On 2012-09-28 15:55, Norbert Thiebaud wrote:
On Fri, Sep 28, 2012 at 7:17 AM, Noel Grandin <[email protected]> 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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