问题
Code:
class Fraction(object):
def __init__(self, num, denom):
self.numerator = num
self.denominator = denom
def main():
f = Fraction(1, 3)
print type(f)
if __name__ == "__main__":
main()
Output:
<class '__main__.Fraction'>
Question:
- Why is the type
__main__.Fraction
instead of justFraction
? Why is there "." between
__main__
andFraction
? "." implies thatFraction
is a sub-class of__main__
. But why? Even if I removeIf __name__ == "__main__"
from the code, I still get the same output:class Fraction(object): def __init__(self, num, denom): self.numerator = num self.denominator = denom f = Fraction(1,3) print type(f) output: <class '__main__.Fraction'>
回答1:
As you have not defined a
__repr__
(or__str__
) on the class, it's inherited the one from the superclass --object
and that how its written there. So, all your class instances are expressed that way. As for the class itself, you need to change the__repr__
/__str__
on the metaclass i.e. the class of which our class in question is an instance of; the default metaclass istype
.__main__
is the name of the module, here as you are directly executing it, its being considered as a script and all scripts have the name__main__
in PythonThere's a
.
in between becauseFraction
is an attribute of the script__main__
, the module; and belongs to the module level scope
Example:
In [47]: class MyMeta(type):
...: def __repr__(cls):
...: return 'Whatever...'
...:
In [48]: class MyClass(metaclass=MyMeta):
...: def __repr__(self):
...: return 'Howdy...'
...:
In [49]: obj = MyClass()
In [50]: print(obj)
Howdy...
In [51]: print(type(obj))
Whatever...
For Python2, you need to define __metaclass__
as a class attribute.
回答2:
The name of the script being run is always __main__
. This is why you check for this specific name, and why classes defined in the script are attributes of it.
来源:https://stackoverflow.com/questions/48845989/understanding-class-type-main-classname