java中try-catch的完整过程是怎样的？

Question

I have been experimenting with try-catch and have been confused on specifically how it works.

try                                     // x is array of 10 doubles set to 0
        {
            for (int a = 0; a < 11; a+= 2)
                x[a] = 5;
        }
        catch(Exception e)
        {
            System.out.println("error");
        }

In this instance, all values in the array can be reached, but the code breaks at x[10]. So, have all the values been set to 5?

try                             // x is array of 10 doubles set to 0
        {
            for (int a = -1; a < 11; a+= 2)
                x[a] = 5;
        }
        catch(Exception e)
        {
            System.out.println("error");
        }

In this instance, it will try x[-1] and catch the error. So, will it not go back and complete the loop (x[0], x[1], ... x[9] )? So, all values still remain 0?

Is this how the try-catch works?
Any help would be appreciated.
Thank you.

Answer 1

Yeah, both examples would throw ArrayIndexOutOfBoundsException .

So, The try block contains a block of program statements within which an exception might occur. A try block is always followed by a catch block , which handles the exception that occurs in associated try block . A try block must followed by a Catch block or Finally block or both .

A catch block must be associated with a try block . The corresponding catch bloc k executes if an exception of a particular type occurs within the try block . For example if an arithmetic exception occurs in try block then the statements enclosed in catch block for arithmetic exception execute.

try
{
     //statements that may cause an exception
}
catch (exception(type) e(object))‏
{
     //error handling code
}

If an exception occurs in try block then the control of execution is passed to the catch block from try block . The exception is caught up by the corresponding catch block . A single try block can have multiple catch statements associated with it, but each catch block can be defined for only one exception class . The program can also contain nested try-catch-finally blocks .

After the execution of all the try blocks , the code inside the finally block executes. It is not mandatory to include a finally block at all, but if you do, it will run regardless of whether an exception was thrown and handled by the try and catch blocks .

Answer 2

Try-Catch

The code within the try block is code that will run but if the code in the try block throws an exception/error then it is caught in catch block.

The issue with your code is you are using the x as an array and you may not be able to set a specific spot in an array, to set that you could do something like x.push or x.put I am not sure on the specific property as I do not know java well. The other issue could be that your index is out of bounds, which means that you may have an array of 10 items that is zero based and you are trying to access an 11th item. so your loop should be something like:

    {
        for (int a = 0; a < x.length; a+= 2)
            x[a] = 5;
    }

Answer 3

When a statement throws an exception inside of a try block, the next statement is the first statement of the catch block.

In the first example, the first few iterations of the for loop happen as normal and elements 0, 2, 4, 6, and 8 are set to 5 . On the iteration where a is 10 , the attempt to access x[10] the runtime throws an ArrayIndexOutOfBoundsException and jumps to the first statement of the catch block. The rest of x is unmodified.

In the second example, the first iteration of the for loop causes the runtime to throw an ArrayIndexOutOfBoundsException (due to the attempt to access x[-1] ) and jump to the first statement of the catch block. The effect is that x is completely unmodified.

For more information about exceptions in Java, read the The Java Tutorials: Exceptions .