Weaker access privileges java

Fragment’s onResume() cannot have weaker access privileges than public

The accessibility of interface methods is inherent, meaning that you cannot decrease the visibility of the method in the class that implements it. This implies that the subclass is more versatile than the parent and can implement additional methods.

Java attempting to assign weaker access privilege error

The two lines below are causing the issue.

new Wait("") >;session().open("/"); new Wait("") >;session().click("id=btnLogin-button"); 

You attempt to substitute the until function, having public rights, in the com.thoughtworks.selenium.Wait category with a until function that is only viewable within the package.

Modifying the visibility of a method through overriding is restricted to increasing it, rather than reducing it. For instance, it is possible to override a method with protected visibility and enhance it to public visibility, but the opposite is impossible.

To resolve the issue, it is necessary to include the public keyword in the affected methods.

new Wait("") >;session().open("/"); new Wait("") >;session().click("id=btnLogin-button"); 

Android AsyncTask error, attempting to assign weaker access, You should make your method processFinish public: new AsyncGETRequestProcess(new AsyncGETRequestProcess.AsyncResponse()< //THE ABOVE ERROR

Cannot override onResume() in Fragment attempting to assign weaker access privileges; was public

In Java inheritance, it is not possible to decrease the access modifiers. For instance, consider a scenario where a base class has a function defined as public and an external user tries to invoke it. However, if the access modifier is changed for the dynamic type, the user will face unexpected behavior as they won’t be able to use the function. This is why Java prohibits such behavior and the solution is to revert the access modifier of onResume back to public.

Java extend static method, can’t compile, why?, the Derived.f()’s modifier must be public, if it’s private ,the compiler say «attempting to assign weaker access privileges; was public».

Why can’t we assign weaker privilege in subclass

It is prohibited to break type substitutability, which is also known as the Liskov Substitution Principle (LSP).

The essence of polymorphism in Java (and other programming languages) is that you can treat a subclass instance as though it were a superclass instance. However, if a method is confined to the subclass, the compiler may not be able to determine whether the access rules permit the method to be invoked.

Suppose your sample code were legitimate, for example.

// Assume this code is in some other class . SuperClass s1 = new SuperClass(); s1.foo(); // OK! SuperClass s2 = new Subclass(); s2.foo(); // What happens now? SuperClass s3 = OtherClass.someMethod(); s3.foo(); // What happens now? 

Allowing a call to a private method from outside the abstraction boundary of Subclass can be the consequence if the decision on whether s2.foo() is permitted is based on the declared type of s2 .

To perform a static access check, it’s impossible to rely on the actual object type referred by s2 . This becomes more evident in the s3 scenario. The compiler lacks the ability to predict the actual type of the object returned by someMethod .

Java applications are susceptible to bugs caused by access checks that may lead to runtime exceptions. However, the discussed language restriction prevents this issue from arising.

It is not possible to limit entry as you have granted additional entry in the superclass as an example.

SuperClass sc = new SubClass(); sc.foo(); // is package local, not private. 

The determination of access for sc is based on the reference type, sc , rather than the type it refers to. The compiler cannot always determine the object’s type during runtime, making this a risky assumption. To ensure safety, the subclass must adhere to the parent’s contract. If the subclass fails to do so, it becomes an invalid subclass. This is analogous to the parent implementing a method while the subclass declares it as inaccessible or not implemented.

One way to address this issue is to restrict access to the sub-class method through the parent class only, rather than directly. However, it can be difficult to predict when a parent class might introduce a new method. This is why it’s important to use the private approach, as it ensures that the method remains private and cannot be accessed through other means.

By using reflection, it is still possible to access private methods. However, this can cause various issues for the JVM, such as having to keep private methods even when it is unlikely that they will be called through normal means.

Essentially, the desired code should have a clear definition and not be ambiguous. It should either be designated as package local or private, without any in-between state. This is not an issue if the subclass is public, as it can simply be utilized in more contexts than its parent class and implement additional methods.

Allowing this could create a means of accessing methods that should remain inaccessible through a backdoor.

Lets say that this is allowed

class Super < public void method() < System.out.println("Super"); >> class Sub extends Super < // This is not allowed, but suppose it was allowed protected void method() < System.out.println("Sub"); >> // In another class, in another package: Super obj = new Sub(); obj.method(); 

It is feasible for obj.method to occur as public is present in the Super class. However, it is not advisable to allow it as the obj refers to an instance of Sub where the method is protected.

