Python Forum

Consider the class Interval below:

class Interval:
    def __init__(self,a,b):
        self.left=a
        self.right=b
    def __repr__(self):
        return '[{},{}]'.format(self.left,self.right)

For given numbers a and b, when printing Interval(a,b) I get:

Output:
[a,b]

However, I also want to be able to get the below output when printing Interval(number):

Output:
[number,number]

What constraints do I need to make on the __init__ method?

a and b are visible only to the __init__ method but you can access left and right outside of the class, but you must know what their values are when you instantiate the class. So I really don't quite understand. please show where you print the values.

class Interval:
    def __init__(self,a,b=None):
        self.left=a
        if b is None:
            self.right = a
        else:
            self.right=b
    def __repr__(self):
        return '[{},{}]'.format(self.left,self.right)

Sorry if I was unclear, but your replies have led me to the following solution.

class Interval:
    def __init__(self,a,b=None):
        self.left=a
        self.right=a
        if b is not None:
            self.left=a
            self.right=b
    def __repr__(self):
        return '[{},{}]'.format(self.left,self.right)
print(Interval(2))

Output:
[2,2]

can be written as

class Interval2:

    def __init__(self, a, b=None):
        self.left = a
        self.right = b or a

    def __repr__(self):
        return f'[{self.left},{self.right}]'


print(Interval2(2))

Output:
[2,2]

if you don't like the else part and don't mind to assign to self.right twice, why not simply

    def __init__(self,a,b=None):
        self.left = a
        self.right = b
        if b is None:
            self.right = a

schniefen

Larz60+

buran

schniefen

Yoriz

buran