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根据javadocs“子类继承其超类的所有成员（字段、方法和嵌套类） ”。此外，Java 通过引用来操作对象 那么为什么这个子类会为 aList[0] 返回错误的值呢？当我希望它们都修改同一个数组时，似乎每个类都在修改自己的数组。

public class mystery {
protected List<String> aList;

public mystery() {
    aList = new ArrayList<String>();
}

public void addToArray() {
    //"foo" is successfully added to the arraylist
    aList.add("foo");
}

public void printArray() {
    System.out.println( "printArray " + aList.get(0) +"" );
}
public static void main(String[] args) {
    mystery prob1 = new mystery();
    mysterySubclass prob2 = new mysterySubclass();
    //add "foo" to array
    prob1.addToArray();
    //add "bar" to array
    prob2.addToArray2();

    //expect to print "foo", works as expected
    prob1.printArray();
    //expect to print "foo", but actually prints "bar"
    prob2.printArray();
    //expect to print "foo", but actually prints "bar"
    prob2.printArray2();
}

}

public class mysterySubclass extends mystery {

public void mysterySubclass() {}
public void addToArray2() {
    aList.add("bar");
}

public void printArray2() {
    System.out.println( "printArray2 " + super.aList.get(0) +"" );
}

}

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3 回答 3

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    mystery prob1 = new mystery();
    mysterySubclass prob2 = new mysterySubclass();

在这里，您正在创建两个新对象，因此它们将拥有自己的列表，并且不会像您期望的那样共享一个。

您正在寻找的是：

    mystery prob1 = new mysterySubclass();
    mysterySubclass prob2 = (mysterySubclass)prob1;

试试这个..

于 2012-11-30T05:07:42.273 回答
0

那是因为您调用prob2.addToArray2();它会将"bar"String 添加到变量中的aList属性。prob2考虑到它们共享相同的属性，而不是相同的引用对象引用，所以每个prob都有prob2自己的aList属性值。使用调试器可以轻松证明此行为。

于 2012-11-30T04:49:44.917 回答
0 
mystery prob1 = new mystery();
mysterySubclass prob2 = new mysterySubclass();
//add "foo" to array
prob1.addToArray();
//add "bar" to array
prob2.addToArray2();

//expect to print "foo", works as expected
prob1.printArray();
//expect to print "foo", but actually prints "bar"
prob2.printArray();
//expect to print "foo", but actually prints "bar"
prob2.printArray2();

你的 prob1 有 aList 你的 prob2 有 aList 是的，它们是不同的。

第 4 行之后 => prob1.aList -- foo

第 6 行之后 => prob2.aList -- bar

所以 prob2.printArray() 只有“bar”要打印，而且 prob2.printArray2() 基本上它们做同样的事情。

于 2012-11-30T05:20:26.870 回答