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Using C# 7.1 default literal in nullable optional argument causes unexpected behavior

C# 7.1 introduces a new feature called "Default Literals" that allows new default expressions.

// instead of writing
Foo x = default(Foo);

// we can just write
Foo x = default;

For Nullable<T> types, the default value is null, and with the usual usage this works as expected:

int? x = default(int?); // x is null

int? x = default; // x is null

However, when I try to use the new default literal as an optional argument (parameter) of a function, it's not working as expected:

static void Foo(int? x = default(int?))
{
    // x is null
}

static void Foo(int? x = default)
{
    // x is 0 !!!
}

To me, this behavior is unexpected and looks like a bug in the compiler.

Can anybody confirm the bug, or explain this behavior?

like image 466
Tom Pažourek Avatar asked Oct 29 '17 07:10

Tom Pažourek


1 Answers

After more research online, I found out that it's a known confirmed bug:

  • https://github.com/dotnet/csharplang/issues/970
  • https://github.com/dotnet/roslyn/issues/22578

It's already fixed and will be part of C# 7.2.

like image 164
Tom Pažourek Avatar answered Sep 23 '22 12:09

Tom Pažourek