The lambda example works, but couldn't you do:
function f(x)
eval(:($x + 1))
end
?
You can still do that in a macro:
macro m()
quote
ex = :($x + 1)
eval(ex)
end
end
function f(x)
@m
end
f(1) == 2
What would be a situation where this wouldn't work?
Em terça-feira, 27 de setembro de 2016 09:36:48 UTC-3, Jussi Piitulainen
escreveu:
>
> You might be able to wrap your expression so as to create a function
> instead, and call the function with the values of the variables that the
> actual expression depends on. In Python, because I haven't learned to
> construct expressions in Julia yet and don't have the time to learn it now:
>
> def f(x): return eval("lambda x: x + 1")(x)
>
>
>
> tiistai 27. syyskuuta 2016 12.28.40 UTC+3 Marius Millea kirjoitti:
>>
>> Hi, is there a way to "eval" something in the current scope? My problem
>> is the following, I've written a macro that, inside the returned
>> expression, builds an expression which I need to eval. It looks like this,
>>
>> macro foo()
>> quote
>> ex = ...
>> eval_in_current_scope(ex)
>> end
>> end
>>
>> Now, you might say I'm using macros wrong and I should just be doing,
>>
>> macro foo()
>> ex = ...
>> end
>>
>>
>> but in this case when I build "ex", it needs to occur at runtime since it
>> depends on some things only available then. So is there any way to go about
>> this? Thanks.
>>
>>
