How do I specify "any non-nullable type" as a generic type parameter constraint?

Question

The post is specific to C# 8. Let's assume I want to have this method:

public static TValue Get<TKey, TValue>(
  this Dictionary<TKey, TValue> src, 
  TKey key, 
  TValue @default
) 
=> src.TryGetValue(key, out var value) ? value : @default;

If my .csproj looks like this (i.e. C# 8 and nullable types are enabled, all warnings are errors):

<Project Sdk="Microsoft.NET.Sdk">
  <PropertyGroup>
    <TargetFramework>netcoreapp3.0</TargetFramework>
    <LangVersion>8</LangVersion>
    <Nullable>enable</Nullable>
    <WarningsAsErrors>true</WarningsAsErrors>
  </PropertyGroup>
  …
</Project>

This code will produce the following build-time error:

DictionaryEx.cs(28, 78): [CS8714] The type 'TKey' cannot be used as type parameter 'TKey' in the generic type or method 'Dictionary'. Nullability of type argument 'TKey' doesn't match 'notnull' constraint.

Is there any way to specify that TKey has to be a non-nullable type?

For me this code compiles without CS8714 or any other warnings. — DK.
– DK., Commented Aug 3, 2019 at 4:50

Alex Yakunin · Accepted Answer · 2019-08-03 04:50:58Z

76

Ok, just found out that you can use notnull constraint:

public static TValue Get<TKey, TValue>(
    this Dictionary<TKey, TValue> src, 
    TKey key, TValue @default)
    where TKey : notnull
    => src.TryGetValue(key, out var value) ? value : @default;

answered Aug 3, 2019 at 4:50

Alex Yakunin

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3 Comments

V0ldek Over a year ago

This gives an IntelliSense error "cannot resolve symbol" in VS 16.2.3, but it does compile and work anyway.

NetherGranite Over a year ago

I couldn't find any documentation on this. Do we know if this is future-proof?

Jeremy Caney Over a year ago

@NetherGranite: A year later, at least, this is now documented as part of Microsoft's Generic Type Constraint documentation in the Language Reference, as well as the Constraints on Type Parameters in the Programming Guide.

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