Why in python if we use 2 bracket within character class without escape them if match [] but not ] or [ :
>>> re.search(r'[[]]','[]').group()
'[]'
>>> re.search(r'[[]]','[').group()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'NoneType' object has no attribute 'group'
But if we escape them it will work as expected :
>>> re.search(r'[\[\]]','[').group()
'['
>>> re.search(r'[\[\]]',']').group()
']'
An opening bracket [ starts a new character class, whose scope ends as soon as it encounters the first closing bracket - ].
Now, consider your regex:
[[]]
The scope of character class started by first [, ends at 3rd position. So, the regex is nothing but, [ (as in character class) followed by ] (The last character). So, clearly it won't match a [.
But, as soon as you escape the first ], it loses its behaviour of closing the character class. So, in the below regex:
[[\]]
the character class gets closed only by the last ]. So, the regex would match either [, or ].
Note: There isn't any need to escape [ inside a character class, as it doesn't have a special behaviour there. It can't start a new character class inside a character class, also, it can't close the existing character class.
], it consider this as a the end of char class. So we have ([[]=[) + (]=]) which inturn matches[]but fails to match[because it expects another][][](it works because empty character class is not allowed and the first closing bracket is seen as a literal character)