Provided the threads are at least started you can call join sequentially on the collection of thread objects...
ArrayList threads; ... foreach (Thread t in threads) t.Join(); Should work though I haven't tried it :) Jim > -----Original Message----- > From: Diego Gonzalez [mailto:[EMAIL PROTECTED]] > Sent: Thursday, April 18, 2002 9:18 AM > To: [EMAIL PROTECTED] > Subject: Re: [DOTNET] Waiting for multiple Threads > > > But WaitAll() receives a WaitHandle[], and i don't know how to get the > Thread hanlde in order to wait for it.Using Join() allows the > current thread > to wait for another thread, i wanted to wait for multiple threads. > > Diego > > ----- Original Message ----- > From: "Nick Wienholt" <[EMAIL PROTECTED]> > To: <[EMAIL PROTECTED]> > Sent: Wednesday, April 17, 2002 11:47 PM > Subject: Re: [DOTNET] Waiting for multiple Threads > > > > Hi Diego, > > > > If you are waiting for the threads to terminate, use Thread.Join. > > Otherwise, WaitHandle.WaitAll is the System.Threading equivalent to > > WaitForMultipleObjects. > > > > Nick > > > > ----- Original Message ----- > > From: "Diego Gonzalez" <[EMAIL PROTECTED]> > > To: <[EMAIL PROTECTED]> > > Sent: Thursday, April 18, 2002 3:35 AM > > Subject: [DOTNET] Waiting for multiple Threads > > > > > > > Is there any way to wait for multiple running Threads? I will call > > > WaitForMultipleObjects, but there is no way to get the > thread handle. > > > > > > Diego > > > > > > You can read messages from the DOTNET archive, > unsubscribe from DOTNET, > or > > > subscribe to other DevelopMentor lists at http://discuss.develop.com. > > > > You can read messages from the DOTNET archive, unsubscribe from DOTNET, or > subscribe to other DevelopMentor lists at http://discuss.develop.com. > You can read messages from the DOTNET archive, unsubscribe from DOTNET, or subscribe to other DevelopMentor lists at http://discuss.develop.com. You can read messages from the DOTNET archive, unsubscribe from DOTNET, or subscribe to other DevelopMentor lists at http://discuss.develop.com.
