is a 'field' but used like a 'type' (it is really a type)

Question

I am getting expectedExceptionType is a 'field' but used like a 'type' error from the if (exception is expectedExceptionType) line. However as you can see it is a Type. What is the problem here?

I am using Visual Studio 2013 with .NET 4.5

public sealed class ExpectedInnerException : ExpectedExceptionBaseAttribute 
{
    private Type expectedExceptionType;
    private string expectedExceptionMessage;

    public ExpectedInnerException(Type expectedExceptionType, string expectedExceptionMessage)
    {
        this.expectedExceptionType = expectedExceptionType;
        this.expectedExceptionMessage = expectedExceptionMessage;
    }

    protected override void Verify(Exception exception)
    {
        if (exception is expectedExceptionType)
        {

        }
        //Some other code
    }
}

MakePeaceGreatAgain · Answer 1 · 2016-03-30T07:58:13.637

5

expectedExceptionType is an instance of the type Type, not a compile-time-type in itself. Thus write this instead:

protected override void Verify(Exception exception)
{
    if (exception.GetType() == expectedExceptionType)
    {

    }
    //Some other code
}

EDIT: If your type is a subclass from that one reflected by expectedExceptionType you can check for IsAssignableFrom:

if (exception.GetType().IsAssignableFrom(expectedExceptionType.GetType()))

edited Mar 30 '16 at 07:58

answered Mar 30 '16 at 07:42

MakePeaceGreatAgain

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In this case if exception is a subclass of expectedExceptionType, it will return false. That's way I wanted to use "is" – Nuri Tasdemir Mar 30 '16 at 07:44
@NuriTasdemir Yes, you´re right, I updated the answer. – MakePeaceGreatAgain Mar 30 '16 at 07:47
1

Your update a bit wierd. `IOException` is not assignable from `Exception` for instance – Diligent Key Presser Mar 30 '16 at 07:48
1

So you should turn it around since the `expectedExceptionType` can be he higher one. – Patrick Hofman Mar 30 '16 at 07:57
@PatrickHofman Yeap, finally noticed, thanks for the hint however :) – MakePeaceGreatAgain Mar 30 '16 at 07:58

score 2 · Answer 2 · edited May 23 '17 at 11:45

2

HimBromBeere told you the reason, but not the best way to solve this, since checking on the type directly won't work with interfaces and derived instances.

This code does:

protected override void Verify(Exception exception)
{
    if (exception.GetType().IsAssignableFrom(expectedExceptionType))
    {

    }
    //Some other code
}

edited May 23 '17 at 11:45

Community

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1

answered Mar 30 '16 at 07:46

Patrick Hofman

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you were faster than me – György Kőszeg Mar 30 '16 at 07:47

score 0 · Accepted Answer · answered Mar 30 '16 at 07:47

The is operator expects a type expression, not a variable/field with System.Type type.

The compatible solution:

protected override void Verify(Exception exception)
{
    if (expectedExceptionType.IsAssignableFrom(exception.GetType()))
    {

    }
    //Some other code
}

is a 'field' but used like a 'type' (it is really a type)

3 Answers3