3 ms·
Yes, because it's checked by the compiler using normal constructs. someNullableInstance.DoSomething(); // compiler error if (someNullableInstance != n
by wvenable 7y ago
Yes, because it's checked by the compiler using normal constructs.
someNullableInstance.DoSomething(); // compiler error
if (someNullableInstance != null)
someNullableInstance.DoSomething(); // not an error
Option<T> is fine but it's a lot of syntax to use correctly.