[Tutor] Pass by reference

[ak@silmarill.org]

On Fri, Jul 13, 2001 at 09:26:56AM +0530, Praveen Pathiyil wrote:
> 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 ?

For performance reasons.

> 
> 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
> >
> >
> > _______________________________________________
> > Tutor maillist  -  Tutor@python.org
> > http://mail.python.org/mailman/listinfo/tutor
> >
> 
> 
> _______________________________________________
> Tutor maillist  -  Tutor@python.org
> http://mail.python.org/mailman/listinfo/tutor

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