CODEWITHSUNDEEP
c#
FosterZ
FosterZ

Reputation: 3911

Creating new a Thread() is asking ParameterizedThreadStart object as a parameter in C sharp

ah i'm very noob in Thread Programming and just started a basic step to create multiple threads, so i googled and got some snippets about creating Thread in c#, here is the snippet i found:

public MyThread(string name) { 
    count = 0; 
    thrd = new Thread(new ThreadStart(this.run)); // here m getting error
    thrd.Name = name; 
    thrd.Start(); 
  } 

  // Entry point of thread. 
  void run() { 
    Console.WriteLine(thrd.Name + " starting."); 

    do { 
      Thread.Sleep(500); 
      Console.WriteLine("In " + thrd.Name + 
                        ", count is " + count); 
      count++; 
    } while(count < 10); 

    Console.WriteLine(thrd.Name + " terminating."); 
  } 
}

The error is The best overloaded method match for System.Threading.Thread.Thread(System.Threading.ParameterizedThreadStart) has some invalid arguments

Why the Thread constructor is asking me for ParameterizedThreadStart object, i want simple ThreadStart object to be passed.

Another Thing is ThreadStart class doesn't have a constructor with 1 argument i.e. it takes 0 arguments, but in snippet they have shown new ThreadStart(this.run) this ? m using C# 2008

Here is the complete code


using System; 
using System.Threading; 

class MyThread { 
  public int count; 
  public Thread thrd; 

  public MyThread(string name) { 
    count = 0; 
    thrd = new Thread(new ThreadStart(this.run)); 
    thrd.Name = name; 
    thrd.Start(); 
  } 

  // Entry point of thread. 
  void run() { 
    Console.WriteLine(thrd.Name + " starting."); 

    do { 
      Thread.Sleep(500); 
      Console.WriteLine("In " + thrd.Name + 
                        ", count is " + count); 
      count++; 
    } while(count

Upvotes: 1

Views: 7791

Answers (3)

Jon Skeet
Jon Skeet

Reputation: 1500505

That should be absolutely fine, because there is an overload taking ThreadStart instead of ParameterizedThreadStart.

I suspect there's something else at play here... could you provide a short but complete example which demonstrates the problem?

Aside from missing the class declaration itself and variable declarations, your code compiles for me with no problem:

using System;
using System.Threading;

class MyThread {

    int count;
    Thread thrd;

    public MyThread(string name) { 
        count = 0; 
        thrd = new Thread(new ThreadStart(this.run)); // here m getting error
        thrd.Name = name; 
        thrd.Start(); 
    } 

    // Entry point of thread. 
    void run() { 
        Console.WriteLine(thrd.Name + " starting."); 

        do { 
            Thread.Sleep(500); 
            Console.WriteLine("In " + thrd.Name + 
                              ", count is " + count); 
            count++; 
        } while(count < 10); 

        Console.WriteLine(thrd.Name + " terminating."); 
    } 
}

Upvotes: 2

Thomas Levesque
Thomas Levesque

Reputation: 292425

Not sure why it doesn't work, but you can try that:

thrd = new Thread(run);

The conversion from the run method group to a ThreadStart delegate is implicit.

I suspect you have a name conflict between System.Threading.ThreadStart and another type defined somewhere else in your code... Try to put the caret on ThreadStart and press F12 to go to the declaration

Upvotes: 2

Dennis
Dennis

Reputation: 2142

I have completed the incomplete example you gave us and I do not encounter the same compiler error.

class Program
{
    static int count;
    static Thread thrd;


    public static void MyThread(string name) { 
        count = 0; 
        thrd = new Thread(new ThreadStart(run)); // here m getting error
        thrd.Name = name; 
        thrd.Start(); 
    } 

  // Entry point of thread. 
  static void  run() { 
    Console.WriteLine(thrd.Name + " starting.");

    do
    {
        Thread.Sleep(500);
        Console.WriteLine("In " + thrd.Name +
                          ", count is " + count);
        count++;
    } while (count == 5);
  }



  static void Main(string[] args)
  {
  }
}

Upvotes: 2

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