Julia macros with multiple return expressions - macros

Question

I got a bit tired of defining functions for different permutations of input types, e.g.
f(x::MyType1, y::MyType2) = x.data + y.data
f(y::MyType2, x::MyType1) = x.data + y.data
so decided to try my hand at a macro that returns both of the above definitions.
I was able to make a macro that switches the argument input, but I can not get it to return more than one function definition.
So this works:
julia> macro argrev(ex)
if (ex.head == :(=)) && (ex.args[1].head == :call)
ex_ = copy(ex)
args = ex_.args[1].args
args[2:3] = args[[3, 2]]
return ex_
end
return ex
end
#argrev (macro with 1 method)
julia> #argrev f(x::Int, y::Float64) = x + y
f (generic function with 1 method)
julia> f(2, 3.5)
MethodError: no method matching f(::Int64, ::Float64)
julia> f(3.5, 2)
5.5
I cannot figure out how to return both ex and ex_, though. This was one of my attempts:
julia> macro argrev1(ex)
if (ex.head == :(=)) && (ex.args[1].head == :call)
ex_ = copy(ex)
args = ex_.args[1].args
args[2:3] = args[[3, 2]]
return quote
ex
ex_
end
else
return ex
end
end
#argrev1 (macro with 1 method)
julia> #argrev1 f(x::Int, y::Float64) = x + y
UndefVarError: ex_ not defined
What's going on with this error, and how can I return two expressions, or somehow else achieve what I'm trying to do here?
Edit: It seems like this is related, but I don't quite see how I should adapt it to my case.

Answer 1

In quote block, you need to interpolate ex and ex_ like
return quote
$ex
$ex_
end