问题
Why can't I use protected constructors outside the package for this piece of code:
package code;
public class Example{
protected Example(){}
...
}
Check.java
package test;
public class Check extends Example {
void m1() {
Example ex=new Example(); //compilation error
}
}
- Why do i get the error even though i have extended the class? Please explain
EDIT:
Compilation error:
The constructor Example() is not visible
回答1:
protected modifier is used only with in the package and in sub-classes outside the package. When you create a object using Example ex=new Example(); it will call parent class constructor by default.
As parent class constructor being protected you are getting a compile time error. You need to call the protected constructor according to JSL 6.6.2.2 as shown below in example 2.
package Super;
public class SuperConstructorCall {
protected SuperConstructorCall() {
}
}
package Child;
import Super.SuperConstructorCall;
public class ChildCall extends SuperConstructorCall
{
public static void main(String[] args) {
SuperConstructorCall s = new SuperConstructorCall(); // Compile time error saying SuperConstructorCall() has protected access in SuperConstructorCall
}
}
Example 2 conforming to JLS 6.6.2.2:
package Super;
public class SuperConstructorCall {
protected SuperConstructorCall() {
}
}
package Child;
import Super.SuperConstructorCall;
public class ChildCall extends SuperConstructorCall
{
public static void main(String[] args) {
SuperConstructorCall s = new SuperConstructorCall(){}; // This will work as the access is by an anonymous class instance creation expression
}
}
回答2:
Usually protected means only accessible to subclasses or classes in the same package. However here are the rules for constructors from the JLS:
6.6.2.2. Qualified Access to a protected Constructor
Let C be the class in which a protected constructor is declared and let S be the innermost class in whose declaration the use of the protected constructor occurs. Then:
If the access is by a superclass constructor invocation super(...), or a qualified superclass constructor invocation E.super(...), where E is a Primary expression, then the access is permitted.
If the access is by an anonymous class instance creation expression new C(...){...}, or a qualified anonymous class instance creation expression E.new C(...){...}, where E is a Primary expression, then the access is permitted.
If the access is by a simple class instance creation expression new C(...), or a qualified class instance creation expression E.new C(...), where E is a Primary expression, or a method reference expression C :: new, where C is a ClassType, then the access is not permitted. A protected constructor can be accessed by a class instance creation expression (that does not declare an anonymous class) or a method reference expression only from within the package in which it is defined.
As an example, this does not compile
public class Example extends Exception {
void method() {
Exception e = new Exception("Hello", null, false, false);
}
}
but this does
public class Example extends Exception {
Example() {
super("Hello", null, false, false);
}
}
and so does this
public class Example {
void method() {
Exception e = new Exception("Hello", null, false, false) {};
}
}
So the rules are clear, but I can't say I understand the reasons behind them!
回答3:
In fact you are already using protected constructor of Example because Check has an implicit constructor and implicit Example constructor call:
public Check() {
super();
}
来源：https://stackoverflow.com/questions/29530199/why-cant-i-use-protected-constructors-outside-the-package