I am trying to do something like this julia> module t a(x)=4 b(x)=x for f in [a,b] export f end end ERROR: syntax: malformed "export" statement Stacktrace: [1] top-level scope at REPL[1]:6245: In my case I have a lot of functions that I have collected in …

You could also consider using nameof to get the name of the function from its type: julia> module t a(x)=4 b(x)=x for f in nameof.([a,b]) @eval export $f end end Main.t julia> using .t julia> a a (generic function with 1 method)

Have you considered this snippet: [image] Exportall? General Usage The following simple snippet seems to work: module Foo foo(x) = x+1 bar(x) = x-1 # exportall: for n in names(current_module(), true) if Base.isidentifier(n) &&…

After some try and error This is considerably less elegant than the previous post, but it goes from [a,b] to [:a,:b] as you planned) module AModule a(x)=4 b(x)=x plusOne(x)=x+1 fn_arr=[a,b,plusOne] fixedLength=length("Main.AModule")+2 for x in fn_arr y=string(x)[fixedLength:end] f=Symbol(y…

Thanks, this is exactly what I wanted. Didn’t know about nameof.

[image] jishnub: Didn’t know about nameof . I learned about it via methodswith: julia> f() = 1 f (generic function with 1 method) julia> typeof(f) typeof(f) julia> typeof(typeof(f)) DataType julia> methodswith(DataType) [1] dump(io::IOContext, x::DataType, n::Int64, indent) in Base at sho…

[image] ffevotte: I learned about it via methodswith : Thank you, thank you so much, I was very annoyed that sometimes I wanted to peruse all methods that took some type of argument and did not believe Julia had a solution like Haskell had (in Haskell case it is a search engine, and so in Jul…

How to export functions from an array in a module?

General Usage

bernhard June 13, 2019, 10:52am 2

Have you considered this snippet:

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How to export functions from an array in a module?

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