Is it possible to do computation before super() in the constructor?

You can define a static supplier lambda which can contain more complicated logic.

public class MyClass {

    private static Supplier<MyType> myTypeSupplier = () -> {
        return new MyType();
    };

    public MyClass() {
        super(clientConfig, myTypeSupplier.get());
    }
}

Yes, there is a workaround for your simple case. You can create a private constructor that takes TextBox as an argument and call that from your public constructor.

public class Simple extends Base {
    private Simple(TextBox t) {
        super(t);
        // continue doing stuff with t here
    }

    public Simple() {
        this(new TextBox());
    }
}

For more complicated stuff, you need to use a factory or a static factory method.

I had the same problem with computation before super call. Sometimes you want to check some conditions before calling super(). For example, you have a class that uses a lot of resources when created. the sub-class wants some extra data and might want to check them first, before calling the super-constructor. There is a simple way around this problem. might look a bit strange, but it works well:

Use a private static method inside your class that returns the argument of the super-constructor and make your checks inside:

public class Simple extends Base {
  public Simple(){
    super(createTextBox());
  }

  private static TextBox createTextBox() {
    TextBox t = new TextBox();
    t.doSomething();
    // ... or more
    return t;
  }
}

It is required by the language in order to ensure that the superclass is reliably constructed first. In particular, "If a constructor does not explicitly invoke a superclass constructor, the Java compiler automatically inserts a call to the no-argument constructor of the superclass."

In your example, the superclass may rely on the state of t at construction time. You can always ask for a copy later.

There's an extensive discussion here and here.