Re: OUString is mutable?

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.

If you tell me that the saving from not having the nCapacity field is 
the motivation, then I get understand, but then we should update the 
docs to reflect that fact, rather than pretending that OUString is 
immutable.

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