Re: OUString is mutable?

Hi,

On Fri, Sep 28, 2012 at 04:05:35PM +0200, 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,
> which makes the difference between it and OUStringBuffer boil down
> to the presence of the nCapacity field.

I find it perfectly reasonable that a variable of a value type (as
opposed to polymorphic type) is assignable. In fact, I would be
surprised if it were not. Value types are supposed to mimic the behavior
of primitive types; that is why copy constructor and operator= are
created by the compiler unless one disables them. You are not surprised
that

int i(1);
i = 2;

or

char const* s("1");
s = "2";

works, are you? So why

rtl::OUString s("1");
s = "2":

should be different? C++ already has a way to express that a variable is
not mutable: the 'const' modifier.

Last, but not least, working assignment is requirement for many, if not
all, STL containers.

D.