Why do I get Code Analysis CA1062 on an out parame

问题:

I have a very simple code (simplified from the original code - so I know it's not a very clever code) that when I compile in Visual Studio 2010 with Code Analysis gives me warning CA1062: Validate arguments of public methods.

public class Foo
{
    protected static void Bar(out int[] x)
    {
        x = new int[1];
        for (int i = 0; i != 1; ++i)
            x[i] = 1;
    }
}

The warning I get:

CA1062 : Microsoft.Design : In externally visible method 'Foo.Bar(out int[])', validate local variable '(*x)', which was reassigned from parameter 'x', before using it.

I don't understand why do I get this warning and how can I resolve it without suppressing it? Can new return null? Is this a Visual Studio 2010 bug?

UPDATE

I've decided to open a bug report on Microsoft Connect.

回答1:

I've reproduced this in Visual Studio 2010 Premium with the code exactly as given and with Microsoft All Rules enabled in the analysis settings.

It looks like this is a bug (see bottom of here: http://msdn.microsoft.com/en-us/library/ms182182.aspx). It is complainng that you are not checking that x is not null before using it, but it's on out parameter so there is no input value to check!

回答2:

It's easier to show than to describe :

public class Program
{
    protected static int[] testIntArray;

    protected static void Bar(out int[] x)
    {
        x = new int[100];
        for (int i = 0; i != 100; ++i)
        {
            Thread.Sleep(5);
            x[i] = 1; // NullReferenceException
        }
    }

    protected static void Work()
    {
        Bar(out testIntArray);
    }

    static void Main(string[] args)
    {
        var t1 = new Thread(Work);
        t1.Start();

        while (t1.ThreadState == ThreadState.Running)
        {
            testIntArray = null;
        }
    }
}

And the correct way is :

    protected static void Bar(out int[] x)
    {
        var y = new int[100];

        for (int i = 0; i != 100; ++i)
        {
            Thread.Sleep(5);
            y[i] = 1;
        }

        x = y;
    }