CODEWITHSUNDEEP
c#visual-studio-2008
James
James

Reputation: 9975

Creation of a new instance of generic type parameter not getting code coverage

I've run my code through code coverage and the line below shows 1 block as not covered.

Can anyone tell me which part of that line isn't executing?

enter image description here

An Example to play with: 

public abstract class Base
{
    public abstract IExample CreateEntity<TExample>() where TExample : IExample, new();
}

public class Class1 : Base
{
    public override IExample CreateEntity<TExample>()
    {
        IExample temp = new TExample();
        return temp;
    }
}

public interface IExample
{

}

public class TEx : IExample
{

}

and the test method

    [TestMethod]
    public void TestMethod1()
    {
        Class1 ex = new Class1();
        ex.CreateEntity<TEx>();
    }

Upvotes: 8

Views: 323

Answers (1)

John Koerner
John Koerner

Reputation: 38077

Change your constraint to force the TExample to be a class:

public abstract IExample CreateEntity<TExample>() where TExample : class, IExample, new();

If you run your compiled code through a tool like ILSpy, you will see the block that is not getting coverage:

TExample temp = (default(TExample) == null) ? Activator.CreateInstance<TExample>() : default(TExample);
return temp;

It is performing a check to see if the type passed to the generic was a reference type or a value type. By forcing it to be a class, this check will be removed. Read more on the default keyword here: http://msdn.microsoft.com/en-us/library/xwth0h0d.aspx

Another way to get complete code coverage would be to use a struct that implements IExample:

public struct S1 : IExample
{ 
}

And then add this test:

[TestMethod]
public void StructTest()
{
    Class1 ex = new Class1();
    ex.CreateEntity<S1>();
}

Upvotes: 6

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