In if-elif-else, the if statement is TRUE yet the else statement is getting executed

Question

string = "aabbcc"
count = 3
while True:
    if string[0] == 'a':
        string = string[2:]
    elif string[-1] == 'b':
        string = string[:2]
    else:
        count += 1
        break
print(string)
print(count)

The output for the program is ‘bbcc’ and 4. But I don’t understand why the else statement is running when the IF statement is true. Please help.

Asked By: Ayush Kumar

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Source

Answer 1

The value of string gets replaced, then the loop continues. The second iteration, the string doesn’t start with "a" anymore. It also doesn’t end with "b". So you get 3+1 and "bbcc".

Answered By: kojiro

Answer 2

you are using while loop remember.

string = "aabbcc"
count = 3
while True:
    if string[0] == 'a': 
        string = string[2:]
        # here the value of string is assigned as bbcc
    elif string[-1] == 'b':
        string = string[:2]
    else:
        count += 1
        break
print(string)
print(count)

After the vaiable string is assigned by bbcc, the loop runs again and this time string[0] != “a” and is equal to “b”. Therefore the statement in else is executed and the count becomes 4 and the loop breaks because of the break statement

There is no bug in the code And the code is running fine as it should be.

Answered By: aditya babar

In if-elif-else, the if statement is TRUE yet the else statement is getting executed

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