Re: OUString is mutable?

On 28/09/12 16:05, Noel Grandin wrote:
> On 2012-09-28 16:00, Caolán McNamara wrote:
> > On Fri, 2012-09-28 at 14:17 +0200, Noel Grandin wrote:
> > > you can do this:
> > >
> > >       void f(OUString s) {
> > >            s = "2";
> > >       }
> > >
> > >       OUString s = "1";
> > >       f(s);
> > >       cout << s; // will print "2"
> > That will print "1" not "2".
> >
> > Maybe you meant
> >
> > void f(OUString& s) {
> >    s = "2";
> > }
> >
> > OUString s = "1";
> > f(s);
> > cout << s; // will print "2"
> >
> > but that's perfectly reasonable.
> >
> > C.
>
>
> Yeah, that's what I meant.
> But that's also what I have a problem with.
> It means that any OUString field or variable is effectively mutable, 

i don't see how that is the case.  the calling code needs to explicitly
pass the string as a parameter to a function for it to be modified, and
it's clear from the parameter non-const reference type that this can
happen.  you can also pass an ordinary pointer by-reference and then the
called function can reset it to a different value, that is just how C++
works...:

void f(void *& p) {
        p = (void*)42;
}

for comparison, in Java you cannot do this directly (because Java does
not have call-by-reference semantics, only call-by-value), you have to
wrap the string in e.g. an array to get something similar:

void f(String[] s) {
        s[0] = "bar";
}