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  • How do I debug this code of the 'TypeError: 'int' object is not callable' error in Python?

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    Hey guys! I want to write a program that finds the sum of cubed terms. i.e. 1**3 + 2**3 + n**3
    I wrote down the following code:
      
       

    // n = input('Give the value of n.')
    n = int (n)
    s = ((n(n+1))/2)**2
    print ('Sum equals =', s)
    The output I got:
       
    raceback (most recent call last):
      File "/Users/apple/Documents/tryin.py", line 15, in <module>
        s = ((n(n+1))/2)**2
    

    TypeError: 'int' object is not callable


    >>>
    I Would really appreciate some help guys.
    :D
    And any suggestions on what to practice to be better at coding and stop getting these errors will be of help.
    :)
    Thanks!

 7 Answer(s)

  • when you get a error like that: "type integer is not callable", it's prabbly a syntax error / typo:

    the syntax for calling a function in Python is to put parens after it:

    fun()

    so look and see if you have put parens after something that is not a function -- and indeed you have:

    (note this UI sucks! I have to write this in a modal dialog, so I can't see the original question!)

    I think you have code like:

    n(1 + n)

    look for the n(...) -- that is math notation for multiply, but it's python notation for call the function "n" -- it should be:

    n * (1 + n)

  • You have to redefined your "sum" function to be an integer data type. So it is rightly telling you that an integer is not something you can pass a range.To fix this, restart your interpreter.Then all is okay.like this n = input('Give the value of n.') n = int (n) s = (((n+1))/2)**2 print ('Sum equals =', s)

  • You have to redefined your "sum" function to be an integer data type. So it is rightly telling you that an integer is not something you can pass a range.To fix this, restart your interpreter.Then all is okay

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