Welcome to the Java Programming Forums
The professional, friendly Java community. 21,500 members and growing!
The Java Programming Forums are a community of Java programmers from all around the World. Our members have a wide range of skills and they all have one thing in common: A passion to learn and code Java. We invite beginner Java programmers right through to Java professionals to post here and share your knowledge. Become a part of the community, help others, expand your knowledge of Java and enjoy talking with like minded people. Registration is quick and best of all free. We look forward to meeting you.
>> REGISTER NOW TO START POSTING
Members have full access to the forums. Advertisements are removed for registered users.
LinkBack URL
About LinkBacks
Unreachable code
When i run the block of code above it says unreachable code till i remove the curly braces as shown belowint var = 2; block1: { if (var != 1) { break block1; System.out.println("variable is one"); // unreachable } }
Why does groupint the code in a block make it unreachable? TIAint var = 2; block1: { if (var != 1) break block1; System.out.println("variable is one"); // reachable }
Re: Unreachable code
The brackets following the if statement delimit a block of code to be executed when the expression in the if statement is true.
A break statement ends the execution of the code within the block. That means the println statement following the break can never be reached.
Without the brackets, the default block of code controlled by the if statement is the single break statement following the if statement.
That leaves the println statement outside of the if statement's control. It will be executed when the if statement's expression is false.
If you don't understand my answer, don't ignore it, ask a question.
Reply With Quote