Julia : pass string to macro - julia

Question

suppose that I have a macro that is defined as :
macro foomacro(ex::Expr)
dump(ex)
ex
end
Currently I would like to pass my expression as a parsed string so that I may pass a rather complicated and case dependent expression that has been obtained via string concatenation.
However, trying :
#foomacro 1+2+3
gives the expected result 6 whereas
#foomacro parse("1+2+3")
returns the parsed expression :(1+2+3) instead of actually parsing it...
As far as I understand this both macros should be receiving the same expression but this is clearly not the case.
How do I get this MWE to work ?
ps: I figured out this fix but I feel like it is very dirty and "incorrect"
macro foomacro(ex::Expr)
if ex.head == :call
#in this case the user is calling the macro via a parsed string
dump(ex)
return ex
end
dump(ex)
ex
end
ps: if this is of any relevance, currently the code is running on 0.6.4 and if possible I'd rather not update to 1.0 yet since this would postpone my actual project to much...

Answer 1

You're mixing up levels. Let's introduce an intermediate function for clarity:
function foomacro_impl(expr)
dump(expr)
expr
end
macro foomacro(expr)
foomacro_impl(expr)
end
If run, the expression #foomacro <someexpr> will be parsed, the <someexpr> part passed to foomacro_impl, and the result treated as an expression and inserted instead of the original expression. That means that writing #foomacro 1+2+3 is equivalent to having written
let expr = :(1+2+3)
dump(expr)
expr
end
which returns
Expr
head: Symbol call
args: Array{Any}((4,))
1: Symbol +
2: Int64 1
3: Int64 2
4: Int64 3
:(1 + 2 + 3)
an Expr that evaluates to 6.
On the other hand, in #foomacro Meta.parse("1+2+3"), the whole argument, parse("1+2+3"), is used as expr:
julia> let expr = :(Meta.parse("1+2+3"))
dump(expr)
expr
end
Expr
head: Symbol call
args: Array{Any}((2,))
1: Expr
head: Symbol .
args: Array{Any}((2,))
1: Symbol Meta
2: QuoteNode
value: Symbol parse
2: String "1+2+3"
:(Meta.parse("1+2+3"))
So the result of the macro call is the expression Meta.parse("1+2+3"), which evaluates to another expression :(1 + 2 + 3), since it is a call to parse. The two forms are thus not receiving the same expression!
But there are ways to manually parse an expression and pass it to a macro:
You can do as I did, and use a separate "macro implementing function". Then, the expression returned by #foomacro bla is equivalent to foomacro_impl(Meta.parse(bla)). (This approach, BTW, is very useful for testing, and I recommend it most of the times.)
You can use the macro #eval to construct an expression, splice into it, and evaluate it immediately:
julia> #eval #foomacro $(Meta.parse("1+2+3"))
Expr
head: Symbol call
args: Array{Any}((4,))
1: Symbol +
2: Int64 1
3: Int64 2
4: Int64 3
6
(Or similarly, use eval and manually constructed Expr values.)