blocks|key|1802125|text|您可以调用您选择的超类构造函数。这可以通过显式调用超类构造函数来实现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1802126|super(para_1,+para_2,........);|code-block|syntax|javascript|1802127|class+BaseA+{
++++BaseA(){
++++++++System.out.println("This+is+BaseA");
++++}
++++BaseA(int+a){
++++++++System.out.println("This+is+BaseA+a");
++++}


}

class+A+extends+BaseA+{

++++A(){
++++++++super(5);
++++++++System.out.println("This+is+A");
++++}

++++public+static+void+main(String[]+args)+{
++++++++A+obj=new+A();

++++}
}|1802128|Output+will+be:
This+is+BaseA+a
This+is+A|1802129|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|D|7|O|8|@]|9|@]|A|$E|F]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

The possibility is that you can call the super class constructor of your choice. That is possible by calling the super class constructor explicitly as :

<pre><code>super(para_1, para_2,........);
</code></pre>

<pre><code>
class BaseA {
 BaseA(){
 System.out.println("This is BaseA");
 }
 BaseA(int a){
 System.out.println("This is BaseA a");
 }


}

class A extends BaseA {

 A(){
 super(5);
 System.out.println("This is A");
 }

 public static void main(String[] args) {
 A obj=new A();

 }
}

</code></pre>

<pre><code>
Output will be:
This is BaseA a
This is A
</code></pre>

blocks|key|2014951|text|不，如果你可以，你的派生对象不会真的是它现在派生的对象，对吧？这将违反is-a原则。所以如果你真的需要它，那么多态性不是你想要的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2014952|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

No and if you could, your derived object wouldn't really be the object it's deriving from now would it? The is-a principle would be violated. So if you really need it, then polymorphism isn't what you're after.

blocks|key|1801364|text|每个超类都需要构造，除了调用构造函数之外，别无他法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1801365|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Every superclass needs to be constructed and there is no other way then calling a constructor.

blocks|key|1801473|text|我认为唯一的办法就是弄乱字节码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1801474|我不确定类加载器或JVM是否会检查是否调用了super()，但是，正如Bozho所写的那样，当这样做时，您可能会得到不一致的对象。|offset|length|style|CODE|1801475|entityMap^0|0|M|7|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

I think the only way to do it is messing up with the byte-code. 
I'm not sure if the Classloader or the JVM checks if <code>super()</code> is being called, but, as Bozho wrote, you probably would end with inconsistent objects when doing so.

blocks|key|2015244|text|不-你不能这样做，为什么你想要这样做呢？这会弄乱你的对象模型。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2015245|无论如何-我相信如果你仍然想这样做，然后你将不得不操作生成的字节代码……有几个库可用来简化字节码的检测。|2015246|强烈建议不要这么做……|2015247|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

Nope - you cannot do it and why would you want to do it anyway? That would mess up your object model.

Anyways - i believe if you still want to do it and then you would have to manipulate the generated byte code.... there are a couple of libraries available that make it easy to instrument the byte code.

Strongly suggest against doing it...

blocks|key|1801668|text|java中的每个对象都是object+(大写为‘O’的对象)的子类。当您创建一个子类的对象时，将调用超类构造函数。即使您的类没有继承任何其他类，它也隐式地继承了Object，因此必须调用Object构造函数。因此，调用super()就是为了达到这个目的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1801669|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Every object in java is a subclass of Object (object with a capital 'O'). when you create an object of a subclass the super class constructor is invoked. Even if your class is not inhereting anyother class, implicitly it is inheriting Object, so the Object constructor has to be called. So super() is invoked for this purpose.

blocks|key|1801741|text|假设你的意思是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1801742|class+B+extends+A+{
+++++public+B()+{+}
}|code-block|syntax|javascript|1801743|那你肯定能做到|1801744|class+B+extends+A+{
+++++public+B()+{
+++++++++this(abort());
+++++}
+++++private+B(Void+dummy)+{
+++++++++/*+super();+*/
+++++}
+++++private+static+Void+abort()+{
+++++++++throw+null;
+++++}
}|1801745|不是很有用。类A的接口not+Java关键字表明您需要运行它的构造函数来构造它，这并不是不合理的。例外情况是，可序列化类是在不调用可序列化类的构造函数的情况下构造的。|offset|length|style|CODE|1801746|entityMap^0|0|0|0|0|7|1|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Assuming you mean

