Re: Are Perl6 threads preemptive or cooperative?

On Fri, 25 Aug 2000, David L. Nicol wrote:

> That's what I was suggesting.  And if you say $a = 1 + foo() you have
> to give up your mutex on $a before calling foo().  So the programmer
> would have to work these things out with the subroutines:

I would think that you'd have already called foo() by the time you
locked $a. 

-- 
Bryan C. Warnock
([email protected])