Additional lessons for Level

Your understanding of the `continue` statement is correct: it indeed causes the loop to immediately jump to the next iteration, skipping any code that follows it within the loop body. However, there is a specific reason your code fails to compile. In Java, the `continue` statement transfers control to the next iteration of the enclosing loop, bypassing any remaining statements in the current iteration. Consequently, any statements placed after `continue` are considered unreachable code because they will never be executed. This is against Java's rules, as the Java compiler flags unreachable code as an error. Let's break down what happens in your code: 1. The `for` loop starts with `i = 0`. 2. The loop condition `i < 10` is checked and found to be true. 3. The `continue` statement is executed, which causes the loop to immediately jump to the next iteration. 4. The `System.out.println(i);` statement is never reached because the `continue` statement has already passed control to the next iteration of the loop. Since the `System.out.println(i);` statement is never reachable due to the `continue` statement, the Java compiler generates an error indicating that there is unreachable code. To fix this issue, you need to remove the unreachable statement after the `continue`. For example:

for (i = 0; i < 10; i++){
    continue;
}

This code will compile and run without errors. The loop will iterate ten times, but nothing will be printed because the `continue` statement causes the loop to skip the `System.out.println(i);` statement (which has been removed). In summary, the problem in your original code is that the `continue` statement makes the subsequent `System.out.println(i);` statement unreachable, which causes a compilation error. Removing the unreachable statement or the `continue` statement will resolve the issue.

Bro in next level(level5) you will learn to use intellij Idea where you will have another option to download the solution so keep it up. oh! sorry you just hung up.

agree

You are absolutely right

round 3.9--> 4 if the number after coma above 5 will be round up if not round down ceil 3.1-->4 always up floor 3.2-->3 always down floor 3.9--> 3

Please read the example belows and you could understand immediately.

public class Main {
    public static void main(String[] args){
        int i = (int) Math.round(-9.9);
        int x = (int) Math.round(9.9);
        System.out.println(i); // -5
        System.out.println(x); // 5

        int y = (int) Math.ceil(-9.9);
        int j = (int) Math.ceil(9.9);
        System.out.println(y); // -4
        System.out.println(j); // 5
    }
}

Because it always skip an iteration before "print", so it will never compile.

I ran this in intellij and the code would not compile. After compiling I got an error saying 'java: unreachable statement'. The unreachable statement is 'System.out.println(i);'. If you get rid of 'System.out.println(i);' the code will run fine. This will not create an endless loop and will still iterate through the for statement as shown below: public class Solution { public static void main(String[] args) throws Exception { for(int i = 0; i < 10; i++) { continue; } System.out.println("hi"); } } This will print out 'hi'.

In Java, the `continue` statement transfers control to the next iteration of the enclosing loop, bypassing any remaining statements in the current iteration. Consequently, any statements placed after `continue` are considered unreachable code because they will never be executed. This is against Java's rules, as the Java compiler flags unreachable code as an error. Let's break down what happens in your code: 1. The `for` loop starts with `i = 0`. 2. The loop condition `i < 10` is checked and found to be true. 3. The `continue` statement is executed, which causes the loop to immediately jump to the next iteration. 4. The `System.out.println(i);` statement is never reached because the `continue` statement has already passed control to the next iteration of the loop. Since the `System.out.println(i);` statement is never reachable due to the `continue` statement, the Java compiler generates an error indicating that there is unreachable code. To fix this issue, you need to remove the unreachable statement after the `continue`. For example:

for (i = 0; i < 10; i++){
    continue;
}

This code will compile and run without errors. The loop will iterate ten times, but nothing will be printed because the `continue` statement causes the loop to skip the `System.out.println(i);` statement (which has been removed). In summary, the problem in your original code is that the `continue` statement makes the subsequent `System.out.println(i);` statement unreachable, which causes a compilation error. Removing the unreachable statement or the `continue` statement will resolve the issue.