Reflection: How to get a generic method? [duplicate]
Possible Duplicates:
How to use reflection to call generic Method? Select Right Generic Method with Reflection
Hi there
Let's say I have two following two methods in a class:
public void MyMethod(object val) {}
public void MyMethod<T>(T val) {}
With reflection, I could get the fir开发者_运维知识库st Method like this:
Type[] typeArray = new Type[1];
typeArray.SetValue(typeof(object), 1);
var myMethod = myInstance.GetType().GetMethod("MyMethod", typeArray);
But how can I get the second, generic method?
var myMethod = myInstance.GetType()
.GetMethods()
.Where(m => m.Name == "MyMethod")
.Select(m => new {
Method = m,
Params = m.GetParameters(),
Args = m.GetGenericArguments()
})
.Where(x => x.Params.Length == 1
&& x.Args.Length == 1
&& x.Params[0].ParameterType == x.Args[0])
.Select(x => x.Method)
.First();
I would do it like this:
var methods = from m in typeof(MyClass).GetMethods()
where m.Name == "MyMethod"
&& m.IsGenericMethodDefinition
let typeParams = m.GetGenericArguments()
let normalParams = m.GetParameters()
where typeParams.Length == 1 && normalParams.Length == 1
&& typeParams.Single() == normalParams.Single().ParameterType
&& !typeParams.Single().GetGenericParameterConstraints().Any()
select m;
var myMethod = methods.Single();
We're looking for a method called "MyMethod" that is a generic method with a single type-parameter having no constraints, and one 'normal' parameter of the same type as the type-parameter.
Obviously, if you're not looking to be very precise, you can just do the bare minimum to disambiguate, such as:
var myMethod = typeof(MyClass)
.GetMethods()
.Single(m => m.Name == "MyMethod" && m.IsGenericMethodDefinition);
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