Why I can't extend bool in Python?
Question:
>>> class BOOL(bool):
... print "why?"
...
why?
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: Error when calling the metaclass bases
type 'bool' is not an acceptable base type
I thought Python trusted the programmer.
Answers:
Here is a post that explains the reasoning behind the decision: http://mail.python.org/pipermail/python-dev/2004-February/042537.html
The idea is that bool has a specific purpose – to be True or to be False, and adding to that would only serve to complicate your code elsewhere.
Because bool is supposed to only have two values — True and False. If you were able to subclass bool, you could define arbitrary numbers of values for it, and that’s definitely not what you want to happen.
A better question is: why do you want to extend bool?
Guido’s take on it:
I thought about this last
night, and realized that you shouldn’t
be allowed to subclass bool at all! A
subclass would only be useful when it
has instances, but the mere existance
of an instance of a subclass of bool
would break the invariant that True
and False are the only instances of
bool! (An instance of a subclass of C
is also an instance of C.) I think
it’s important not to provide a
backdoor to create additional bool
instances, so I think bool should not
be subclassable.
Reference: http://mail.python.org/pipermail/python-dev/2002-March/020822.html
If you are using Python 3, and you want to have a class that can be evaluated as a boolean, but also contain other functionality, implement __bool__ in your class.
In Python 2 the same effect can be achieved by implementing __nonzero__ or __len__ (if your class is a container).
Since the OP mentions in a comment:
I want 1 and 2 to return an instance
of my class.
I think it’s important to point out that this is entirely impossible: Python does not let you alter built-in types (and, in particular, their special methods). Literal 1 will always be an instance of built-in type int, and in any case the basic semantics of the and operator are not overridable anyway — a and b is always identical to b if a else a for any a and b (no bool coercion involved, even though the OP appears to mistakenly believe one is happening).
Restating this crucial point: the value of a and b is always, unchangeably either a or b — there is no way to break this semantic constraint (even if a and b were instances of your own peculiar classes — even less so of course when they’re constrained to be instances of Python’s built-in int!-).
>>> class BOOL(bool):
... print "why?"
...
why?
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: Error when calling the metaclass bases
type 'bool' is not an acceptable base type
I thought Python trusted the programmer.
Here is a post that explains the reasoning behind the decision: http://mail.python.org/pipermail/python-dev/2004-February/042537.html
The idea is that bool has a specific purpose – to be True or to be False, and adding to that would only serve to complicate your code elsewhere.
Because bool is supposed to only have two values — True and False. If you were able to subclass bool, you could define arbitrary numbers of values for it, and that’s definitely not what you want to happen.
A better question is: why do you want to extend bool?
Guido’s take on it:
I thought about this last
night, and realized that you shouldn’t
be allowed to subclass bool at all! A
subclass would only be useful when it
has instances, but the mere existance
of an instance of a subclass of bool
would break the invariant that True
and False are the only instances of
bool! (An instance of a subclass of C
is also an instance of C.) I think
it’s important not to provide a
backdoor to create additional bool
instances, so I think bool should not
be subclassable.
Reference: http://mail.python.org/pipermail/python-dev/2002-March/020822.html
If you are using Python 3, and you want to have a class that can be evaluated as a boolean, but also contain other functionality, implement __bool__ in your class.
In Python 2 the same effect can be achieved by implementing __nonzero__ or __len__ (if your class is a container).
Since the OP mentions in a comment:
I want
1 and 2to return an instance
of my class.
I think it’s important to point out that this is entirely impossible: Python does not let you alter built-in types (and, in particular, their special methods). Literal 1 will always be an instance of built-in type int, and in any case the basic semantics of the and operator are not overridable anyway — a and b is always identical to b if a else a for any a and b (no bool coercion involved, even though the OP appears to mistakenly believe one is happening).
Restating this crucial point: the value of a and b is always, unchangeably either a or b — there is no way to break this semantic constraint (even if a and b were instances of your own peculiar classes — even less so of course when they’re constrained to be instances of Python’s built-in int!-).