质疑代码执行顺序

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英文:

Questioning the order in which the code is executed

问题

我发现了这段代码(https://stackoverflow.com/questions/25340310/java-constructor-does-not-explicitly-invoke-a-superclass-constructor-java-doe)...

输出结果是:a2 a1 b2 b3 b1 c1 a2 a1 b2 b3 b1 c1 c2

我预期输出应该只有 c1 c2,但是我无法理解代码执行的顺序。

我希望有一个详细的解释,只是为了确保我足够清楚地理解事情。谢谢!

英文:

So I stumbled across this piece of code (https://stackoverflow.com/questions/25340310/java-constructor-does-not-explicitly-invoke-a-superclass-constructor-java-doe)...

public class App {
    public static void main(String[] args){
        new C();
        new C(1.0);
    }
}

class A {
    public A(){
        this(5);
        System.out.println("a1");
    }

    public A(int x){
        System.out.println("a2");
    }
}

class B extends A {
    public B(){
        this(false);
        System.out.println("b1");
    }

    public B(int x){
        super();
        System.out.println("b2");
    }

    public B(boolean b){
        this(2);
        System.out.println("b3");
    }
}

class C extends B {
    public C(){
        System.out.println("c1");
    }

    public C(double x){
        this();
        System.out.println("c2");
    }
}

And the output that I got is a2 a1 b2 b3 b1 c1 a2 a1 b2 b3 b1 c1 c2

I was expecting it to be c1 c2 only but I just cannot wrap my head around the sequence in which the code is executed.

I would love a detailed explanation, just to make sure I understand things clearly enough. Thank you!

答案1

得分: 1

继承类在构造函数的开始处隐式调用super()

因此,执行顺序如下:

C() --> B() --> A() --> A(5) --> print('a2') --> print('a1') --> B(false) --> B(2) --> print('b2') --> print('b3') --> print('b1') --> print('c1')(这完成了对C()的第一次调用)

C(1.0) --> 对于a2, a1, b2, b3, b1, c1,执行完全相同的轨迹,最后执行print('c2')来完成函数。

英文:

Inheriting classes call super() implicitely at the beginning of the constructor.

Thus, the execution is as follows:

C() --> B() --> A() --> A(5) --> print('a2') --> print('a1') --> B(false) --> B(2) --> print('b2') --> print('b3') --> print('b1') --> print('c1') (this finishes the very first call to C())

C(1.0) --> does the exact same trajectory for a2, a1, b2, b3, b1, c1, and then finally print('c2') to finish the function.

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  • 本文由 发表于 2023年2月27日 18:38:28
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