Convert between nullable int and int

Question

I would like to do something like this :

int? l = lc.HasValue ? (int)lc.Value : null; 

where lc is a nullable enumeration type, say EMyEnumeration?. So I want to test if lc has a value, so if then give its int value to l, otherwise l is null. But when I do this, C# complains that 'Error type of conditional expression cannot be determined as there is no implicit conversion between 'int' and ''.

How can I make it correct? Thanks in advance!!

Answer 1

You have to cast null as well:

int? l = lc.HasValue ? (int)lc.Value : (int?)null; 

By the way, that's one of the differences between the conditional operator and an if-else:

if (lc.HasValue)
    l = (int)lc.Value;
else
    l = null;  // works

The relevant C# specification is 7.14 Conditional operator:

The second and third operands, x and y, of the ?: operator control the type of the conditional expression.

• If x has type X and y has type Y then
  • If an implicit conversion (§6.1) exists from X to Y, but not from Y to X, then Y is the type of the conditional expression.
  • If an implicit conversion (§6.1) exists from Y to X, but not from X to Y, then X is the type of the conditional expression.
  • Otherwise, no expression type can be determined, and a compile-time error occurs.

Since int is not convertible to null implicitly and vice-versa you get a compiler error. But you could also cast the enum to int? instead of int which is convertible. Then the compiler can derive the type and it'll compile:

int? l = lc.HasValue ? (int?)lc.Value : null; // works also

or, as Sergey has mentioned in his answer, directly:

int? l = (int?)lc;