Explain the output of the program in java ? [closed]

1
votes

In this program Is it possible to use the access specifier inside the method 

class AccessTest{
int i;

   public static void main (String... str) 
  {
   int i;
   private int a = 1;
   protected int b = 1;
   public int c = 1;
   System.out.print (a+b+c);
  }

}

what is the final output can anybody explain this ?

5
javamethodsaccess-specifier
To begin with, why would you want private, protected or public variables inside a method? What would be the meaning of having this? - Luiggi Mendoza
"what is the final output can anybody explain this ?" just run it and see for yourself - Marco Forberg
"What is the final output".. what did it output when you tried it? - Nanne
Have you tried to compile this? It does not even compile. What do you expect to get as result? - Ean V
If I used the public only at the place of the private or protected then also I have got the error ... - Shivam

5 Answers

2
votes

The access modifiers specify the visibility of one class' fields for other classes. Since local variables (those declared inside methods) are never exposed, it doesn't make sense to set a certain access for them. Actually it's a compilation error, if you try it.

1
votes

No it is not possible. As there are no use of it so it is restricted.

Local veriable's scope are restricted with in body so there would be no use of modying there access.

0
votes

Variables declared in a method are local to the method; i.e. they can't be accessed outside the method.

0
votes

You can't use private, protected, public modifiers inside a method. Final output is compilation error.

0
votes

You cannot set visibility scopes (private,...) to local variables. because The Local variables scope is already well defined : within the scope of the method it lives in.