How do I check if a type provides a parameterless constructor?
C#.NetReflectionConstructorTypesC# Problem Overview
I'd like to check if a type that is known at runtime provides a parameterless constructor. The Type
class did not yield anything promising, so I'm assuming I have to use reflection?
C# Solutions
Solution 1 - C#
The Type
class is reflection. You can do:
Type theType = myobject.GetType(); // if you have an instance
// or
Type theType = typeof(MyObject); // if you know the type
var constructor = theType.GetConstructor(Type.EmptyTypes);
It will return null if a parameterless constructor does not exist.
If you also want to find private constructors, use the slightly longer:
var constructor = theType.GetConstructor(
BindingFlags.Instance | BindingFlags.Public | BindingFlags.NonPublic,
null, Type.EmptyTypes, null);
There's a caveat for value types, which aren't allowed to have a default constructor. You can check if you have a value type using the Type.IsValueType
property, and create instances using Activator.CreateInstance(Type)
;
Solution 2 - C#
type.GetConstructor(Type.EmptyTypes) != null
would fail for struct
s. Better to extend it:
public static bool HasDefaultConstructor(this Type t)
{
return t.IsValueType || t.GetConstructor(Type.EmptyTypes) != null;
}
Succeeds since even enum
s have default parameterless constructor. Also slightly speeds up for value types since the reflection call is not made.
Solution 3 - C#
Yes, you have to use Reflection. But you already do that when you use GetType()
Something like:
var t = x.GetType();
var c = t.GetConstructor(new Type[0]);
if (c != null) ...
Solution 4 - C#
This should work:
myClass.GetType().GetConstructors()
.All(c=>c.GetParameters().Length == 0)
Solution 5 - C#
Depending on your situation, you could also use a generic type restriction:
public void DoSomethingWith<T>(T myObject) where T:new() {...}
The above method declaration will restrict the parameter type to any Object that can be instantiated with a parameterless constructor. The advantage here is that the compiler will catch any attempt to use the method with a class that doesn't have a parameterless constructor, so as long as the type is known SOMEWHERE at compile-time, this will work and will alert you to a problem earlier.
Of course if the type really is known only at runtime (i.e. you're using Activator.CreateInstance() to instantiate an object based on a string or a constructed Type) then this won't help you. I generally use reflection as the absolute last option, because once you've gone to dynamic land you pretty much have to stay in dynamic land; it's usually difficult or even messier to dynamically instantiate something and then start dealing with it statically.
Solution 6 - C#
I needed to count constructors with only optional parameters the same as true parameter-less constructors. To do this:
myClass.GetType().GetConstructors()
.All(c => c.GetParameters().Length == 0 || c.GetParameters().All(p => p.IsOptional))
Solution 7 - C#
If anyone is interested in an "official" version, the following was found via .NET Reflector:
from: System.Activities.Presentation.TypeUtilities
in System.Activities.Presentation.dll, Version=4.0.0.0
public static bool CanCreateInstanceUsingDefaultConstructor(this Type t) =>
t.IsValueType || !t.IsAbstract && t.GetConstructor(Type.EmptyTypes) != null;
Notice the check for t.IsAbstract
, which is not mentioned elsewhere on this page.
You can also expand the GetConstructor
call out as follows, if you feel like micro-optimizing away one stack frame:
…t.GetConstructor(BindingFlags.Public|BindingFlags.Instance, null, Type.EmptyTypes, null)…
Solution 8 - C#
Yes, you have to use reflection.
object myObject = new MyType();
Type type = myObject.GetType();
ConstructorInfo conInfo = type.GetConstructor(new Type[0]);
Solution 9 - C#
To get the one that has more optional parameters or an empty constructor at all, use:
typeof(myClass)
.GetConstructors()
.OrderBy(x => x.GetParameters().Length - x.GetParameters().Count(p => p.IsOptional))
.FirstOrDefault();