Thread: What is the difference between a statement surrounded by curly brackets and one that isn't?

Question is a little bit elementary, but i just need a programmer to programmer explanation.
For example:

if (number < 1 || number > 100 ) 
 throw new InvalidInputException();
 System.out.println("You entered " + number);

In this example, the statement

  System.out.println("You entered " + number);

gets executed if the number entered is within range or an exception is thrown if it isn't.
But if i change the if statement to:

if (number < 1 || number > 100 ) {
 throw new InvalidInputException();
 System.out.println("You entered " + number);
}

it says the code

System.out.println("You entered " + number);

is unreachable. Why is this? and of what significance are curly brackets in control statements generally?

The if() {} statement is not inside a try {} block, so there is nothing to catch that exception, and the program will end.

If a number out of range is entered, the program stops running.

If a number in range is entered, the if() {} block doesn't even execute, so in no case will the println() statement ever be executed.

-summit45

Yeah, the if () {} statement is inside a try catch statement, i just forgot to include it in my question.

Originally Posted by ojonugwa

Yeah, the if () {} statement is inside a try catch statement, i just forgot to include it in my question.

Well if it throws that exception, that breaks out of the try {} block, and executes the catch statement. And if the number is in range, the if() {} block doesn't execute, so either way, that println() will never run.

-summit45

Originally Posted by KevinWorkman

I don't think you quite understand the question.

In the first code example, the if statement does not have any curly brackets. That means that when the if statement evaluates to true, only the very next statement will be run, which is the throw statement. If the if statement evaluates to false, the println() will indeed run.

However, the next code example does use curly brackets, but it puts the end curly bracket after the print statement. Since the throw always happens right before the print statement, the print statement is never run, hence the compiler error.

The solution is to properly use curly brackets for every if statement and loop.

Yeah I got him, that's what I said, or so I thought lol. Well I didn't talk about that first if() without braces but yeah exactly, that's just it. If you put that println() inside the if(), and the person enters a right number (within range), the if() block doesn't execute, and it won't print. If they enter a wrong number (out of range), the if() block throws an exception, which stops it from executing further code, and so println() still doesn't run.

-summit45