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Home/ Questions/Q 9266091
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Editorial Team
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Editorial Team
Asked: June 18, 20262026-06-18T14:14:17+00:00 2026-06-18T14:14:17+00:00

my code is:- class Test { static int a = 11; static { System.out.println(Hello

  • 0

my code is:-

class Test
{
  static int a = 11;
  static
  {
    System.out.println("Hello static! " + main() + a);
  }

  public static void main(String[]args)
  {
    System.out.println("Hello String!");
  }

  public static char main()
  {
    System.out.println("Hello char!");
    return 'H';
  }
}

Output:-

Hello char!
Hello static! H11
Hello String!

Why “Hello char!” is printed before “hello static!”?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-18T14:14:18+00:00Added an answer on June 18, 2026 at 2:14 pm

    Why “Hello char!” is printed before “hello static!”?

    Because, main() method is invoked and executed before the completion of the sysout statement printing Hello static.

    Here’s the execution order: –

    • Class is loaded.
    • static variable a is loaded and initialized to 11.
    • static block is executed.
    • From sysout statement there, main() static method is invoked. At this point of time Hello static has not been printed yet, because the sysout statement is not complete.
    • Execution control goes to main() method. Because without that the current sysout statement cannot complete.
    • main() method prints "Hello char", and returns 'H'.
    • Execution goes back to static block, inside sysout.
    • sysout completes execution, and prints "Hello static! H11"

    Note: – sysout above means – System.out.println().

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