CODEWITHSUNDEEP
c#begininvoke
nirmus
nirmus

Reputation: 5093

beginInvoke, GUI and thread

I have application with two thread. One of them (T1) is main GUI form, another (T2) is function working in loop. When T2 gets some information must call function with GUI form. I'm not sure that I do it right.

T2 call function FUNCTION, which update something in GUI form.

  public void f() {
        // controler.doSomething();
  }


 public void FUNCTION() {

    MethodInvoker method = delegate {
            f();
    };

    if ( InvokeRequired ) {
        BeginInvoke( method );
    } else {
            f();
    }
 }

But now I must declare two function. How does it using only one function? Or how does it right.

Upvotes: 6

Views: 15132

Answers (2)

Reed Copsey
Reed Copsey

Reputation: 564333

You can do this in a single method by calling invoking yourself:

public void Function()
{
     if (this.InvokeRequired)
     {
         this.BeginInvoke(new Action(this.Function));
         return;
     }

     // controller.DoSomething();         
}

Edit in response to comments:

If you need to pass additional arguments, you can do it by using a lambda expression as follows:

public void Function2(int someValue)
{
     if (this.InvokeRequired)
     {
         this.BeginInvoke(new Action(() => this.Function2(someValue)));
         return;
     }

     // controller.DoSomething(someValue);         
}

Upvotes: 16

KeithS
KeithS

Reputation: 71565

Looks good to me. You may be able to change the anonymous delegate to a lambda, which is a little cleaner. To get rid of the f() method declaration, you can inline its code into the delegate, then either Invoke the delegate as a MethodInvoker or simply call it like you would any other method:

 public void FUNCTION() {

    MethodInvoker method = ()=> controller.doSomething();

    if ( InvokeRequired ) {
        BeginInvoke( method );
    } else {
            method();
    }
 }

Upvotes: 3

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