Created on 2019-01-12 21:33 by terry.reedy, last changed 2020-12-25 21:18 by serhiy.storchaka.
| Pull Requests | |||
|---|---|---|---|
| URL | Status | Linked | Edit |
| PR 23885 | merged | desmondcheongzx, 2020-12-22 05:36 | |
| Messages (2) | |||
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| msg333532 - (view) | Author: Terry J. Reedy (terry.reedy) *  |
Date: 2019-01-12 21:33 | |
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.
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| msg383763 - (view) | Author: Serhiy Storchaka (serhiy.storchaka) * |
Date: 2020-12-25 21:18 | |
New changeset 36a779e64c580519550aa6478c5aa8c58b8fa7b6 by Desmond Cheong in branch 'master': bpo-35728: Add root parameter to tkinter.font.nametofont() (GH-23885) https://github.com/python/cpython/commit/36a779e64c580519550aa6478c5aa8c58b8fa7b6 |
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| History | |||
|---|---|---|---|
| Date | User | Action | Args |
| 2020-12-25 21:18:09 | serhiy.storchaka | set | messages: + msg383763 |
| 2020-12-22 05:36:13 | desmondcheongzx | set | keywords:
+ patch nosy: + desmondcheongzx pull_requests: + pull_request22744 stage: needs patch -> patch review |
| 2020-12-21 09:25:23 | serhiy.storchaka | set | keywords:
+ easy versions: + Python 3.10, - Python 3.8 |
| 2019-01-12 21:33:32 | terry.reedy | create | |