module A
foo(???) = println("I've been called from $???")
end
module B
import A
A.foo() # should pring "I've been called from B"
end
Is it possible to get this behavior, i.e. find out the caller’s module without requiring it to explicitly pass @__MODULE__? I tired:
foo(; mod=@__MODULE__)
But for obvious reasons it returns the module of foo itself instead of the caller.
julia> module A
function foo()
s = stacktrace()
get(s[2].linfo).def.module
end
end
A
julia> module B
using A
foo() = A.foo()
end
B
julia> B.foo()
B
Obviously needs error handling, probably only works on 0.6, doesn’t quite do what you want (since it needs to be called in a function defined in B) and is generally bad since it reaches into internals, but that is at least one way to get what you want 
Interesting solution, thanks!
For reference, here’s Julia 0.7 version:
module A
function foo()
s = stacktrace()
s[2].linfo.def.module
end
end
module B
using Main.A
foo() = A.foo()
end
B.foo() # ==> Main.B
This does not seem to work for macros, does somebody know how to access this from a macro?
module A
macro foo()
s = stacktrace()
s[end].linfo.def.module
end
end
module B
using Main.A
foo() = A.@foo()
end
B.foo() # WRONG: ==> Main.A
it returns Main.A instead of Main.B, is that a bug?
No, that’s just a consequence of the differences between macros and functions. You want
julia> module A
macro foo()
__module__
end
end
Main.A
julia> module B
using Main.A
foo() = A.@foo()
end
Main.B
julia> B.foo()
Main.B
instead. See here for why that works.