在C＃中，您可以从main返回默认的int值吗？

Question

I want to return a default int value from main.

Consider the following:

using System;
class Class1
{
    static int Main(string[] args)
    {
        int intReturnCode = 1;
        int intRandNumber;

        Random myRandom = new Random();
        intRandNumber = myRandom.Next(0,2);
        if(intRandNumber ==1)
        {
            throw new Exception("ErrorError");      
        }
        return intReturnCode;
    }
}

When the exception is reached I don't get to set the returncode.

Is it possible to have a default return code inside of main?

Clarification: I have a program that is throwing Unhandled Exceptions. I have the application inside a try catch, however some errors (probably out of memory, stackoverflow etc) are still bubling up and causing my application to fail in production.

To fix this I've added code to capture unhandled exceptions.

This was added to main:

AppDomain.CurrentDomain.UnhandledException += new UnhandledExceptionEventHandler(OnUnhandledException);

And now I have this method that is reached when an unhandled exception occurs.

public static void OnUnhandledException(object sender, UnhandledExceptionEventArgs e)   
{ 
    //here's how you get the exception  

    Exception exception = (Exception)e.ExceptionObject;                 

    //bail out in a tidy way and perform your logging
}

The prblem is that I'm no longer in Main and I want to exit with a non-zero exit code.

Answer 1

An unhandled exception is implementation defined behaviour . Anything can happen; the CLR can decide to set the return code of the process as it sees fit, it can start a debugger, it can do whatever it wants. You cannot rely on any behaviour of any program that contains an unhandled exception.

If you want to have predictable behaviour, like determining what the return code is when the process ends, then there must be a total of zero unhandled exceptions.

If you have a third party component that is throwing unhandled out of memory exceptions then your best bet is: fix the bug in that component . If you can't do that then isolate the component into its own process or its own appdomain .

Answer 2

The question is really why you are throwing an exception in main instead of providing a return code that indicates an error? Instead of what you're doing, my code would look as follows:

static int Main(string[] args)
{
    int intRandNumber;

    try
    {
        Random myRandom = new Random();
        intRandNumber = myRandom.Next(0,2);
        if(intRandNumber ==1)
        {
            Console.WriteLine("Got invalid random number!");
            return 0;
        }
    }
    catch (Exception exp)
    {
        Console.WriteLine("Strange ... an error occurred! " + exp.ToString());
        return -1;
    }

    return 1;
}

As a rule of thumb you should never throw exceptions to control program flow. If you can, handle conditions like oops, I got the wrong number without throwing an exception.

Answer 3

Throwing an exception in the main thread ends execution without reaching the return : that's when you get the "Console application has stopped working, would you like to debug?" dialog from the operating system. The Main cannot return anything under these conditions, because there is nothing to return.

If you would like to return something when you get an exception, code your program like this:

// Put your implementation into a private method
private static int DoWork(string[] args) {
    ... // Do actual work, and throw exceptions if you need to
    return intReturnCode;
}
// The Main method calls the implementation, and returns the value
// returned from the implementation if everything goes well.
// If an exception is thrown, however, Main could return another value
// of its choice.
public static int Main(string[] args) {
    try {
        return DoWork(args);
    } catch {
        return myDefaultReturnCode;
    }
}

Answer 4

catch your exception and set return statement in finally block

using System;
class Class1
{
    static int Main(string[] args)
    {
        try
        {
            int intReturnCode = 1;
            int intRandNumber;

            Random myRandom = new Random();
            intRandNumber = myRandom.Next(0,2);
            if(intRandNumber ==1)
            {
                throw new Exception("ErrorError");      
            }
        }
        catch(Exception e)
        {
          // ...
        }
        finally
        {
            return intReturnCode;
        }
    }
}