问题
I have a method that takes as an input parameter an object of type IEnumerable. I want to enumerate the enumeration and for each item use reflection to get the value for each property.
I have the following code:
protected void WriteData(IEnumerable data)
{
var enumerationTypeInfo = data.GetType();
var itemTypeInfo = enumerationTypeInfo.GetElementType();
...
}
The problem is enumerationTypeInfo.GetElementType() always returns null. In particular, I'm passing in a List<Entry> into WriteData, where Entry is a class I created. When I use the debugger and set a breakpoint I can see that enumerationTypeInfo correctly shows that it's a List of type Entry, but why does GetElementType return null?
Thanks
回答1:
GetElementType is for use with arrays, not other generic classes. To get a generic type's generic parameters, you can use Type.GetGenericArguments.
回答2:
GetElementType() returns the element type of arrays. List<T> is not an array type, and therefore has no "element type."
If you want to get the type of elements a random IEnumerable<T> produces, try something like this:
public static Type GetEnumerableType(Type type)
{
if (type == null)
throw new ArgumentNullException("type");
if (type.IsGenericType && type.GetGenericTypeDefinition() == typeof(IEnumerable<>))
return type.GetGenericArguments()[0];
var iface = (from i in type.GetInterfaces()
where i.IsGenericType && i.GetGenericTypeDefinition() == typeof(IEnumerable<>)
select i).FirstOrDefault();
if (iface == null)
throw new ArgumentException("Does not represent an enumerable type.", "type");
return GetEnumerableType(iface);
}
Note that types can implement more than one version of the same generic interface; a type can implement both IEnumerable<int> and IEnumerable<string> for example. How you handle that case is up to you. The method I provide will take whichever interface type the runtime hands it first.
See an example using the above method on ideone.
来源:https://stackoverflow.com/questions/7072088/why-does-type-getelementtype-return-null