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

在调用子对象的构造函数之前,可以使用匿名初始化块初始化子对象中的字段。这个例子将演示:

public class Test {
    public static void main(String[] args) {
        new Child();
    }
}

class Parent {
    public Parent() {
        System.out.println("In parent");
    }
}

class Child extends Parent {

    {
        System.out.println("In initializer");
    }

    public Child() {
        super();
        System.out.println("In child");
    }
}

这将输出:

在父 在初始化 在儿童

其他回答

在调用子对象的构造函数之前,可以使用匿名初始化块初始化子对象中的字段。这个例子将演示:

public class Test {
    public static void main(String[] args) {
        new Child();
    }
}

class Parent {
    public Parent() {
        System.out.println("In parent");
    }
}

class Child extends Parent {

    {
        System.out.println("In initializer");
    }

    public Child() {
        super();
        System.out.println("In child");
    }
}

这将输出:

在父 在初始化 在儿童

你问为什么,其他的答案,在我看来,并没有真正说为什么它是可以调用你的超级构造函数,但只有当它是第一行。原因是您并没有真正调用构造函数。在c++中,等价的语法是

MySubClass: MyClass {

public:

 MySubClass(int a, int b): MyClass(a+b)
 {
 }

};

When you see the initializer clause on its own like that, before the open brace, you know it's special. It runs before any of the rest of the constructor runs and in fact before any of the member variables are initialized. It's not that different for Java. There's a way to get some code (other constructors) to run before the constructor really starts, before any members of the subclass are initialized. And that way is to put the "call" (eg super) on the very first line. (In a way, that super or this is kind of before the first open brace, even though you type it after, because it will be executed before you get to the point that everything is fully constructed.) Any other code after the open brace (like int c = a + b;) makes the compiler say "oh, ok, no other constructors, we can initialize everything then." So it runs off and initializes your super class and your members and whatnot and then starts executing the code after the open brace.

如果几行之后,它遇到一些代码说“哦,是的,当你构造这个对象时,这是我想让你传递给基类构造函数的参数”,那就太晚了,这没有任何意义。所以你会得到一个编译器错误。

Tldr:

其他的答案都解决了这个问题的“为什么”。我将提供一个关于这个限制的hack:

基本思想是用嵌入式语句劫持超级语句。这可以通过将语句伪装成表达式来实现。

Tsdr:

假设我们想在调用super()之前执行Statement1()到Statement9():

public class Child extends Parent {
    public Child(T1 _1, T2 _2, T3 _3) {
        Statement_1();
        Statement_2();
        Statement_3(); // and etc...
        Statement_9();
        super(_1, _2, _3); // compiler rejects because this is not the first line
    }
}

编译器当然会拒绝我们的代码。所以,我们可以这样做:

// This compiles fine:

public class Child extends Parent {
    public Child(T1 _1, T2 _2, T3 _3) {
        super(F(_1), _2, _3);
    }

    public static T1 F(T1 _1) {
        Statement_1();
        Statement_2();
        Statement_3(); // and etc...
        Statement_9();
        return _1;
    }
}

唯一的限制是父类必须有一个构造函数,该构造函数必须至少接受一个参数,以便我们可以将语句作为表达式潜入。

这里有一个更详细的例子:

public class Child extends Parent {
    public Child(int i, String s, T1 t1) {
        i = i * 10 - 123;
        if (s.length() > i) {
            s = "This is substr s: " + s.substring(0, 5);
        } else {
            s = "Asdfg";
        }
        t1.Set(i);
        T2 t2 = t1.Get();
        t2.F();
        Object obj = Static_Class.A_Static_Method(i, s, t1);
        super(obj, i, "some argument", s, t1, t2); // compiler rejects because this is not the first line
    }
}

改写成:

