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Python

abhishek krishna vandadi
abhishek krishna vandadi
878 Points
Posted by abhishek krishna vandadi
abhishek krishna vandadi
878 Points

I did not understand the exception handling code which was given as quiz question

print("A") try: result = "test" + 5 print("B") except ValueError: print("C") except TypeError: print("D") else: print("E") print("F")

1 Answer

Cruze Howard
Cruze Howard
444 Points
Cruze Howard
Cruze Howard
444 Points

The code prints "A" as there are no parameters around it's execution. It then attempts to try and assign "test" + 5 to the variable result. This results in a TypeError as you can not add a string to an integer, thus skipping over the attempt at printing B. Since it is not a ValueError C is also not printed. The exception of a TypeError is occurring because of the attempt at "test" + 5 which then prints "D". Since an except was ran we skip over the else statement, as it only runs if none of the errors come up. "F" is then printed like "A" because of its lack of any parameters. Hope this helps! 

print("A")
try:
    result = "test" + 5
    print("B")
except ValueError:
    print("C")
except TypeError:
    print("D")
else:
    print("E")
print("F")

RESULTS A D F