beginInvoke, GUI and thread

I have application with two thread. One of them (T1) is main GUI form, another (T2) is func开发者_运维知识库tion 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.

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);         
}

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();
    }
 }