Java要求,如果在构造函数中调用this()或super(),它必须是第一条语句。为什么?

例如:

public class MyClass {
    public MyClass(int x) {}
}

public class MySubClass extends MyClass {
    public MySubClass(int a, int b) {
        int c = a + b;
        super(c);  // COMPILE ERROR
    }
}

Sun编译器说,调用super必须是构造函数中的第一条语句。Eclipse编译器说,构造函数调用必须是构造函数中的第一个语句。

然而,你可以通过稍微重新安排代码来解决这个问题:

public class MySubClass extends MyClass {
    public MySubClass(int a, int b) {
        super(a + b);  // OK
    }
}

下面是另一个例子:

public class MyClass {
    public MyClass(List list) {}
}

public class MySubClassA extends MyClass {
    public MySubClassA(Object item) {
        // Create a list that contains the item, and pass the list to super
        List list = new ArrayList();
        list.add(item);
        super(list);  // COMPILE ERROR
    }
}

public class MySubClassB extends MyClass {
    public MySubClassB(Object item) {
        // Create a list that contains the item, and pass the list to super
        super(Arrays.asList(new Object[] { item }));  // OK
    }
}

因此,它不会阻止您在调用super()之前执行逻辑。它只是阻止您执行无法放入单个表达式中的逻辑。

调用this()也有类似的规则。编译器说,调用this必须是构造函数中的第一条语句。

为什么编译器有这些限制?你能给出一个代码例子,如果编译器没有这个限制,就会发生不好的事情吗?


当前回答

实际上,super()是构造函数的第一个语句,因为要确保在构造子类之前父类已经完全形成。即使在第一个语句中没有super(),编译器也会为你添加它!

其他回答

在构造子对象之前,必须先创建父对象。 如你所知,当你这样写类时:

public MyClass {
        public MyClass(String someArg) {
                System.out.println(someArg);
        }
}

它转向下一个(extend和super只是被隐藏了):

public MyClass extends Object{
        public MyClass(String someArg) {
                super();
                System.out.println(someArg);
        }
}

First we create an Object and then extend this object to MyClass. We can not create MyClass before the Object. The simple rule is that parent's constructor has to be called before child constructor. But we know that classes can have more that one constructor. Java allow us to choose a constructor which will be called (either it will be super() or super(yourArgs...)). So, when you write super(yourArgs...) you redefine constructor which will be called to create a parent object. You can't execute other methods before super() because the object doesn't exist yet (but after super() an object will be created and you will be able to do anything you want).

So why then we cannot execute this() after any method? As you know this() is the constructor of the current class. Also we can have different number of constructors in our class and call them like this() or this(yourArgs...). As I said every constructor has hidden method super(). When we write our custom super(yourArgs...) we remove super() with super(yourArgs...). Also when we define this() or this(yourArgs...) we also remove our super() in current constructor because if super() were with this() in the same method, it would create more then one parent object. That is why the same rules imposed for this() method. It just retransmits parent object creation to another child constructor and that constructor calls super() constructor for parent creation. So, the code will be like this in fact:

public MyClass extends Object{
        public MyClass(int a) {
                super();
                System.out.println(a);
        }
        public MyClass(int a, int b) {
                this(a);
                System.out.println(b);
        }
}

正如其他人所说,你可以这样执行代码:

this(a+b);

你也可以像这样执行代码:

public MyClass(int a, SomeObject someObject) {
    this(someObject.add(a+5));
}

但是你不能像这样执行代码,因为你的方法还不存在:

public MyClass extends Object{
    public MyClass(int a) {

    }
    public MyClass(int a, int b) {
        this(add(a, b));
    }
    public int add(int a, int b){
        return a+b;
    }
}

此外,在this()方法链中必须有super()构造函数。你不能像这样创建一个对象:

public MyClass{
        public MyClass(int a) {
                this(a, 5);
        }
        public MyClass(int a, int b) {
                this(a);
        }
}

我知道我有点晚了,但我已经用过几次这个技巧了(我知道这有点不寻常):

我用一个方法创建了一个泛型接口InfoRunnable<T>:

public T run(Object... args);

如果我需要在把它传递给构造函数之前做一些事情,我只需要这样做:

super(new InfoRunnable<ThingToPass>() {
    public ThingToPass run(Object... args) {
        /* do your things here */
    }
}.run(/* args here */));

这是因为你的构造函数依赖于其他构造函数。要使你的构造函数正常工作,其他构造函数正常工作是必要的,这是依赖的。这就是为什么有必要首先检查由this()或super()在构造函数中调用的依赖构造函数。如果由this()或super()调用的其他构造函数有问题,那么什么点执行其他语句,因为如果被调用的构造函数失败,所有的构造函数都会失败。

构造函数按照的顺序完成执行是有意义的 推导。因为父类不知道任何子类,任何 它需要执行的初始化与可能的初始化是分开的 子类执行任何初始化的先决条件。 因此,它必须首先完成它的执行。

一个简单的演示:

class A {
    A() {
        System.out.println("Inside A's constructor.");
    }
}

class B extends A {
    B() {
        System.out.println("Inside B's constructor.");
    }
}

class C extends B {
    C() {
        System.out.println("Inside C's constructor.");
    }
}

class CallingCons {
    public static void main(String args[]) {
        C c = new C();
    }
}

这个程序的输出是:

Inside A's constructor
Inside B's constructor
Inside C's constructor

在子类构造函数中添加super()的主要目标是编译器的主要工作是将所有类与Object类建立直接或间接的连接,这就是为什么编译器检查我们是否提供了super(参数化),然后编译器不承担任何责任。 这样所有的实例成员从Object初始化为子类。