OUString is mutable?

Lubos Lunak l.lunak at suse.cz
Wed Oct 3 09:19:56 PDT 2012

On Monday 01 of October 2012, Michael Stahl wrote:
> On 01/10/12 11:40, Noel Grandin wrote:
> > On 2012-10-01 10:15, Stephan Bergmann wrote:
> >> While that argument is irrelevant in a C/C++ context, immutability is
> >> also an important concept when reasoning about multi-threaded code.
> >> Therefore, the distinction between OUString and OUStringBuffer IMO
> >> does make sense after all.
> >
> > I don't see how the design helps you in a multithreaded context.
> > If you share an OUString instance between two threads, either thread
> > could assign to it, replacing it's contents, and invalidating what the
> > other thread sees.
> but if you share 2 copies of an OUString instance in 2 threads, they
> could access the one underlying buffer concurrently without issue, and
> when they want to point their copy at a different value, they can do
> that independently of the other -- again, OUString works in exactly the
> same way as a pointer to an immutable object here, if you copy the
> pointer you can dereference it without issue and point your copy to a
> different object; you only need a lock if you don't copy the pointer and
> share a single pointer that one thread may update (which depending on
> the situation may be necessary, say you don't just want to save memory
> but actually always want the threads to point to the same value).

 Either I don't get what immutable actually means in this context (in which 
case I'd like to be enlightened), or you are talking about things that 
actually have nothing to do with immutability. OUString is not the same like 
a pointer to an immutable object, because this object contains a reference 
count (that obviously does change, for example when you point your OUString 
copy to it). What you describe is possible with OUString, but not because of 
immutability, but because OUString (or the internal object rather) is 
ref-counted copy-on-write. It should be perfectly possible to have a normally 
mutable OUString with the same abilities.

 As such, as far as I can tell, immutable OUString is only a complication with 
no practical advantages.

> > So it's really not any safer than using an OUStringBuffer.
> > It just gives the illusion of safety.
> doesn't seem any more illusory than other things C++ programmers are
> used to.

[citation needed]

 Lubos Lunak
 l.lunak at suse.cz

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