while calling the function, the same function_name and parameter_name in function creates obstruct! Can anyone explain whats happening?

Question

#In simple form 

def name(name):
    print(name)

name='arjun'

name(name) #In this line, what's happening?

#Error 'str' object is not callable

You've renamed `name` from a function to a `str`. Call that string something else and it will fix your problem — C.Nivs, Jun 02 '20 at 15:24
You replaced the function `name` with a string, and you can't call a string as if it is a function. — kindall, Jun 02 '20 at 15:24
`name` is no more a function name but a string instead with the value `'arjun'` — Saadi Toumi Fouad, Jun 02 '20 at 15:25
Does this answer your question? [Python Variable Declaration](https://stackoverflow.com/questions/11007627/python-variable-declaration) — khachik, Jun 02 '20 at 15:26

score 2 · Accepted Answer · answered Jun 02 '20 at 15:42

There are not separate namespaces for functions and other objects. If you write

def name1(name2):
    print(name2)

you create an instance of type function, then assign that instance to the name name1. (A def statement is just a very fancy kind of assignment statement.)

Parameter names are local variables, belonging to the scope defined by the function, not the scope in which the function is defined. That means you can reuse the name, though it isn't recommended as it can be confusing.

   belongs to the scope in which the function is defined
      |
      |    +--- belongs to the scope inside the function
      |    |
      v    v
def name(name):
    print(name)

However, the following assignment

name = 'arjun'

is in the same scope as the function definition, so this makes name refer to a str object instead of the function object it used to refer to.

while calling the function, the same function_name and parameter_name in function creates obstruct! Can anyone explain whats happening?

1 Answers1