New submission from Terry J. Reedy <tjre...@udel.edu>:
font.Font.__init__, font.families, and font.names have a 'root=None' argument
and start with
if not root:
root = tkinter._default_root
But font.nametofont does not, and so it calls Font without passing a root
argument:
return Font(name=name, exists=True)
Font fails if there is no default root. There cannot be one if, as
recommended, one disables it.
import tkinter as tk
from tkinter import font
tk.NoDefaultRoot()
root = tk.Tk()
font.nametofont('TkFixedFont')
# AttributeError: module 'tkinter' has no attribute '_default_root'
Proposed fix: add 'root=None' parameter to nametofont (at end, to not break
code) and 'root=root' to Font call.
----------
components: Tkinter
messages: 333532
nosy: serhiy.storchaka, terry.reedy
priority: normal
severity: normal
stage: needs patch
status: open
title: Tkinter font nametofont requires default root
type: behavior
versions: Python 3.8
_______________________________________
Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue35728>
_______________________________________
_______________________________________________
Python-bugs-list mailing list
Unsubscribe:
https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
