Beginnger Question

[Torben Lauritzen]

Matthias Hanitzsch <Matthias.Hanitzsch at dynamic.de> writes:

> Dennis wrote:
> 
> > x = 2
> >
> > def wrong(num):
> >     return num + 1
> >
> > print wrong(x)
> > print wrong(x)
> >
> > I cannot figure out why this (simplified) program won't increment x
> > and produce 3 and 4.
> 
> Hi,
> this is "call by value", that means that the parameter num of your
> function will get the VALUE of the variable x.
> Then num is incremented and returned.
> The variable x is not changed, as you never write to it. You just
> take its value.
> 
> HTH,
> Matthias
> 

As I see it, num is not incremented, before it is returned. So, even
if it had been a 'call by reference', x would not increase.

-- 
- Torben