[FIXED] How to override the an inherent property when using its related method?

Issue

class foo:

   @property
   def nums(self):
       return [1, 2, 3, 4, 5]

   def gt(self, x):
       return [num for num in self.nums if num > x] 

class bar(foo):

    @property
    def nums(self):
        return super().gt(3)

f = foo()
b = bar()

print(f.nums)
print(b.nums)

The above code will have infinite recursion.

The desired result is to print [4, 5] when I call b.nums. How may I have the desired result? Thank you.

Solution

You are generating a recursion because the duplicate nums. You have two ways to solve it:

• Rename the nums inside the bar class with another name
• Don't use super() but specify the class name foo()

Option 1

class foo:
    @property
    def nums(self):
       return [1, 2, 3, 4, 5]

    def gt(self, x):
       return [num for num in self.nums if num > x] 

class bar(foo):
    @property
    def diffrent_nums(self):
        return super().gt(3)

f = foo()
b = bar()

print(f.nums)
print(b.diffrent_nums)

Having the duplicate property name nums generate the recursion, because you call:

print(b.nums) --> return super().gt(3) --> return [num for num in self.nums if num > x] --> return super().gt(3) --> ....

In your for loop you call self.nums that is the bar().nums that calls again the for that calls the nums again and again

Option 2

class foo:
    @property
    def nums(self):
       return [1, 2, 3, 4, 5]

    def gt(self, x):
       return [num for num in self.nums if num > x] 

class bar(foo):
    @property
    def nums(self):
        return foo().gt(3)

f = foo()
b = bar()

print(f.nums)
print(b.nums)

This will generate a new instance of the class foo.

Answered By - Carlo Zanocco