Python IntFlags behavior for 0 item

Question

In the below example codes which uses IntFlag of enum in Python3, I can understand the first 5 behaviors. But I cannot understand the last 3 behaviors. I think these results should be False . Can anyone elaborate on these behaviors? And any idea how can I define the CLEAR which makes the 3 results 'False'?

In [3]: from enum import IntFlag, auto

In [4]: class Functions(IntFlag):
   ...:     CLEAR=0
   ...:     FUNC1=auto()
   ...:     FUNC2=auto()
   ...:     FUNC3=auto()
   ...:     FUNC12=FUNC1|FUNC2
   ...:     ALL=FUNC1|FUNC2|FUNC3
   ...:

In [5]: Functions.FUNC1 in Functions.FUNC12 #OK
Out[5]: True

In [6]: Functions.FUNC2 in Functions.FUNC12 #OK
Out[6]: True

In [7]: Functions.FUNC3 in Functions.FUNC12 #OK
Out[7]: False

In [8]: Functions.FUNC12 in Functions.ALL #OK
Out[8]: True

In [9]: Functions.ALL in Functions.FUNC12 #OK
Out[9]: False

In [10]: Functions.CLEAR in Functions.ALL #?
Out[10]: True

In [11]: Functions.CLEAR in Functions.FUNC12 #??
Out[11]: True

In [12]: Functions.CLEAR in Functions.FUNC1 #???
Out[12]: True

Answer 1

CLEAR represents the empty set of flags.

The documentation never seems to explicitly mention this, but as far as I understand it, the in operator tests whether the flags on the left-hand side are a subset of the right-hand side.

However, the source code for the Flag.__contains__ method (which is inherited by IntFlag ) supports this understanding – this method implements the operation other in self :

 def __contains__(self, other): """ Returns True if self has at least the same flags set as other. """ if not isinstance(other, self.__class__): raise TypeError( "unsupported operand type(s) for 'in': '%s' and '%s'" % ( type(other).__qualname__, self.__class__.__qualname__)) return other._value_ & self._value_ == other._value_

The empty set is a subset of any other set, therefore Functions.CLEAR in Functions.x is true for any x .

Or, in terms of the concrete bitwise operations that are performed, CLEAR is 0, so CLEAR in x evaluates 0 & x == 0 which is always true.