Documentation

The claim at:

Consider the following example:

>>> bool(re.search(string=b"a\\b",pattern=b"[\\\n\r]"))
False

My expectation would be that after backslash-unescaping the b"…"-string, pattern is assigned the sequence of:

literal \, the line-feed "character", the carriage-return "character"

If it would be true, that "Special characters lose their special meaning inside sets.", then the resolved \ in the unescaped pattern should match the one in my test string b"a\\b", however it does not.

I guess what Python actually "sees" is:

backslash-escaped line-feed "character", the carriage-return "character"

which probably effectively yields:

the line-feed "character", the carriage-return "character"

Now you could argue that the \ is not considered a special-character for the terms of the regular expression syntax... but it is, at least already because of:

Also, even the section that explains […] mentions the escaping functionality of it:

I think:

• \ is exempt from this
• whether or this is only the case for characters that are actually special with respect to the RE bracket expression, i.e. [0\-9] is 0, - and 9, because the - was special in that position. But what about [\-9]? Here, the - would not have been special, so it the result \, - and 9 or just - and 9?
• or whether this is simply the case for any character following the \ ... ones that are special outside and RE bracket expression, like \$, \D. \w or \number... and/or ones that are never special, like \ü.

Thanks,
Chris.

Linked PRs

• gh-106482: Clarify documentation of character set in RE #106517
• [3.13] gh-106482: Clarify documentation of character set in RE (GH-106517) #132365