Given the following: a = 23 b = 45 c = 16 round ( (a/b)*0.9*c) Running the above outputs an error: TypeError: 'int' object is not callable. How can I round the output to an integer?

Jan 24, 2014 · 11 The action occurs when you attempt to call an object which is not a function, as with (). For instance, this will produce the error:

Apr 23, 2010 · When you set self.numerator = n, you're overwriting the reference to the method, and so when you call f.numerator(2), it's trying to do a method call on the member variable, which is an int, …

Jun 25, 2012 · 12 input () is a Python function, and you are using it both as the function and an identifier. Using input as a variable name will work the first time, but the 2nd time through the loop there won't …

Jun 26, 2012 · In Python a "callable" is usually a function. The message means you are treating a number (an >"int") as if it were a function (a "callable"), so Python doesn't know what to do, so it >stops.

Aug 16, 2021 · The issue is data types (hence TypeError). A data type may be a string, integer, or even a function (a function is something that is called, such as print(): notice the brackets).

Nov 11, 2020 · 1 if len (inputs [key].size ()) == 1: TypeError: 'int' object is not callable intのオブジェクトは呼び出しできない、と言うエラーなので、 その行を見直してみればいいかと。 inputs [key].size …

May 15, 2013 · Python interprets that as a call expression; i must be a callable, passing in i+1 as the single argument. Since i is an integer instead, which is not callable, you get an exception.

You no doubt used the identifier sum previously in your code as a local variable name, and the last value you bound to it was an int. So, in that code snippet, you're trying to call the int. print sum just before …

That is not okay :) Python thinks that the total you are referring to in the final line is the integer variable total and not the function. It is considered a good practice to name the function total as get_total instead