defining style via python with different parent than style.default

[svatem1]

I have the following code:

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init offset = -1

style default:
    properties gui.text_properties()
    language gui.language
    color "#ff0000"

style whatever:
    properties gui.text_properties()
    language gui.language
    color "#ff0000"

init 1 python:
    style.example = Style(style.default)

This code works without any problems. If i want to use style.whatever as the parent of my example style:

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init 1 python:
    style.example = Style(style.whatever)

, the code fails with "style whatever doesn't exist" traceback. In the documentation, I found, that style statement is run during init 0 phase, if not stated otherwise. Also, if I delete style default definition and use style.default as the parent, it still works. What exactly is going on here and how can I use my own style as parent?

[Ocelot]

What happens if you use 5 as init priority for Python code?

[enaielei]

Don't do it in python.

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style example is whatever:
   # properties

[svatem1]

What happens if you use 5 as init priority for Python code?

same error

[svatem1]

Don't do it in python.

Code: Select all

style example is whatever:
   # properties

I'll try to stay out of python as much as possible, I just found random example of something that looked like I could use, started modifying it and this was the error I ran into

[Ocelot]

Try Style("whatever") instead. If it fails too, use default style and call .set_patent later.