Busy means would-block, yes.
Burak thanks, but that doesn't work for read-lock.
On Tuesday, December 3, 2019 at 5:39:48 PM UTC-8, Robert Engels wrote:
>
> It depends then, because technically the Go RW lock queues readers behind
> a waiting writer so “busy” is somewhat undefined. If “busy” means “would
> block” you can still do it - I’ll post the code tonight.
>
> > On Dec 3, 2019, at 6:49 PM, Robert Engels <ren...@ix.netcom.com
> <javascript:>> wrote:
> >
> > I would use an atomic and a lock instead of two locks.
> >
> >>> On Dec 3, 2019, at 6:35 PM, burak serdar <bse...@computer.org
> <javascript:>> wrote:
> >>>
> >>> On Tue, Dec 3, 2019 at 5:21 PM Liam Breck <networ...@gmail.com
> <javascript:>> wrote:
> >>>
> >>> I have a problem that is trivially solved via
> >>>
> >>> door sync.RWMutex
> >>>
> >>> func Reader() T {
> >>> if !door.TryRLock() { // missing in stdlib :-(
> >>> return busy
> >>> }
> >>> defer door.RUnlock()
> >>> ...
> >>> }
> >>>
> >>> func Writer() {
> >>> door.Lock()
> >>> defer door.Unlock()
> >>> ...
> >>> }
> >>>
> >>> How does one achieve this in Go?
> >>
> >> Two locks and a bool?
> >>
> >> var door=sync.Mutex{}
> >> var x=sync.Mutex{}
> >> var b bool
> >>
> >> func trylock() bool {
> >> x.Lock()
> >> if b {
> >> x.Unlock()
> >> return false
> >> }
> >> b=true
> >> door.Lock()
> >> x.Unlock()
> >> return true
> >> }
> >>
> >> unlock:
> >>
> >> x.Lock()
> >> b=false
> >> door.Unlock()
> >> x.Unlock()
> >>
> >>
> >>
> >>
> >>>
> >>> --
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>
>
> >>
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>
>
>
>
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