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Python Python Basics Functions and Looping Create a Function

Nicolas Villarreal
Nicolas Villarreal
282 Points

I am not sure why I am getting the error: 'int' object is not callable based on my code.

I thought I passed the argument of 3 into my square function I created and set that equal to my new variable called result and I then return the answer. I am not sure why I am getting this error.

squaring.py
def square(number):
    square = number*number
    result = square(3)
    return result

2 Answers

Steven Parker
Steven Parker
231,275 Points

When you put parentheses after a name, it's normally to invoke the function of that name. But "square" is not a function here but an integer value created on the line before.

Extra hints:

  • don't use the same names for functions and variables
  • calling a function from inside itself (unconditionally) creates an "infinite loop"
Susan krystof
PLUS
Susan krystof
Courses Plus Student 2,662 Points

What you should do is return the square of the number that will be passed in, which is number * number, What you did is call the function square inside itself. You can't call a function inside itself

def square(number):
    return number*number

OR

def square(number):
    result = number*number
    return result
Nicolas Villarreal
Nicolas Villarreal
282 Points

Yes I had that part down but I need to call my function and pass in the argument of 3. I should store this as the result variable

Steven Parker
Steven Parker
231,275 Points

Sure, but that task should be done after the function definition, not inside it.