> Expr tree, `fn` is given as a Symbol (which makes sense).
>>>>>
>>>>> However, if I try to turn `fn` into a function I run into the
>>>>> namespace issue I've had previously. If `fn` is defined in my module, it
>>>>> works without problem. If it's defined in the code that imports the
>>>>> module,
>>>>> it will not work because that function doesn't exist in the namespace of
>>>>> the module.
>>>>>
>>>>> I'm guessing there isn't an easy solution to fix this problem, but I
>>>>> thought I'd check to see if someone had an idea.
>>>>>
>>>>> Thanks!
>>>>>
>>>>>
