I have a user defined module which invokes a method defined in another module from a package. How can I get the name of the invoking user-defined module, so that my package can include/eval code back into the user’s module? Thanks!

[image] kristoffer.carlsson: This is for macros and can be used to see what module the macro is expanded in. Right, and in the rare cases where you need this information in a function, you should often have the caller invoke the function via a macro. For example, define a function foo(m::Mod…

Hard to say more without context, but this is almost certainly a horrible idea, even if implementable. But I would imagine that the answer is not even well defined: if H.h calls G.g, which calls f, should f think it is called from H or G? What if the compiler inlines G.g, which it is free to do by …

I think __module__ is what you are looking for: module M macro g() __module__ end end julia> M.@g() Main current_module vs `@__MODULE__ opened 09:44PM - 27 Jun 18 UTC (UTC) closed 10:11PM - 27 …

[image] jonathanBieler: I think __module__ is what you are looking for: This is for macros and can be used to see what module the macro is expanded in.

Well if you want a function you can just pass the calling module as an argument: module M g(m) = @info "module $m is calling" end module N import ..M f() = M.g(@__MODULE__) end N.f()

[image] Get module of a caller General Usage 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 r…

Thank you all for the feedback! For a bit of context, trying to abstract the use case as much as possible: there’s a user module UserModule and a package Package Package exposes a function x. When Package.x() is invoked from UserModule, Package does a lot of complex processing and generates a…

@kristoffer.carlsson and @stevengj - yes, that’s most likely the way to go about it, I’ll give it a try, thanks for your help!

[image] essenciary: a new function y which is tailor-made for UserModule - and executes the y function (basically x is a wrapper for y ). So y is a function specialised for and determined by the state of UserModule . This function y is also cached in a .jl file and loaded from memory or…

@stevengj In order to avoid that the user does x()() :see_no_evil: I feel that if I force the user to use x()() I expose the implementation. The user shouldn’t care that I return a function which then she needs to invoke. The user wants the result of invoking x, without caring about what I need to …

[image] essenciary: I feel that if I force the user to use x()() I expose the implementation. The user shouldn’t care that I return a function which then she needs to invoke But isn’t that the whole point of x(), to return a function? Can you summarize what you are actually trying to do?

Get the name of the invoking module

General Usage

pfitzseb April 3, 2019, 1:57pm 7

But it’s not a good idea to do this. If you want to do something as intrusive as eval’ing code into a module that’s not yours then do at least require that to be explicitly passed.

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Get the name of the invoking module

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