Java 隐式超级构造函数 Person() 未定义。必须显式调用另一个构造函数?
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implicit super constructor Person() is undefined. Must explicitly invoke another constructor?
提问by sirnomnomz
I am working on a project and i am getting the error "implicit super constructor Person() is undefined. Must explicitly invoke another constructor" and i don't quite understand it.
我正在处理一个项目,但收到错误“隐式超级构造函数 Person() 未定义。必须显式调用另一个构造函数”,我不太明白。
Here is my person class:
这是我的个人课程:
public class Person {
public Person(String name, double DOB){
}
}
And my student class when trying to implement the person class, and giving it an instructor variable.
和我的学生类在尝试实现人类时,并给它一个讲师变量。
public class Student extends Person {
public Student(String Instructor) {
}
}
采纳答案by pxm
If a constructor does not explicitly invoke a superclass constructor, the Java compiler automatically inserts a call to the no-argument constructor of the superclass.
If the super class does not have a no-argument constructor, you will get a compile-time error. Object does have such a constructor, so if Object is the only superclass, there is no problem.
如果构造函数没有显式调用超类构造函数,Java 编译器会自动插入对超类的无参数构造函数的调用。
如果超类没有无参数构造函数,您将收到编译时错误。Object 确实有这样的构造函数,所以如果 Object 是唯一的超类,就没有问题。
Reference: http://docs.oracle.com/javase/tutorial/java/IandI/super.html: (See under section 'SubClass Constructors')
参考:http://docs.oracle.com/javase/tutorial/java/IandI/super.html :(参见“子类构造函数”部分)
So whenever dealing with parameterized constructors make a super(parameter1, parameter2 ..)call to the parent constructor.
Also this super() call should be the FIRST line in your constructor block.
因此,每当处理参数化构造函数时,都会 super(parameter1, parameter2 ..)调用父构造函数。
此外,这个 super() 调用应该是构造函数块中的第一行。
回答by rgettman
When creating a subclass constructor, if you don't explicitly call a superclass constructor with super, then Java will insert an implicit call to the no-arg "default" superclass constructor, i.e. super();.
创建子类构造函数时,如果您没有使用 显式调用超类构造函数super,那么Java 将插入对无参数“默认”超类构造函数的隐式调用,即super();.
However, your superclass Persondoesn't have a no-arg constructor. Either supply an explicit no-arg constructor in Personor explicitly call the existing superclass constructor in the Studentconstructor.
但是,您的超类Person没有无参数构造函数。在构造函数中提供显式无参数构造函数Person或显式调用现有超类构造Student函数。
回答by Rogue
You need to make a supercall to your defined constructor:
您需要super调用您定义的构造函数:
public Student(String instructor) {
super(/* name */, /* date of birth */);
}
You can't just call super()because that constructor is not defined.
您不能super()因为未定义该构造函数而只是调用。
回答by EasterBunnyBugSmasher
you can't create an instance without calling a constructor of its super class. And the jvm doesn't know how to call Person(String, double) from your Student(String) constructor.
您不能在不调用其超类的构造函数的情况下创建实例。并且 jvm 不知道如何从您的 Student(String) 构造函数调用 Person(String, double) 。
回答by Jaime Montoya
This is how I achieved it (in my case the superclass was Team and the subclass was Scorer):
这就是我实现它的方式(在我的例子中,超类是 Team,子类是 Scorer):
// Team.java
public class Team {
String team;
int won;
int drawn;
int lost;
int goalsFor;
int goalsAgainst;
Team(String team, int won, int drawn, int lost, int goalsFor, int goalsAgainst){
this.team = team;
this.won = won;
this.drawn = drawn;
this.lost = lost;
this.goalsFor = goalsFor;
this.goalsAgainst = goalsAgainst;
}
int matchesPlayed(){
return won + drawn + lost;
}
int goalDifference(){
return goalsFor - goalsAgainst;
}
int points(){
return (won * 3) + (drawn * 1);
}
}
// Scorer.java
public class Scorer extends Team{
String player;
int goalsScored;
Scorer(String player, int goalsScored, String team, int won, int drawn, int lost, int goalsFor, int goalsAgainst){
super(team, won, drawn, lost, goalsFor, goalsAgainst);
this.player = player;
this.goalsScored = goalsScored;
}
float contribution(){
return (float)this.goalsScored / (float)this.goalsFor;
}
float goalsPerMatch(){
return (float)this.goalsScored/(float)(this.won + this.drawn + this.lost);
}
}

