Curious null-coalescing operator custom implicit conversion behaviour

Note: this appears to have been fixed in Roslyn

This question arose when writing my answer to this one, which talks about the associativity of the null-coalescing operator.

Just as a reminder, the idea of the null-coalescing operator is that an expression of the form

x ?? y

first evaluates x, then:

• If the value of x is null, y is evaluated and that is the end result of the expression
• If the value of x is non-null, y is not evaluated, and the value of x is the end result of the expression, after a conversion to the compile-time type of y if necessary

Now usually there’s no need for a conversion, or it’s just from a nullable type to a non-nullable one – usually the types are the same, or just from (say) int? to int. However, you can create your own implicit conversion operators, and those are used where necessary.

For the simple case of x ?? y, I haven’t seen any odd behaviour. However, with (x ?? y) ?? z I see some confusing behaviour.

Here’s a short but complete test program – the results are in the comments:

using System;

public struct A
{
    public static implicit operator B(A input)
    {
        Console.WriteLine("A to B");
        return new B();
    }

    public static implicit operator C(A input)
    {
        Console.WriteLine("A to C");
        return new C();
    }
}

public struct B
{
    public static implicit operator C(B input)
    {
        Console.WriteLine("B to C");
        return new C();
    }
}

public struct C {}

class Test
{
    static void Main()
    {
        A? x = new A();
        B? y = new B();
        C? z = new C();
        C zNotNull = new C();

        Console.WriteLine("First case");
        // This prints
        // A to B
        // A to B
        // B to C
        C? first = (x ?? y) ?? z;

        Console.WriteLine("Second case");
        // This prints
        // A to B
        // B to C
        var tmp = x ?? y;
        C? second = tmp ?? z;

        Console.WriteLine("Third case");
        // This prints
        // A to B
        // B to C
        C? third = (x ?? y) ?? zNotNull;
    }
}

So we have three custom value types, A, B and C, with conversions from A to B, A to C, and B to C.

I can understand both the second case and the third case… but why is there an extra A to B conversion in the first case? In particular, I’d really have expected the first case and second case to be the same thing – it’s just extracting an expression into a local variable, after all.

Any takers on what’s going on? I’m extremely hesistant to cry “bug” when it comes to the C# compiler, but I’m stumped as to what’s going on…

EDIT: Okay, here’s a nastier example of what’s going on, thanks to configurator’s answer, which gives me further reason to think it’s a bug. EDIT: The sample doesn’t even need two null-coalescing operators now…

using System;

public struct A
{
    public static implicit operator int(A input)
    {
        Console.WriteLine("A to int");
        return 10;
    }
}

class Test
{
    static A? Foo()
    {
        Console.WriteLine("Foo() called");
        return new A();
    }

    static void Main()
    {
        int? y = 10;

        int? result = Foo() ?? y;
    }
}

The output of this is:

Foo() called
Foo() called
A to int

The fact that Foo() gets called twice here is hugely surprising to me – I can’t see any reason for the expression to be evaluated twice.