Cannot convert null to type parameter because it could be a non nullable type when expecting "T?"

MrD at KookerellaLtd Published at Dev

8

MrD at KookerellaLtd

this code complains

public static T? Foo<T>()
{
   return null;
}

with

Error CS0403
Cannot convert null to type parameter 'T' because it could be a non-nullable value type. Consider using 'default(T)' instead.

But aren't I telling the compiler it can be null?

If T is int, then I want the answer to be null, not 0.

Guru Stron

Because T is unconstrained in your case, and for unconstrained (neither to struct nor to class) generic type parameters T - T? is handled differently for value types and reference types. From the docs:

If the type argument for T is a reference type, T? references the corresponding nullable reference type. For example, if T is a string, then T? is a string?.

If the type argument for T is a value type, T? references the same value type, T. For example, if T is an int, the T? is also an int.

If the type argument for T is a nullable reference type, T? references that same nullable reference type. For example, if T is a string?, then T? is also a string?.

If the type argument for T is a nullable value type, T? references that same nullable value type. For example, if T is a int?, then T? is also a int?.

So for Foo<int> actual signature will be int Foo<int>, hence the error.

If it is suitable for you - constrain T to be either struct or class.

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edited at2023-04-29

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