Pausing in Parallel.ForEach. How to pause and continue? C #

Look, if you use such a code, it should work. However, there is one stream, and the whole point of using Parallel.ForEach is lost (MaxDegreeOfParallelism = 1). Otherwise, the blocking will work for the new stream that Parallel.ForEach creates and for the stream in which it works, and this is not exactly what is needed.

 using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading; using System.Threading.Tasks; namespace parallelOutput { class Program { public static void MainTask(ManualResetEvent re) { string[] recs = { "str_01", "str_02", "pause", "str_03" }; ParallelLoopResult plr = Parallel.ForEach(recs, new ParallelOptions() { MaxDegreeOfParallelism = recs.Count() }, (string rec, ParallelLoopState pls) => { if (rec != "pause") { Console.WriteLine("{0} : {1} ", DateTime.Now, rec); } else { re.WaitOne(); } }); } public static void StartTask(ManualResetEvent re) { Thread.Sleep(10000); re.Set(); } static void Main(string[] args) { ManualResetEvent re = new ManualResetEvent(false); Thread st = new Thread(() => StartTask(re)); st.Start(); Thread mt = new Thread(() => MainTask(re)); mt.Start(); } } } 

The easiest, probably, is to stop / start the code like this:

Declare a public object:

 object mutex = new object(); 

In the place that should be stopped (for example, at each iteration of the cycle):

 lock (mutex) { /* ничего не делать, сразу освободить */ } 

In the place where it is necessary to prohibit the execution:

 lock (mutex) { // тут выполнение потоков остановлено Thread.Sleep(1000); } // а тут снова разрешено 

Well, or if it is necessary to prohibit execution in one place and allow it in another, then

 Monitor.Enter(mutex); // запретить // ... Monitor.Exit(mutex); // разрешить 

But if you need to start / stop the execution of code, Parallel.ForEach probably not the best idea: after all, the internal scheduler in it assumes that it needs to execute everything as quickly as possible, and does not know that your code will stop.