java - 创建子类时是否需要 super ？

Question

在此程序中，不需要 super 来访问父类(super class)的构造函数:

class Base{

    Base(){

        System.out.println("Base");
    }
}

public class test2 extends Base{

    test2() {

        //super();
        System.out.print("test2"); 
    }

    public static void main(String argv[]){

        test2 c = new test2();
    }    
}

但是这个程序需要 super 并在 quest1 构造函数中给出错误

constructor quest can't be applied to given types: required int, found no arguments

class Quest {

    Quest(int y){

        System.out.print("A:"+y);
    }
}

class Quest1 extends Quest {

    Quest1(int x){

        //super(x+1); 
        System.out.print("B:"+x);
    }
}

class Test {

    public static void main(String argv[]){

        Quest1 q = new Quest1(5); 
    }
}

Answer 1

JLS 8.8.7. Constructor Body

If a constructor body does not begin with an explicit constructor invocation and the constructor being declared is not part of the primordial class Object, then the constructor body implicitly begins with a superclass constructor invocation "super();", an invocation of the constructor of its direct superclass that takes no arguments.

在

class Base {

    Base() {

    }
}

public class test2 extends Base {

    test2() {

        //super();
        System.out.print("test2"); 
    }  
}

注释掉的行是自动添加的，并且由于定义了父类(super class)的无参构造函数，因此不会出现错误。

如果是

class Quest {

    Quest(int y) {

        System.out.print("A:"+y);
    }
}

class Quest1 extends Quest {

    Quest1(int x) {

        //super(x+1); 
        System.out.print("B:"+x);
    }
}

隐式调用super()试图调用父类(super class)中未定义的构造函数，这会导致错误

Implicit super constructor Quest() is undefined. Must explicitly invoke another constructor.

取消注释显式构造函数调用将替换隐式调用，从而解决问题。或者，在父类(super class)中定义一个无参构造函数。