为什么要将null值转换为可以为null的struct

Question

I have a function which gives back a nullable struct . I noticed two similar cases

First: works well:

public static GeometricCircle? CircleBySize(GeometricPoint point, double size)
{
    if (size >= epsilon)
        return null;

    return new GeometricCircle(point.Position, new Vector(1, 0, 0), size, true);
}

Second: needs to convert the null value to GeometricCircle?

public static GeometricCircle? CircleBySize(GeometricPoint point, double size)
{
    return size > epsilon ? new GeometricCircle(point.Position, new Vector(1, 0, 0), size, true) : (GeometricCircle?)null;
}

Does anybody know what is the difference?

Answer 1

In your first example, you are returning null when size >= epsilon . The compiler knows that null is a valid value for a nullable type.

In your second example, you are using the ?: ternary operator , which comes with its own set of rules.

condition ? first_expression : second_expression;

MSDN tells us (my emphasis)...

Either the type of first_expression and second_expression must be the same, or an implicit conversion must exist from one type to the other.

The key difference here is that null cannot be implicitly converted into a GeometricCircle , (the type of your first_expression ).

So you have to do it explicity , using a cast to GeometricCircle? , which is then implicitly convertible to GeometricCircle .