On 2012-09-28 14:09, Michael Stahl wrote:
On 28/09/12 13:46, Noel Grandin wrote:
Hi
I don't really understand the point of not being able to mutate OUString
in-place, but being able to overwrite it via the assignment operator?
It seems to me it should be one or the other.
Either OUString is immutable, and it cannot be assigned to, except via a
constructor, or it is mutable, and we can pretty much throw away
OUStringBuffer.
[accidentally sent this first to Noel only instead of the list]
the point you are missing is that OUString is a sort of "smart pointer"
to an immutable buffer. it works similarly to "String" in Java, e.g. in
Java "String s = "foo"; s = "bar"" is legal but you cannot modify the
content of whatever value is assigned to "s".
that's why OUString has an operator= that makes it point at a different
buffer but offers no way to write into the buffer (at least i hope it
doesn't...).
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.
Also, in Java, this
String s = "1";
is really syntactic sugar for this:
String s = new String("1");
So it's not a real assignment operation.
It just seems to me that we could easily enough steal a bit from the
length field to indicate that the buffer is immutable, and then we could
reduce our complexity by eliminating OUStringBuffer.
Disclaimer: http://www.peralex.com/disclaimer.html
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