На самом деле .NET Framework и Windows довольно хорошо управляют потоками, равномерно распределяя их по каждому процессору. Однако распределением потоков можно манипулировать вручную, используя Process
и ProcessThread
.
using System;
using System.Diagnostics;
using System.Threading;
namespace ThreadTest
{
class Program
{
static void Main(string[] args)
{
//Get the our application's process.
Process process = Process.GetCurrentProcess();
//Get the processor count of our machine.
int cpuCount = Environment.ProcessorCount;
Console.WriteLine("CPU Count : {0}", cpuCount);
//Since the application starts with a few threads, we have to
//record the offset.
int offset = process.Threads.Count;
Thread[] threads = new Thread[cpuCount];
Console.WriteLine(process.Threads.Count);
LogThreadIds(process);
//Create and start a number of threads that equals to
//our processor count.
for (int i = 0; i < cpuCount; ++i)
{
Thread t = new Thread(new ThreadStart(Calculation))
{ IsBackground = true };
t.Start();
}
//Refresh the process information in order to get the newest
//thread list.
process.Refresh();
Console.WriteLine(process.Threads.Count);
LogThreadIds(process);
//Set the affinity of newly created threads.
for (int i = 0; i < cpuCount; ++i)
{
//process.Threads[i + offset].ProcessorAffinity = (IntPtr)(1L << i);
//The code above distributes threads evenly on all processors.
//But now we are making a test, so let's bind all the threads to the
//second processor.
process.Threads[i + offset].ProcessorAffinity = (IntPtr)(1L << 1);
}
Console.ReadLine();
}
static void Calculation()
{
//some extreme loads.
while (true)
{
Random rand = new Random();
double a = rand.NextDouble();
a = Math.Sin(Math.Sin(a));
}
}
static void LogThreadIds(Process proc)
{
//This will log out all the thread id binded to the process.
//It is used to test whether newly added threads are the latest elements
//in the collection.
Console.WriteLine("===Thread Ids===");
for (int i = 0; i < proc.Threads.Count; ++i)
{
Console.WriteLine(proc.Threads[i].Id);
}
Console.WriteLine("===End of Thread Ids===");
}
}
}
Теперь проверьте диспетчер задач, мы видим, что второй процессор принимает на себя все рабочие нагрузки.
Окно диспетчера задач