<pre><code>class B extends A {
 public B() { }
}
</code></pre>

then sure you can

<pre><code>class B extends A {
 public B() {
 this(abort());
 }
 private B(Void dummy) {
 /* super(); */
 }
 private static Void abort() {
 throw null;
 }
}
</code></pre>

Not very useful. The interface [not Java keyword] to class <code>A</code> says that you need to run its constructor in order to construct it, not unreasonably. The exception is that serialisable classes are constructed without calling the constructors of the serialisable classes.

blocks|key|2015513|text|我有一个类似的需求，我需要我的子类不通过超类的构造函数，我想要超类的其余好处。既然超级类也是我的，下面是我所做的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2015514|class+SuperClass+{
++++protected+SuperClass()+{
++++++++init();
++++}

++++//+Added+for+classes+(like+ChildClassNew)+who+do+not+want+the+init+to+be+invoked.
++++protected+SuperClass(boolean+doInit)+{
++++++++if+(doInit)
++++++++++++init();
++++}

++++//
}

class+ChildClass1+extends+SuperClass+{
++++ChildClass1()+{
++++++++//+This+calls+default+constructor+of+super+class+even+without+calling+super()+explicitly.
++++++++//+...
++++}
++++//+....
}

class+ChildClass2+extends+SuperClass+{
++++ChildClass2()+{
++++++++//+This+calls+default+constructor+of+super+class+even+without+calling+super()+explicitly.
++++++++//+...
++++}
++++//+....
}

class+ChildClassNew+extends+SuperClass+{
++++ChildClassNew()+{
++++++++/*
+++++++++*+This+is+where+I+didn't+want+the+super+class'+constructor+to+
+++++++++*+be+invoked,+because+I+didn't+want+the+SuperClass'+init()+to+be+invoked.
+++++++++*+So+I+added+overloaded+the+SuperClass'+constructor+where+it+diesn;t+call+init().
+++++++++*+And+call+the+overloaded+SuperClass'+constructor+from+this+new+ChildClassNew.
+++++++++*/
++++++++super(false);
++++++++//+
++++++++//+...
++++}
++++//+....
}|code-block|syntax|javascript|2015515|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I had a similar requirement where I needed my child class NOT to go through the super class' constructor, and I wanted the rest of the benefits of the super class. Since super class is also mine, here's what I did.

<pre><code>class SuperClass {
 protected SuperClass() {
 init();
 }

 // Added for classes (like ChildClassNew) who do not want the init to be invoked.
 protected SuperClass(boolean doInit) {
 if (doInit)
 init();
 }

 //
}

class ChildClass1 extends SuperClass {
 ChildClass1() {
 // This calls default constructor of super class even without calling super() explicitly.
 // ...
 }
 // ....
}

class ChildClass2 extends SuperClass {
 ChildClass2() {
 // This calls default constructor of super class even without calling super() explicitly.
 // ...
 }
 // ....
}

class ChildClassNew extends SuperClass {
 ChildClassNew() {
 /*
 * This is where I didn't want the super class' constructor to 
 * be invoked, because I didn't want the SuperClass' init() to be invoked.
 * So I added overloaded the SuperClass' constructor where it diesn;t call init().
 * And call the overloaded SuperClass' constructor from this new ChildClassNew.
 */
 super(false);
 // 
 // ...
 }
 // ....
}
</code></pre>

If I have a class:

<pre><code>class A {
 public A() { }
}
</code></pre>

and another

<pre><code>class B extends A {
 public B() { }
}
</code></pre>

is there any way to get <code>B.B()</code> not to call <code>A.A()</code>?

Any way to _not_ call superclass constructor in Java?

Java

如果我有一个类：class A {   public A() { }}还有另一个class B extends A {   public B() { }}有没有办法让B.B()不调用A.A()

问有没有办法在Java中用_not_调用超类构造函数？
EN

回答 8

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问有没有办法在Java中用_not_调用超类构造函数？EN

回答 8

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问有没有办法在Java中用_not_调用超类构造函数？
EN