为什么方法 addMouseListener 不需要 super？

Question

我的问题是，当我创建一个从 JPanel 继承的 class 时，为什么不使用 super.addMouseListener() 添加监听器？我认为这个方法在超级 class 中，也就是 JPanel。这是代码：

private class DrawPanel extends JPanel
{
    private int prefwid, prefht;

    // Initialize the DrawPanel by creating a new ArrayList for the images
    // and creating a MouseListener to respond to clicks in the panel.
    public DrawPanel(int wid, int ht)
    {
        prefwid = wid;
        prefht = ht;

        chunks = new ArrayList<Mosaic>();

        // Add MouseListener to this JPanel to respond to the user
        // pressing the mouse.  In your assignment you will also need a
        // MouseMotionListener to respond to the user dragging the mouse.
        addMouseListener(new MListen());
    }

Answer 1

因为没有必要。

您没有在 DrawPanel class 中声明方法 addMouseListener，因此编译器会检查 superclasses 中是否存在这样的方法，并在java.awt.Component。因为这个方法是被DrawPanel class继承的，所以在这里调用就可以了。

如果您想深入了解原因，您需要阅读 JLS Sec 15.12, "Method Invocation Expressions"。然而，这并不完全是轻松阅读。

我认为关键的句子是：

Sec 15.12.1

For the class or interface to search, there are six cases to consider, depending on the form that precedes the left parenthesis of the MethodInvocation:

If the form is MethodName, that is, just an Identifier, then:

If the Identifier appears in the scope of a visible method declaration with that name (§6.3, §6.4.1), then:

If there is an enclosing type declaration of which that method is a member, let T be the innermost such type declaration. The class or interface to search is T.

...

所以 T 是 DrawPanel。

Sec 15.12.2.1

The class or interface determined by compile-time step 1 (§15.12.1) is searched for all member methods that are potentially applicable to this method invocation; members inherited from superclasses and superinterfaces are included in this search.

因此在 DrawPanel 及其所有超classes 中搜索名为 addMouseListener 的方法。

为什么方法 addMouseListener 不需要 super？

Why method addMouseListener doesn't need a super?

java

swing

mouselistener

super