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Start your free trialMatt Duarte
2,497 PointsTrying to turn an argument into an int
Hey I am trying to turn the argument into an int, but it keeps saying syntax errordef squarerd("5"). Is my code right?
def squared("5"):
try:
arg2 = "This"
if arg == int("5"):
return arg**2
else:
return arg2*len()
# EXAMPLES
# squared(5) would return 25
# squared("2") would return 4
# squared("tim") would return "timtimtim"
def squared("5"):
try:
arg = "This"
if arg == int("5"):
return arg**2
else:
return arg*len()
1 Answer
Steven Parker
231,275 PointsHere's a few hints:
- a function parameter name can't be a number or have quotes around it
- a "try" must have a matching "except"
- you won't need to create any new string values in this challenge
- the code inside a try must do something that might cause an exception (not just an assignment)
- the len function must have an argument