ct answerWhat will be the output of the following Java program? class scopevariable { public static void main(String args[]) { int c; c = 15; { int d = 36; System.out.print(c + " " + d); } System.out.println(c + " " + d); } }

The Java program will not compile successfully. The variable 'd' is declared inside a block and it is not accessible outside of that block. So, when the program tries to print the value of 'd' outside of the block, it will give a compile-time error saying that it cannot find the symbol 'd'.

If you want to print the value of 'd' outside of the block, you should declare 'd' at the same level as 'c'. Here is the corrected code:

class scopevariable {
    public static void main(String args[]) {
        int c;
        int d; // Declare 'd' here
        c = 15;
        {
            d = 36; // Assign value to 'd' here
            System.out.print(c + " " + d);
        }
        System.out.println(c + " " + d);
    }
}

This program will output:

15 36
15 36

The Java program you provided will not compile successfully. The reason is that the variable 'd' is declared inside a block (between the inner set of curly braces {}). This means 'd' is only visible and accessible within this block.

When you try to print the value of 'd' outside of this block (in the last line of the main method), the compiler doesn't recognize 'd' as it's out of its scope, hence it will throw a compilation error.

So, there is no output for this program as it fails during compilation. If you want to print 'd' outside of the block, you should declare it in the same scope as 'c'.

The Java program you provided will not compile successfully. The reason is that the variable 'd' is declared inside a block (between the inner set of curly braces {}). This means 'd