Re: OUString is mutable?

On 2012-09-28 15:55, Norbert Thiebaud wrote:
> On Fri, Sep 28, 2012 at 7:17 AM, Noel Grandin <noel@peralex.com> wrote:
> > That is exactly what makes it weird - it looks like a Java String, but it's
> > not, because you can do this:
> >
> >      void f(OUString s) {
> >           s = "2";
> >      }
> >
> >      OUString s = "1";
> >      f(s);
> >      cout << s; // will print "2"
> >
> > ie. the modification inside the method is visible outside the method.
> Really ? it does that ?
> Whoaa, that is unexpected, and way wrong. That should be considered
> 'entrapment' :-/
>
> Norbert

Sorry, that example code should read (note that the parameter is now a 
reference param)

    void f(OUString & s) {
         s = "2";
    }

    OUString s = "1";
    f(s);
    cout << s; // will print "2"



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