a wierd parameter passing behavior

POYEN OP Olivier (DCL) Olivier.POYEN at clf-dexia.com
Wed Jun 4 15:59:49 CEST 2003

> -----Message d'origine-----
> De : Torvo Sollazzo [mailto:mpradella at yahoo.it]
> Envoyé : mercredi 4 juin 2003 15:30
> À : python-list at python.org
> Objet : a wierd parameter passing behavior
> Hi!
> I found a strange parameter passing behavior in Python (v. 2.2.2).
> Look at this piece of code:

One mutates l in place, while the other affect a new local variable : 

> def p(l) :
>     l += [3]
>     l[0] = 2

Here, you change a mutable object in place (ensured byt the += stuff) 

> def p1(l) :
>     l = l + [3]
>     l[0] = 2

Here you create a local variable l, assigning the value l+[3] to it. Then you change this local variable first element.
But outside the scope of the p1 function, this "new" l does not exists, ence l is still [1]

On the contrary, since list are mutable objects in python, if you change list *in place* with l.__iadd__(stuff) (or l += stuff), it change l without creating a new one, a local one.

> def f() :
>     l = [1]
>     p(l)
>     print l
>     l = [1]
>     p1(l)
>     print l
> If I call f(), I obtain the following results:
> [2,3]
> [1]
> Why p1() is different from p()? The list l should be passed by
> reference in both cases, while it seems that "l = l + [3]" creates a
> new local variable, also called l.

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