What will be the result of executing the following code? class I: def __init__(self): self.s = ‘abc’ self.i = 0 def __iter__(self): return self def __next__(self): if self.i == len(self.s): raise StopIteration v = self.s[self.i] self.i += 1 return v for x in I(): print(x,end=”)

class I:
    def __init__(self):
        self.s = 'abc'
        self.i = 0
 
    def __iter__(self):
        return self
 
    def __next__(self):
        if self.i == len(self.s):
            raise StopIteration
        v = self.s[self.i]
        self.i += 1
        return v
 
 
for x in I():
    print(x,end='')

• it will print abc
• it will print 0
• it will raise an handled exception
• it will raise an unhandled exception

Explanation: Let’s analyze this code snippet:

• First, a for loop using the x variable iterates through the already defined I() generator.
• The generator class initializes the self.s = ‘abc’ and self.i = 0 variables within the constructor.
• The __next__ method compares if the value in i is equal to the length of s.
• Since it is not, v is assigned the first character of s, which is a, i is incremented by one, v is returned and printed in the console.
• The end=’’ argument in the print function will prevent new lines, so all the outputs will be shown on the same line.
• The iteration continues for the remaining characters in s. The b and c characters are also printed in the console.
• When the condition self.i == len(self.s) becomes true, a StopIteration is raised. This terminates the execution.

More Questions: Python Essentials 2 – Module 3 Test