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

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    Default 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?


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

    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

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

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

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

    This code:

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

    Should be re-written as this:

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

    Which hopefully clears up your question. Always use curly brackets for if statements and loops!
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    ojonugwa (August 30th, 2013)

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

    Quote Originally Posted by ojonugwa View Post
    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

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

    Quote Originally Posted by summit45 View Post
    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
    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.
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    Default Re: What is the difference between a statement surrounded by curly brackets and one that isn't?

    Quote Originally Posted by KevinWorkman View Post
    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

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

    Quote Originally Posted by summit45 View Post
    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
    The point is that the println() *does* run in the first example when the if statement evaluates to false. It's still a reachable statement. In the second example, there's no way that the code could execute to run the println() statement, hence the error.
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