// This compiles fine:

public class Child extends Parent {
    public Child(int i, String s, T1 t1) {
        super(Arg1(i, s, t1), Arg2(i), "some argument", Arg4(i, s), t1, Arg6(i, t1));
    }

    private static Object Arg1(int i, String s, T1 t1) {
        i = Arg2(i);
        s = Arg4(s);
        return Static_Class.A_Static_Method(i, s, t1);
    }

    private static int Arg2(int i) {
        i = i * 10 - 123;
        return i;
    }

    private static String Arg4(int i, String s) {
        i = Arg2(i);
        if (s.length() > i) {
            s = "This is sub s: " + s.substring(0, 5);
        } else {
            s = "Asdfg";
        }
        return s;
    }

    private static T2 Arg6(int i, T1 t1) {
        i = Arg2(i);
        t1.Set(i);
        T2 t2 = t1.Get();
        t2.F();
        return t2;
    }
}

事实上,编译器可以为我们自动化这个过程。他们只是选择不这么做。

我猜他们这样做是为了让那些编写处理Java代码的工具的人更轻松,在某种程度上也让阅读Java代码的人更轻松。

如果允许super()或this()调用移动,则需要检查更多的变化。例如,如果您将super()或this()调用移动到条件if()中,它可能必须足够聪明,将隐式super()插入到else中。如果调用super()两次,或者同时使用super()和this(),它可能需要知道如何报告错误。它可能需要禁止在接收器上调用方法,直到super()或this()被调用,并计算出何时会变得复杂。

让每个人都做额外的工作似乎是弊大于利。

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?

class Good {
    int essential1;
    int essential2;

    Good(int n) {
        if (n > 100)
            throw new IllegalArgumentException("n is too large!");
        essential1 = 1 / n;
        essential2 = n + 2;
    }
}

class Bad extends Good {
    Bad(int n) {
        try {
            super(n);
        } catch (Exception e) {
            // Exception is ignored
        }
    }

    public static void main(String[] args) {
        Bad b = new Bad(0);
//        b = new Bad(101);
        System.out.println(b.essential1 + b.essential2);
    }
}

An exception during construction almost always indicates that the object being constructed could not be properly initialized, now is in a bad state, unusable, and must be garbage collected. However, a constructor of a subclass has got the ability to ignore an exception occurred in one of its superclasses and to return a partially initialized object. In the above example, if the argument given to new Bad() is either 0 or greater than 100, then neither essential1 nor essential2 are properly initialized.

你可能会说忽略异常总是一个坏主意。好的,这里还有一个例子:

class Bad extends Good {
    Bad(int n) {
        for (int i = 0; i < n; i++)
            super(i);
    }
}

很有趣,不是吗?在这个例子中我们创建了多少个对象?一个?两个?或者什么都没有……

允许在构造函数中间调用super()或this()将打开一个令人讨厌的构造函数的潘多拉盒子。


另一方面,我理解在调用super()或this()之前经常需要包含一些静态部分。这可能是任何不依赖于此引用的代码(实际上,它已经存在于构造函数的最开始,但在super()或This()返回之前不能有序使用),并且需要进行这样的调用。此外,像在任何方法中一样,在调用super()或this()之前创建的一些局部变量可能会在调用super()或this()之后被需要。

在这种情况下,你有以下机会:

Use the pattern presented at this answer, which allows to circumvent the restriction. Wait for the Java team to allow pre-super() and pre-this() code. It may be done by imposing a restriction on where super() or this() may occur in a constructor. Actually, even today's compiler is able to distinguish good and bad (or potentially bad) cases with the degree enough to securely allow static code addition at the beginning of a constructor. Indeed, assume that super() and this() return this reference and, in turn, your constructor has

return this;

最后。以及编译器拒绝代码

public int get() {
    int x;
    for (int i = 0; i < 10; i++)
        x = i;
    return x;
}

public int get(int y) {
    int x;
    if (y > 0)
        x = y;
    return x;
}

public int get(boolean b) {
    int x;
    try {
        x = 1;
    } catch (Exception e) {
    }
    return x;
}

with the error "variable x might not have been initialized", it could do so on this variable, making its checks on it just like on any other local variable. The only difference is this cannot be assigned by any means other than super() or this() call (and, as usual, if there is no such call at a constructor, super() is implicitly inserted by compiler in the beginning) and might not be assigned twice. In case of any doubt (like in the first get(), where x is actually always assigned), the compiler could return an error. That would be better than simply return error on any constructor where there is something except a comment before super() or this().