C.
As @ Tim pointed out, you won't get an error for the following code: long foo = 42; if (foo = null) You'll get a warning instead: The result of the expression is always 'false' since a value of type 'long' is never equal to 'null' of type 'long?'. This gives a warning instead of an error because of lifted operators, defined in the C language specification as such: Lifted operators permit predefined and user-defined operators that operate on non-nullable value types to also be used with nullable forms of those types. [.] For the equality operators =!= a lifted form of an operator exists if the operand types are both non-nullable value types and if the result type is bool. The lifted form is constructed by adding a single? modifier to each operand type. The lifted operator considers two null values equal, and a null value unequal to any non-null value. If both operands are non-null, the lifted operator unwraps the operands and applies the underlying operator to produce the bool result. The underlying operator in this case is the predefined value type long's = operator: For predefined value types, the equality operator (=) returns true if the values of its operands are equal, false otherwise. Because foo is implicitly converted ( Predefined implicit conversions that operate on non-nullable value types can also be used with nullable forms of those types. ) and the null literal is also implicitly converted ( An implicit conversion exists from the null literal to any nullable type. the expression: (long)foo = null Becomes: (long?)foo = (long?)null Which, given foo is of type long and thus always has a value, always returns false and won't even apply long's = operator. I'm not entirely sure, but I suspect this to exist to enable comparison between nullable and non-nullable values without explicit casting: long? foo = 42; long bar = 42; Console. Write Line(foo =.
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