Why can't I set a nullable int to null in a ternary if statement? [duplicate]

Solution 1:

The compiler tries to evaluate the right-hand expression. null is null and the 0 is an int literal, not int?. The compiler is trying to tell you that it can't determine what type the expression should evaluate as. There's no implicit conversion between null and int, hence the error message.

You need to tell the compiler that the expression should evaluate as an int?. There is an implicit conversion between int? and int, or between null and int?, so either of these should work:

int? x = true ? (int?)null : 0;

int? y = true ? null : (int?)0;

Solution 2:

You need to use the default() keyword rather than null when dealing with ternary operators.

Example:

int? i = (true ? default(int?) : 0);

Alternately, you could just cast the null:

int? i = (true ? (int?)null : 0);

Personally I stick with the default() notation, it's just a preference, really. But you should ultimately stick to just one specific notation, IMHO.

HTH!

Solution 3:

The portion (true ? null : 0) becomes a function in a way. This function needs a return type. When the compiler needs to figure out the return type it can't.

This works:

int? i;
i = (true ? null : (int?)0);