t 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); } }Options5 6 5 6Runtime error5 6 5Compilation error
Question
t 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); } }Options5 6 5 6Runtime error5 6 5Compilation error
Solution
The output of the Java program will be a Compilation error.
Here's why:
In Java, the scope of a variable is the region of the code where the variable can be accessed. In this program, the variable 'd' is declared inside a block of code (inside the curly braces {}). This means 'd' is only accessible within this block.
When the program tries to print the value of 'd' outside of this block (in the line "System.out.println(c + " " + d);"), it will not be able to access 'd' because 'd' is out of scope. This will cause a Compilation error.
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