Behavior of default parameter in a function

[Shashank Singh]

quoting from docs:http://docs.python.org/reference/compound_stmts.html

*Default parameter values are evaluated when the function definition is
executed.* This means that the expression is evaluated once, when the
function is defined, and that that same “pre-computed” value is used for
each call. This is especially important to understand when a default
parameter is a mutable object, such as a list or a dictionary: if the
function modifies the object (e.g. by appending an item to a list), the
default value is in effect modified

so default object for x in foo is precomputed once and is modified with each
call.

hth
--shashank

On Thu, Mar 11, 2010 at 3:02 PM, jitendra gupta <jitu.icfai at gmail.com>wrote:

>
> def foo(x = [0]):
> x[0] = x[0] + 1
>  return x[0]
>
> def soo(x = None):
> if x is None:
> x = [0]
> x[0] = x[0] + 1
>  return x[0]
>
> >>> foo()
> 1
> >>>foo()  #See the behavior incremented by one
> 2
> >>>foo([1]) # but here based on given number
> 2
> >>>foo()
> 3
> >>>foo([1])
> 2
> >>>foo()
> 4
>
> >>>soo()
> 1
> >>>soo()
> 1
> >>>soo([1])
> 2
> >>>soo()
> 1
>
> Why foo() is incremented by 1 always when we are not passing any argument,
>  but this is not happening in soo() case, In which scenario we will use
> these type of  function.'
>
> Thanks
> Jitendra Kumar
>
>
> --
> http://mail.python.org/mailman/listinfo/python-list
>
>


-- 
Regards
Shashank Singh
Senior Undergraduate, Department of Computer Science and Engineering
Indian Institute of Technology Bombay
shashank.sunny.singh at gmail.com
http://www.cse.iitb.ac.in/~shashanksingh
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