To prevent external access to a method in class Sub, a restriction is implemented.

Cannot set method in inner class as private after extends from local, ‘innerMethod()’ in ‘InnerClassSubclass’ clashes with ‘innerMethod()’ in ‘InnerClass’; attempting to assign weaker access privileges (‘private’);

Why the overridden method should be public in this program?

The int divisor_sum(int n) method implements an interface method, which has implicit public access. Therefore, it is not possible to decrease the visibility of this method in the implementing class, even if it is not explicitly specified.

MyCalculator mc = new MyCalculator(); AdvancedArithmetic aa = mc; 

To be able to call a method in a certain class through a class reference ( mc.divisor_sum(4) ), the access level of that method ( divisor_sum() ) in the class ( MyCalculator ) must be provided a certain level of access ( public ). If the access level is not provided, the method can still be called using an interface reference ( aa.divisor_sum(4) ), but this is not allowed as it does not make sense.

One approach to consider is programming with interfaces, which ensure the availability of a method with a default access modifier of public .

public void doSomething(AdvancedArithmetic implementation) < int test = implementation.divisor_sum(5); // continue >

Allowing any implementation of interface in Java to narrow its access modifier compared to interface would result in significant issues and software breakage. This is because such an implementation would not adhere to the interface contract.

Java — Why can I not call the interface method?, The OnClickListener method onClick must be overridden like this: chatButton.setOnClickListener(new View.OnClickListener() < @Override public void onClick

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Weaker access privileges java

As per the rule in overriding , you cannot apply weaker access specifier over a stronger access specifier.Suppose , your parent class has a method Display() with stronger access specifier like Public, then you cannot use weaker access specifier like private or public when you are overriding the Display() method.

Public is the 1st stronger access specifier
Protected is the 2nd stronger access specifier
Default is the 3rd stronger access specifier
Private is the most weaker access specifier

Always keep in mind about the access specifiers has vital role in inheritance (OOPS concept).

class AccessTest protected void display() < //Stronger Access specifier
System.out.println(«Hello AccessTest:Display»);
>
>

class TestWithMain extends AccessTest // I am trying to override with weaker specifier , it show error
// You can use public or protected (higher or same level of access specifier)
private void display() <
System.out.println(«Hello TestWithMain:display»);
>
public static void main(String str[]) <
AccessTest acObj=new TestWithMain();
acObj.display();
>
>

TestWithMain.java:8: display() in TestWithMain cannot override display() in AccessTest;
attempting to assign weaker access privileges; was protected
private void display()

So, now if you change the access specifier to protected or public then it will work properly.

public void display() <
System.out.println(«Hello TestWithMain:display»);
>

protected void display() <
System.out.println(«Hello TestWithMain:display»);
>

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Weaker Access Privilege

posted 13 years ago

when we override why cant we grant weaker access to overridden methods. i need ans in object oriented way. if weaker access to granted in subclass what are all possible problems that could come up..ans with example would be more elaborative

Sheriff

posted 13 years ago

SCJP 1.4 — SCJP 6 — SCWCD 5 — OCEEJBD 6 — OCEJPAD 6
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posted 13 years ago

In what sense do you mean «weaker»? I mean, you can change a protected superclass method to public. Overriding a private superclass method obviously doesn’t make sense.

Sheriff

posted 13 years ago

SCJP 1.4 — SCJP 6 — SCWCD 5 — OCEEJBD 6 — OCEJPAD 6
How To Ask Questions How To Answer Questions

posted 13 years ago

posted 13 years ago

posted 13 years ago

Java Cowboy

posted 13 years ago

Because there is an is a relationship between the superclass and the subclass (the Liskov substitution principle that Rob mentioned). If this were allowed, there would be a backdoor through which you could call methods that should not be accessible.

You can assign an instance of a subclass to a variable that has the type of the superclass:

The call in line 17 would be possible, because method() is public in class Super. But it should not be allowed, because obj is really referring to an instance of Sub, and in that class, the method is protected! So, there is a backdoor through which you could now suddenly call a method in class Sub that should not be accessible from the outside.

To prevent this, it’s not allowed to give an overriding method more restricted access than the method it is overriding.

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