How to patch another package?

General Usage
1

Hi there,

I would like to patch another package. This worked perfectly fine up to julia 1.9 but with 1.10 it broke. See this thread ERROR: Method overwriting is not permitted during Module precompilation. Use `__precompile__(false)` to opt-out of precompilation

Concretely I would like to patch PythonCall’s serialize_py and deserialize_py methods to use dill instead of pickle.

It seems the only option is to make the package __precompile__(false), which is just a bit too much for me. The alternative sometimes suggested is to use __init__, but this does not work because @eval is buggy inside __init__, aka does not work, and you also aren’t allowed to define top-level functions inside __init__.

This is very crucial for the project I am working on. Any help is highly appreciated.

2

Fork and use your own fork?

1 Like
3

I was not thinking about this - yes, sounds like a workaround.

I still would prefer to stay on top of the original package for easier updates.

4

Can you contribute something to the underlying package (a keyword argument or something) that lets you choose the format?

2 Likes
5

Why is it “too much”? Overwriting a method works reliably in Julia, packages with the overwrite just cannot be precompiled.

And btw, even without this __precompile__ line, everything should just work! A scary error is printed, but the methods get overwritten properly. Not sure why it is printed as “error” instead of info or warning…

6

The new function Base.generating_output() in Julia 1.11 will help with this. Base.generating_output() is true if generating a pkgimage cache. You do not watn to perform patching then.

if !Base.generating_output()
    # hacky patching
end

If Julia 1.10, you can define it as follows:

generating_output() = ccall(:jl_generating_output, Cint, ()) != 0
7

__precompile__(false) regards the entire package as far as I understand it, well as all packages dependening on it, which results in bad user experience as the package time to first anything is quite slow

8

Thank you, this sounds interesting. Would this if !Base.generating_output() block be executed?

I mean is it like __init__ in that we have a guarantee that it will run? Or is it more like a normal if in the file where depending one the if clause on part is executed and the other maybe never.

9

Julia is “generating output” when it is “precompiling”. This helps us distinguish between the precompiling state and other states on module loading.

1 Like