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])
- Re: Are Perl6 threads preemptive or co... David L. Nicol
- Re: Are Perl6 threads preemptive ... Steven W McDougall
- Re: Are Perl6 threads preempt... Markus Peter
- Re: Are Perl6 threads preempt... Chaim Frenkel
- Re: Are Perl6 threads preempt... Steven W McDougall
- Re: Are Perl6 threads preempt... Chaim Frenkel
- Re: Are Perl6 threads preempt... Steven W McDougall
- Re: Are Perl6 threads preemptive or cooperative? Markus Peter
- Re: Are Perl6 threads preemptive or cooperative? Markus Peter
- Re: Are Perl6 threads preemptive or cooperative? David L. Nicol
- Re: Are Perl6 threads preemptive or cooperativ... Bryan C . Warnock
- Re: Are Perl6 threads preemptive or cooper... David L. Nicol
- Re: Are Perl6 threads preemptive or co... Dan Sugalski
- Re: Are Perl6 threads preemptive or co... Chaim Frenkel
- Re: Are Perl6 threads preemptive ... David L. Nicol
- Re: Are Perl6 threads preempt... Chaim Frenkel
- Re: Are Perl6 threads preempt... Markus Peter