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Why are Java rules for protected access such as they are?


Why are Java rules for protected access such as they are?

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Why are Java rules for protected access such as they are?
Tag : java , By : jbcrail
Date : November 28 2020, 03:01 PM

With these it helps protected is used mainly for utility and helper methods inside a class. They are not meant to be accessible from outside, like private methods.
However it could be necessary for an extending class to access those methods or even to modify them. So basically it is like private, whose right to exist should be clear, but you want to give extending classes the chance to also access it.
            | Class | Package | Subclass | Subclass | World
            |       |         |(same pkg)|(diff pkg)| 
————————————+———————+—————————+——————————+——————————+————————
public      |   +   |    +    |    +     |     +    |   +     
————————————+———————+—————————+——————————+——————————+————————
protected   |   +   |    +    |    +     |     +    |         
————————————+———————+—————————+——————————+——————————+————————
no modifier |   +   |    +    |    +     |          |    
————————————+———————+—————————+——————————+——————————+————————
private     |   +   |         |          |          |    

+ : accessible
blank : not accessible

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Java Protected Access Not Working


Tag : java , By : Daniel Halsey
Date : March 29 2020, 07:55 AM
help you fix your problem In java, there's three levels of access: , Actually it should be:

Java access a protected attribute


Tag : java , By : OllieDoodle
Date : March 29 2020, 07:55 AM
I hope this helps you . "Unreachable statement" means that there is no path in your code that can get you to this line in your program. It has nothing to do with protected fields. Check, if there is a return statement before your a+=1 that always exits your method before this line is ever reached. Or, if your a+=1 is in an if-block that is never executed, because the condition you specified always evaluates to false.

Java protected variable access and modify rules


Tag : java , By : todor
Date : March 29 2020, 07:55 AM
will be helpful for those in need Protected variable can accessible in subclass when it is in the same package.If BFrag is in different package then you can not access the super class protected variable in subclass.

Access to a protected constructor in Java


Tag : java , By : Antony Briggs
Date : March 29 2020, 07:55 AM
hop of those help? Seems the problem is that you have your classes in different packages. Java doc says:

C++ nested struct inheritance rules (access to protected member)


Tag : cpp , By : cautionsign
Date : March 29 2020, 07:55 AM
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