CODEWITHSUNDEEP
c#genericstypesgeneric-type-argument
aleemb
aleemb

Reputation: 32105

How to turn a Type instance into a generic type argument

I basically have something like this:

void Foo(Type ty)
{
    var result = serializer.Deserialize<ty>(inputContent);
}

Foo(typeof(Person));

The Deserialize<ty> doesn't work because it expects Deserialize<Person> instead. How do I work around this?

I'd also like to understand how generics work and why it won't accept ty which is typeof(Person).

EDIT: I ought to have mentioned that this is a contrived example. I cannot actually change the signature of the function because it implements an interface.

EDIT: serializer is a JavascriptSerializer and implemented as an action filter here. It is called thusly:

[JsonFilter(Param="test", JsonDataType=typeof(Person))]

Solution

Based on Marc and Anton's answers:

var result = typeof(JavaScriptSerializer).GetMethod("Deserialize")
                 .MakeGenericMethod(JsonDataType)
                 .Invoke(serializer, new object[] { inputContent });

Upvotes: 10

Views: 1648

Answers (5)

Rytmis
Rytmis

Reputation: 32067

Like Lucero said,

void Foo<ty>()
{
    var result = serializer.Deserialize<ty>(inputContent);
}

Foo<Person>();

typeof(Person) is not the same thing as Person. Person is a compile-time type, whereas typeof(Person) is an expression that returns a Type instance representing the runtime type information of Person.

Upvotes: 1

Anton Gogolev
Anton Gogolev

Reputation: 115897

If ty is known at compile-time, why don't just

void Foo<T>()
{
    var result = serializer.Deserialize<T>(inputContext);
}

Otherwise, 

MethodInfo genericDeserializeMethod = serializer.GetType().GetMethod("Deserialize");
MethodInfo closedDeserializeMethod = genericDeserializeMethod.MakeGenericMethod(ty);
closedDeserializeMethod.Invoke(serializer, new object[] { inputContext });

Upvotes: 6

Lucero
Lucero

Reputation: 60276

In this case, just do this:

void Foo<ty>()
{
    var result = serializer.Deserialize<ty>(inputContent);
}

Foo<Person>();

Otherwise, you need to call the generic method late-bound, since you have to get the correct generic method for it first (it is not known at compile time). Have a look at the MethodInfo.MakeGenericMethod method.

Upvotes: 1

Marc Gravell
Marc Gravell

Reputation: 1064244

Which serializer is that? If you only know the Type at runtime (not compile time), and it doesn't have a non-generic API, then you might have to use MakeGenericMethod:

void Foo(Type ty)
{
    object result = typeof(ContainingClass).GetMethod("Bar").
        .MakeGenericMethod(ty).Invoke(null, new object[] {inputContent});
}
public static T Bar<T>(SomeType inputContent) {
    return serializer.Deserialize<T>(inputContent);
}

Upvotes: 7

stevehipwell
stevehipwell

Reputation: 57568

Use

void Foo<T>(){ var result = serializer.Deserialize<T>(inputContent); }

With the following call

Foo<Person>();

Upvotes: 2

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