在 roslyn 分析期间从 switch 语句中的可为空的枚举类型获取实际的枚举类型

Question

I would like to analyse a switch statement, that uses a "nullable enum" to decide. I have the following class that I would like to analyze:

namespace Common.Model.Schema
{
    public enum ModuleType
    {
        Case1,
        Case2,
        Case3
    }   
}

namespace Analyzer.Test
{
    using Common.Model.Schema;

    public class Test
    {
        private static void GetSelectedAblageOrdner()
        {
            ModuleType? moduleType = null;
            switch (moduleType)
            {
                case ModuleType.Case1:
                {
                    break;
                }
            }
        }
    }
}

When the input of the switch is not nullable, I can use the following code and I have the correct type to do my analysis on.

TypeInfo typeInfo = context.SemanticModel.GetTypeInfo(expression);
ITypeSymbol expressionType = typeInfo.ConvertedType;
if (!(expressionType is INamedTypeSymbol namedType))
{
    return;
}

switch (namedType.EnumUnderlyingType.Name)
{
      // do stuff
}

But with nullable enum the convertedType is Nullable<ModuleType> (or in other words ModuleType? ). This makes the property EnumUnderlyingType etc. NULL. I need the actual enum so I can continue.

How do I get to the ModuleType , so I can continue my default algorithm I have for non-nullable enums?

Answer 1

Nullable<> is just a generic. Check that it is generic and fetch its argument.

if (typeof(Nullable<>) == typeInfo.ConvertedType.GetGenericTypeDefinition())
{
   var actualType = typeInfo.ConvertedType.GetGenericArguments()[0];
}

Answer 2

Quercus pointed me in the right direction. Here is my solution:

if (namedType.IsGenericType)
{
    INamedTypeSymbol typeSymbol = namedType.TypeArguments.FirstOrDefault() as INamedTypeSymbol;
    if (typeSymbol == null)
    {
        return;
    }

    expressionType = typeSymbol;
    namedType = typeSymbol;
}