[issue27800] Regular expressions with multiple repeat codes

[R. David Murray]

R. David Murray added the comment:

It seems perfectly logical and consistent to me.  {4} is a repeat count, as is *.  You get the same error if you do 'a?*', and the same bypass if you do '(a?)*' (though I haven't tested if that does anything useful :).  You don't need the ?:, as far as I can tell, you just need to have the * modifying a group, making the group the "preceding regular expression".

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nosy: +r.david.murray

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Python tracker <report at bugs.python.org>
<http://bugs.python.org/issue27800>
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