OUString is mutable?

[Norbert Thiebaud]

On Fri, Sep 28, 2012 at 8:58 AM, Noel Grandin <noel at peralex.com> wrote:
> 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"
>

ah!! then, that's pretty much 'normal'... no surprise there.


Norbert