如何在Java和Android中使用构造函数？

Question

I have a short question about the following code from

http://www.androidhive.info/2013/09/android-sqlite-database-with-multiple-tables/

Here are used two Constructors, one with the id, and the other without - I don't understand why. What's the benefit?

I already read this thread:

Why does this class have two constructors?

The answer I could understand is, that I can create a Tag with id and not, but I'm trying to understand, how to know which constructor it shall use? Is it just by the number of parameters?

    public class Tag {

    int id;
    String tag_name;

    // constructors
    public Tag() {

    }

    public Tag(String tag_name) {
        this.tag_name = tag_name;
    }

    public Tag(int id, String tag_name) {
        this.id = id;
        this.tag_name = tag_name;
    }

    // ...     
}

Answer 1

Yes, only by its amount of parameters.

It's called "overloading" of functions. You can overload a function by providing the same signature with different parameters (according to their type and order).

The JVM will then decide which method to use in a certain situation.

Please note: If you provide a constructor the JVM won't provide a default constructor any more.

class Foo{

private int x;
private String name;

    Foo(int x){      //constructor 1
        this(x, "Default Name");
    }
    Foo(String name){  //constructor 2
        this(0, name);
    }
    Foo(int x, String name){  //constructor 3
        this.x = x;
        this.name = name;
    }
}

Foo f1 = new Foo(9); //calls constructor 1, who will call constructor 3
                     //f1.x = 9, f1.name = "Default Name"
Foo f2 = new Foo("Test"); // calls constructor 2, who will call constructor 3
                          // f2.x = 0; f2.name = "Test"
Foo f3 = new Foo(3, "John"); //calls constructor 3
                             // f3.x = 3; f3.name = "John"

Foo f4 = new Foo()  // This won't work! No default Constructor provided!

Answer 2

**which constructor it shall use? only of its amount of paremeters? ** Yes, for example If you call

Tag newTag = new Tag();

it will call

 public Tag() {

}

but if you call

 Tag newTag = new Tag("Name");

it will call

 public Tag(String tag_name) {

}

and so on

by how many arguments you pass to the constructor that way it will know which one to call

Answer 3

The compiler knows "which constructor it shall use" by a rule, in the Java Language Specification .

A resume: Is by signature of the method (the type and order of arguments --- the exceptions dont effect the signature, and the return type too). This is not restricted only by the constructor, but any method, properly Overloaded; You can study these by topic of "Overloading". The reason to Overload a method --- or a constructor too ---, is to give more flexibility.