Explanation of static block output required in Java

Question

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!"?

user207421 · Accepted Answer · 2013-02-10 08:50:55Z

3

The static block executes first.
It calls main() with no arguments while assembling a String to print.
main() prints some output.
The static block prints its output.
The static block exits.
Your main(String[]) method executes and prints its output.

answered Feb 10, 2013 at 8:50

user207421

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2 Comments

Himanshu Aggarwal Over a year ago

can u please explain me the working of static block to make it more clear why the main() is called first before the first argument of println(), i.e., "Hello static!" ?

user207421 Over a year ago

@HimanshuAggarwal I've already done that. See #2: 'while assembling a String to print'. The argument to printn() has to be evaluated before println() can be called. Part of that evaluation consists of calling main().

Rohit Jain · Accepted Answer · 2013-02-10 09:00:51Z

1

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().

edited Feb 10, 2013 at 9:00

answered Feb 10, 2013 at 8:47

Rohit Jain

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3 Comments

user207421 Over a year ago

There is no such thing as a 'sysout statement'. Do you mean the System.out.println method invocation?

Himanshu Aggarwal Over a year ago

it was clear from the text. no problem. thanx for the answer.

user207421 Over a year ago

@RohitJain You shouldn't talk about a statement when you mean a method call. It's clear enough what you actually meant, but that doesn't excuse the misuse of standard terminology.

Patashu · Accepted Answer · 2013-02-10 08:48:19Z

1

Let's break it down.

The very first thing that runs is the static initializer block. So we run this line:

System.out.println("Hello static! " + main() + a);

But before we can call that method, we have to call main(), because we have to fully evaluate all arguments first.

So this gets called before that println in the static initializer:

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

And that prints the Hello char! we see in the first line.

It returns H, which gets printed alongside Hello static!

Finally, main(string[]) is called to begin the program. And there we see Hello String!

answered Feb 10, 2013 at 8:48

Patashu

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Comments

Sirko · Accepted Answer · 2013-02-10 08:49:01Z

1

The general execution logic here is as follows:

The class is loaded, hence the static constructor is executed.
Inside the constructor, the String (parameter to System.out.println()) is evaluated, which calls the char main() method.
The char main() method is evaluated.
Inside the char main() the string is printed: 1st output line
The String in the static constructor has been build and thus can be printed: 2nd output line
The actual void main() can be executed, which prints the third line: 3rd output line

answered Feb 10, 2013 at 8:49

Sirko

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Comments

nmjohn · Accepted Answer · 2013-02-10 08:49:38Z

1

In your line

System.out.println("Hello static! " + main() + a);

main() is executed before the System.out.println, so it makes sense that the print statement within your char main() would execute before the outer print statement "Hello Static".

answered Feb 10, 2013 at 8:49

nmjohn

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3 Comments

Himanshu Aggarwal Over a year ago

but why is the main() called first? that is what my doubt is.

nmjohn Over a year ago

it's like if you have the statement a = a + func(); func() needs to fully evaluate before you can add and assign to a.

Himanshu Aggarwal Over a year ago

ohh! is this the case? then its fine. thanx.:-)

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