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Home/ Questions/Q 8921799
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T06:39:06+00:00 2026-06-15T06:39:06+00:00

I’ve a problem with a Java Project, using SWT library. It’s my first time

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I’ve a problem with a Java Project, using SWT library. It’s my first time trying with SWT, so don’t throw stones to me please.

Environment: Eclipse – Juno

Class and functions:

1º Client -> Where’s “main”. Creates a object of “GUI_Creator” class (mine).

2º GUI_Creator -> Has 2 functions, in order to “desing” my forms.

2.1 -> MakeLogIn -> Void method, wich adds the controls i need to the Shell i give him as parameter. I use it, to desing the first form i show to the user.

This function also link the “click” event of a button of this shell, with the second function of GUI_Creator class, that creates a new one shell. Then, this shell would be open, and first one would be closed.

2.2 -> MakeClient -> This function returns a shell var. For me, this returns the second shell of my program, even with the desing, etc.

Problem:

I’m using, at the end of “main” this code:

while (!Conexion.isDisposed())
{
   if (!ourDisplay.readAndDispatch())
      ourDisplay.sleep();
}

ourDisplay.dispose(); 

If i’m not wrong, this code is doing that when i try to close my first shell, and then leave only opened the second shell, my program closes completely, because i’m disposing the parent display.

However, i can’t delete this code, because if i do this, when i run my program, it doesn’t “wait” an action, and just executes all “main” code and then closes the GUI and the program finalizes.

So, i really don’t know what would i have to do, in order to fix this situation.

Thanks,
Btc

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-15T06:39:08+00:00Added an answer on June 15, 2026 at 6:39 am

    I’m not sure if I got your problem right, but you could try:

    while (!(shell1.isDisposed() && shell2.isDisposed())) {
    ...
    }
    
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