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David PCourses Plus Student 1,162 Points
i keep getting a syntax error....what am I doing wrong?
This challenge is similar to an earlier one. Remember, though, I want you to practice! You'll probably want to use try and except on this one. You might have to not use the else block, though. Write a function named squared that takes a single argument. If the argument can be converted into an integer, convert it and return the square of the number (num ** 2 or num * num). If the argument cannot be turned into an integer (maybe it's a string of non-numbers?), return the argument multiplied by its length. Look in the file for examples.
def squared(5): try: int(5) except: return int(5) ** 2 else: return 5 * len(5)
Krahr ChaudhuryCourses Plus Student 139 Points
There's a LOT of mistakes in this code - I suggest you go through the videos/tutorials again.
When defining a function you should leave the argument to be a variable - like x. Not a constant value - like 5.
After defining the function, THEN you call it passing a constant or a variable that has a defined value as the variable.
Inside 'try', you should be converting the input variable (in the code below, I've used x) to an integer. Inside 'except', you include a fallback code for when int() doesn't work on the input (i.e. when the input is a string that doesn't read a number). So don't call int() again! We know it doesn't work. Instead call len() on the input to get its string length and return the square of that. Inside 'else' (which is where the program goes when the bit in try succeeds), you return the square of x. No need to call len().
def squared(x): try: x = int(x) # Ensure x is converted to an integer. except: return len(x) ** 2 # if x is a string that can't be translated to a number, then square its length else: return x * x # otherwise square the integer print squared(5) # x=5, int(5) = 5, 5*5 = 25 print squared("3") #x="3", int("3") = 3, 3*3 = 9 print squared("boo") #x="boo", int("boo") doesn't work! so do len("boo") = 3, 3**2 = 9