On 2023-03-26 19:43:44 +0200, Jen Kris via Python-list wrote: > The base class: > > > class Constraint(object): [...] > def satisfy(self, mark): > global planner > self.choose_method(mark) > > The subclass: > > class UrnaryConstraint(Constraint): [...] > def choose_method(self, mark): > if self.my_output.mark != mark and \ > Strength.stronger(self.strength, self.my_output.walk_strength): > self.satisfied = True > else: > self.satisfied = False > > The base class Constraint doesn’t have a "choose_method" class method, > but it’s called as self.choose_method(mark) on the final line of > Constraint shown above. > > My question is: what makes "choose_method" a method of the base > class,
Nothing. choose_method isn't a method of the base class.
> called as self.choose_method instead of
> UrnaryConstraint.choose_method? Is it super(UrnaryConstraint,
> self).__init__(strength) or just the fact that Constraint is its base
> class?
This works only if satisfy() is called on a subclass of Constraint which
actually implements this method.
If you do something like
x = UrnaryConstraint()
x.satisfy(whatever)
Then x is a member of class UrnaryConstraint and will have a
choose_method() method which can be called.
> Also, this program also has a class BinaryConstraint that is also a
> subclass of Constraint and it also has a choose_method class method
> that is similar but not identical:
...
> When called from Constraint, it uses the one at UrnaryConstraint. How
> does it know which one to use?
By inspecting self. If you call x.satisfy() on an object of class
UrnaryConstraint, then self.choose_method will be the choose_method from
UrnaryConstraint. If you call it on an object of class BinaryConstraint,
then self.choose_method will be the choose_method from BinaryConstraint.
hp
PS: Pretty sure there's one "r" too many in UrnaryConstraint.
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_ | Peter J. Holzer | Story must make more sense than reality.
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