[Tutor] Pass by reference

[Praveen Pathiyil]

I would like to thank Michael, Rob, D-Man, Wesley, Ibraheem, Remco and
everybody else who helped me clarify my doubts ...

I am continuing with my doubts :-))
Can some one explain the differece between the allocation of variables or
objects on the heap and that on the stack ? (May be from the python
perspective itself).
Also the differece this makes on the scope or life of that variable.

One un-related question --> Why was strings made as immutable in python ?

Once again thanks to all,
Praveen.

----- Original Message -----
From: "D-Man" <dsh8290@rit.edu>
To: <tutor@python.org>
Sent: Wednesday, July 11, 2001 10:27 PM
Subject: Re: [Tutor] Pass by reference


> On Wed, Jul 11, 2001 at 07:44:24PM +0530, Praveen Pathiyil wrote:
> | Hi,
> |
> | Can we do a "pass by reference" in python ?
>
> Sort of -- Python always does call-by-value, but the value is always a
> reference to an object on the heap.  If you assign to the argument,
> then, no that won't be seen by the caller (just like C, C++ and Java).
> If, instead, the reference refers to a mutable object and you modify
> that object then you have the same effect (this is like using pointers
> in C/C++ or a reference to a mutable object in Java).
>
> Don't get bogged down in the implementation details and terminology
> though (there can be huge flamewars as to what "pass-by-value" and
> "pass-by-reference" really mean).
>
> | Ex:
> |
> | ( I would like to know whether there is a way to have the modified
> | ex_dict to be visible in modified form in func1 with out explicitly
> | returning that )
> |
> | def func2(x, ex_dict):
> |     ex_dict[x] = x*x
> |
> | def func1(a):
> |     ex_dict = {}
> |     func2(a, ex_dict)
> |
> | func1(2)
>
> Lets try it :
>
> >>> def func( x , ex_dict ) : ex_dict[x] = x*x
> ...
> >>> ex_dict = {}
> >>> func( 2 , ex_dict )
> >>> print ex_dict
> {2: 4}
> >>>
>
> It works for dictionaries because you didn't modify the reference but
> instead modified the referred-to object on the heap.
>
> If you try and do it with, say, integers it won't work because you
> can't modify the integer object on the heap :
>
> >>> def badfunc( i ) :
> ...     i = i * i
> ...     print "in badfunc, i = %d" % i
> ...
> >>> a = 5
> >>> badfunc( a )
> in badfunc, i = 25
> >>> print a
> 5
> >>>
>
> -D
>
>
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