The key here is that you want your macro to return an expression, not
necessarily do the computation itself. So instead of
macro my(exp)
for i in exp.args
println("the arg is ", eval(i))
end
end
You want something like:
macro my(exp)
quote
for i in $(exp.args)
println("the arg is ", i)
end
end
end
Which gives you:
In [21]: @my quote
a = 2
b = 3
end
the arg is begin # In[21], line 2:
a = 2 # line 3:
b = 3
end
On Tue, Sep 9, 2014 at 12:42 PM, <[email protected]> wrote:
> Just puzzling over this simple problem I'm having while learning about
> macros. Here's an expression:
>
> julia> e = quote
> a = 2
> b = 3
> end
>
> quote # none, line 2:
> a = 2 # line 3:
> b = 3
> end
>
>
> If I go through this simply, I'll get a crack at each element of the args
> array:
>
>
>
> julia> for i in e.args
> println("the arg is ", i)
> end
>
>
>
> the arg is # none, line 2:
> the arg is a = 2
> the arg is # line 3:
> the arg is b = 3
>
>
> If I try to write a macro:
>
>
> julia> macro my(exp)
> for i in exp.args
> println("the arg is ", eval(i))
> end
> end
>
> and call it like this:
>
>
> julia> @my :e
> the arg is begin # none, line 2:
> a = 2 # line 3:
> b = 3
> end
>
>
> it does all the elements at once.
>
> It's probably a simple thing, but I could do with a hint!
>
> cheers
>
>
