Issue 33410: Using type in a format with padding causes TypeError (original) (raw)

Issue33410

Created on 2018-05-02 14:22 by alexomics, last changed 2022-04-11 14:59 by admin. This issue is now closed.

Messages (3)
msg316072 - (view)	Author: Alex (alexomics)	Date: 2018-05-02 14:22
When trying to print a type in a formatted string with padding TypeError is raised. See examples below. These work: >>> a = 'abc' >>> print('{a}'.format(a=type(a))) <class 'str'> >>> print('{a}'.format(a=str(type(a)))) <class 'str'> These don't: >>> print('{a: >10}'.format(a=type(a))) Traceback (most recent call last): File "", line 1, in TypeError: unsupported format string passed to type.__format__ >>> t = type(a) >>> print('{a: >10}'.format(a=t)) Traceback (most recent call last): File "", line 1, in TypeError: unsupported format string passed to type.__format__
msg316074 - (view)	Author: Anilyka Barry (abarry) *	Date: 2018-05-02 14:33
`type` has a default `__format__` implementation which doesn't accept any formatting options. This is expected behaviour.
msg316081 - (view)	Author: Eric V. Smith (eric.smith) *	Date: 2018-05-02 17:06
The problem is that type.__format__ doesn't exist, so object.__format__ is being called, and it throws an error if you provide a format spec. This is done for future expansion: if we do want to add type.__format__ in the future, we don't have to worry about existing cases that are using format specs that might not work with the new type.__format__. No format spec is the same as calling str() on the argument and returning that, which is what is happening in your working examples. If you want to apply a str formatting spec, you should covert the argument to a str first, using either !s or str(): >>> print('{a!s: >10}'.format(a=type(a))) <class 'str'> >>> print('{a: >10}'.format(a=str(type(a)))) <class 'str'>