Cannot Have Nullable Return of Async Generic Method

[kalaskarsanket]

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Issue moved from microsoft/dotnet#1249

• Please respond to @mysteryx93.

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From @mysteryx93 on Thursday, September 10, 2020 8:17:31 PM

With C# 8 Nullables, you can define a nullable generic like this

[return: MaybeNull]
public T Get<T>() => default;

However, I'm running into a problem with a Generic Async method.

[return: MaybeNull]
public async Task<T> GetAsync<T>() => default;

This would tell that Task is nullable, not T.

The only solution for now would be to add ! after default, but then the IDE will wrongly indicate to the users that the type is not null when it can be. This can work for private use (and can easily lead to errors), but not for a public API.

Is there any other solution?

Actually what I'd really need is a generic constraint "where T : nullable" because I need to document to pass in nullable types.

Currently, the only way of doing so is to have 2 copies, one with "where T : class" and one with "where T : struct", and using T? where ? is interpreted differently in each copy (Nullable vs nullable attribute). That can lead to doubling a lot of code.

[kalaskarsanket]

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Issue moved from microsoft/dotnet#1249

• Please respond to @mysteryx93.

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From @mysteryx93 on Saturday, September 19, 2020 5:51:30 PM

Since there has been no response, here's my take on this.

The .NET team could fix it in 2 ways.

1. Add a "nullable" constraint

2. Tweaking MaybeNull attribute to work in the above case

On a different topic, a 3rd suggestion to help working with Generics

if (typeof(T) == typeof(JsonElement))
{
    return (T)(object)doc.RootElement;
}

This works but has unnecessary boxing/unboxing, and disables type-checking

The compiler should recognize this as valid IMO

if (typeof(T) == typeof(JsonElement))
{
    return (T)doc.RootElement;
}

[333fred]

In C# 9, we've enabled unconstrained T?, so you can simply write Task<T?>. Note that this still behaves like regular generics: int does not become int? here, but rather just becomes int. If the type you substitute is a reference type, then the return will be nullable.

[RikkiGibson]

I'm using C# 10 with .NET Framework 4.8. Maybe what you're missing here is where T : default. I thought that only mattered for overrides though.

We haven't introduced an actionable diagnostic for this yet? 😕

[zzzachzzz]

I was going to test with <LangVersion> set to 10.0 but it's not available for me:

$ dotnet /usr/share/dotnet/sdk/3.1.417/Roslyn/bincore/csc.dll /langversion:\?
Supported language versions:
default
1
2
3
4
5
6
7.0
7.1
7.2
7.3
8.0 (default)
9.0
latestmajor
preview
latest

[zzzachzzz]

When changing the method signature to the following:
async Task<T?> ICache.Get<T>(CacheEntry<T> cacheEntry) where T : default

Lines where I return null still yield:

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

[jnm2]

@zzzachzzz You need to install the .NET 6 SDK to compile C# 10. That's not the same thing as needing your app to run on .NET 6; the .NET 6 SDK can build .NET Core 3.1 and even .NET Framework apps.

[jnm2]

@zzzachzzz You simply can't convert null to T in any version of the C# language. Unless you have where T : struct or where T : class, T? means "defaultable" rather than "nullable" which is what @333fred was saying above.

If T appears in the parameter section of a method signature, you can define overloads side by side with where T : struct and where T : class. I do this sometimes.