java question about try catch statement with finally

Here is a code written by java.

public class work2{
public static void main(String args[]) {
    try {
        int x = 0;
        for (x=1; x<4; x++); 
        System.out.println(x);
    } catch(Exception e) {}
    finally {
        System.out.println("Error");
    }
}

}

And the output is

> 4 Error

Would you explain why the output is looks like this?

Always remember in a try catch block - "finally always Run", irrespective of whether there is an exception or not. Now coming to your code. Inside the try block you have below line:

 try {
    int x = 0;
    for (x=1; x<4; x++); 
    System.out.println(x);
}

Here the for loop ends in a semicolon, which means it doesn't have a body.
Now, the for loop will run from i=1 till i<4 and then will come out of loop with value of i as 4.

Now the next line is a print statement which prints the current value of i as 4.

After that the controls goes to the finally block and prints "Error"

That's how you get the output as

4
Error

The for - Loop you're using is only increase the value of x but doesn't print it out. So you increase x to 4 and than print it out, the finally part will be printed anyway.
If you want to put out every value of x, you had to add { } after the for - Loop. <br>
It will look like:<br>

public class Main{
    public static void main(String args[]) {
        try {
            int x = 0;
            for (x=1; x<4; x++) {
                System.out.println(x);
            }
        } catch(Exception e) {}
        finally {
            System.out.println("Error");
        }
    }
}

try {
        int x = 0;
        for (x=1; x<4; x++); 
        System.out.println(x);
    }

the try block didn't catch any exception, so the code were run as it is.
and in

finally {
        System.out.println("Error");
    }

> A finally block contains all the crucial statements that must be
> executed whether exception occurs or not. The statements present in
> this block will always execute regardless of whether exception occurs
> in try block or not such as closing a connection, stream etc.

it will print "Error" regardless what happen.
So, Try block printed 4, and then Finally block print "Error"

You write the statements which cause exception in try block. And these exceptions are handled using catch block. finally block is executed even if exception is not triggered.

In your question, x=0; initially. Since there is a semicolon ; after for loop, the loop will be executed 3 times and when x=4, the condition fails and it prints the value of x that is 4. And finally block is executed.

    public class Main{
public static void main(String args[]) {
    try {
        int x = 0;
        for (x=1; x<4; x++)
        System.out.println(x);
    } catch(Exception e) {}
    finally {
        System.out.println("Error");
    }
}
}

The output should be
1
2
3
error

you should remove the ; after for loop .

A finally block contains all the crucial statements that must be executed whether exception occurs or not. ... The statements present in this block will always execute regardless of whether exception occurs in try block or not such as closing a connection, stream etc.