help explaining default function arg weirdness

Robert Brewer fumanchu at amor.org
Mon Oct 11 05:52:13 CEST 2004


John M. Gabriele wrote:
> This is almost straight out of the Python tutorial,
> section 4.7.1
> 
> ------------------------code ---------------------
> #!/usr/bin/python
> 
> def func( append_this, default_list=[] ):
>      default_list.append( append_this )
>      return default_list
> 
> print func( "foo" )
> print func( "bar" )
> print func( "baz" )
> ------------------------/code-----------------
> 
> And running it gives me this:
> 
> ['foo']
> ['foo', 'bar']
> ['foo', 'bar', 'baz']
> 
> which looks wrong to me. The explanation in the tutorial
> says "The default value is evaluated only once. This makes a
> difference when the default is a mutable object such as a
> list, dictionary, or instances of most classes."
> 
> I don't get it: isn't default_list a local to func()?

The name is. The value, however, is just another spot on the heap.

> Doesn't it get created/destroyed with each call to func()
> so we'd get a fresh empty one with each function call?

Nope.

> What's the rationale for having the function remember
> default_list across calls?

http://www.python.org/doc/faq/general.html#why-are-default-values-shared
-between-objects

The setting of default values is performed when the function is defined
(only once), not when it is called (each time).


Robert Brewer
MIS
Amor Ministries
fumanchu at amor.org



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