Most efficient way to get default constructor of a Type.

Question

What is the most efficient way to get the default constructor (i.e. instance constructor with no parameters) of a System.Type?

I was thinking something along the lines of the code below but it seems like there should be a simplier more efficient way to do it.

Type type = typeof(FooBar)
BindingFlags flags = BindingFlags.Public | BindingFlags.NonPublic | BindingFlags.Instance;
type.GetConstructors(flags)
    .Where(constructor => constructor.GetParameters().Length == 0)
    .First();

Answer 1

type.GetConstructor(Type.EmptyTypes)

Answer 2

If you actually need the ConstructorInfo object, then see Curt Hagenlocher's answer.

On the other hand, if you're really just trying to create an object at run-time from a System.Type, see System.Activator.CreateInstance -- it's not just future-proofed (Activator handles more details than ConstructorInfo.Invoke), it's also much less ugly.

Answer 3

you would want to try FormatterServices.GetUninitializedObject(Type) this one is better than Activator.CreateInstance

However , this method doesn't call the object constructor , so if you are setting initial values there, this won't work Check MSDN for this thing http://msdn.microsoft.com/en-us/library/system.runtime.serialization.formatterservices.getuninitializedobject.aspx

there is another way here http://www.ozcandegirmenci.com/post/2008/02/Create-object-instances-Faster-than-Reflection.aspx

however this one fails if the object have parametrize constructors

Hope this helps