Curious null-coalescing operator custom implicit conversion behaviour

Question

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 n

Answer 1

This is most definitely a bug.

public class Program {
    static A? X() {
        Console.WriteLine("X()");
        return new A();
    }
    static B? Y() {
        Console.WriteLine("Y()");
        return new B();
    }
    static C? Z() {
        Console.WriteLine("Z()");
        return new C();
    }

    public static void Main() {
        C? test = (X() ?? Y()) ?? Z();
    }
}

This code will output:

X()
X()
A to B (0)
X()
X()
A to B (0)
B to C (0)

That made me think that the first part of each ?? coalesce expression is evaluated twice. This code proved it:

B? test= (X() ?? Y());

outputs:

X()
X()
A to B (0)

This seems to happen only when the expression requires a conversion between two nullable types; I've tried various permutations with one of the sides being a string, and none of them caused this behaviour.

Answer 2

If you take a look at the generated code for the Left-grouped case it actually does something like this (csc /optimize-):

C? first;
A? atemp = a;
B? btemp = (atemp.HasValue ? new B?(a.Value) : b);
if (btemp.HasValue)
{
    first = new C?((atemp.HasValue ? new B?(a.Value) : b).Value);
}

Another find, if you use first it will generate a shortcut if both a and b are null and return c. Yet if a or b is non-null it re-evaluates a as part of the implicit conversion to B before returning which of a or b is non-null.

From the C# 4.0 Specification, §6.1.4:

• If the nullable conversion is from S? to T?:
  • If the source value is null (HasValue property is false), the result is the null value of type T?.
  • Otherwise, the conversion is evaluated as an unwrapping from S? to S, followed by the underlying conversion from S to T, followed by a wrapping (§4.1.10) from T to T?.

This appears to explain the second unwrapping-wrapping combination.

---

The C# 2008 and 2010 compiler produce very similar code, however this looks like a regression from the C# 2005 compiler (8.00.50727.4927) which generates the following code for the above:

A? a = x;
B? b = a.HasValue ? new B?(a.GetValueOrDefault()) : y;
C? first = b.HasValue ? new C?(b.GetValueOrDefault()) : z;

I wonder if this is not due to the additional magic given to the type inference system?

Answer 3

Thanks to everyone who contributed to analyzing this issue. It is clearly a compiler bug. It appears to only happen when there is a lifted conversion involving two nullable types on the left-hand side of the coalescing operator.

I have not yet identified where precisely things go wrong, but at some point during the "nullable lowering" phase of compilation -- after initial analysis but before code generation -- we reduce the expression

result = Foo() ?? y;

from the example above to the moral equivalent of:

A? temp = Foo();
result = temp.HasValue ? 
    new int?(A.op_implicit(Foo().Value)) : 
    y;

Clearly that is incorrect; the correct lowering is

result = temp.HasValue ? 
    new int?(A.op_implicit(temp.Value)) : 
    y;

My best guess based on my analysis so far is that the nullable optimizer is going off the rails here. We have a nullable optimizer that looks for situations where we know that a particular expression of nullable type cannot possibly be null. Consider the following naive analysis: we might first say that

result = Foo() ?? y;

is the same as

A? temp = Foo();
result = temp.HasValue ? 
    (int?) temp : 
    y;

and then we might say that

conversionResult = (int?) temp 

is the same as

A? temp2 = temp;
conversionResult = temp2.HasValue ? 
    new int?(op_Implicit(temp2.Value)) : 
    (int?) null

But the optimizer can step in and say "whoa, wait a minute, we already checked that temp is not null; there's no need to check it for null a second time just because we are calling a lifted conversion operator". We'd them optimize it away to just

new int?(op_Implicit(temp2.Value)) 

My guess is that we are somewhere caching the fact that the optimized form of (int?)Foo() is new int?(op_implicit(Foo().Value)) but that is not actually the optimized form we want; we want the optimized form of Foo()-replaced-with-temporary-and-then-converted.

Many bugs in the C# compiler are a result of bad caching decisions. A word to the wise: every time you cache a fact for use later, you are potentially creating an inconsistency should something relevant change. In this case the relevant thing that has changed post initial analysis is that the call to Foo() should always be realized as a fetch of a temporary.

We did a lot of reorganization of the nullable rewriting pass in C# 3.0. The bug reproduces in C# 3.0 and 4.0 but not in C# 2.0, which means that the bug was probably my bad. Sorry!

I'll get a bug entered into the database and we'll see if we can get this fixed up for a future version of the language. Thanks again everyone for your analysis; it was very helpful!

UPDATE: I rewrote the nullable optimizer from scratch for Roslyn; it now does a better job and avoids these sorts of weird errors. For some thoughts on how the optimizer in Roslyn works, see my series of articles which begins here: https://ericlippert.com/2012/12/20/nullable-micro-optimizations-part-one/

Answer 4

I am not a C# expert at all as you can see from my question history, but, I tried this out and I think it is a bug.... but as a newbie, I have to say that I do not understand everything going on here so I will delete my answer if I am way off.

I have come to this bug conclusion by making a different version of your program which deals with the same scenario, but much less complicated.

I am using three null integer properties with backing stores. I set each to 4 and then run int? something2 = (A ?? B) ?? C;

(Full code here)

This just reads the A and nothing else.

This statement to me looks like to me it should:

1. Start in the brackets, look at A, return A and finish if A is not null.
2. If A was null, evaluate B, finish if B is not null
3. If A and B were null, evaluate C.

So, as A is not null, it only looks at A and finishes.

In your example, putting a breakpoint at the First Case shows that x, y and z are all not null and therefore, I would expect them to be treated the same as my less complex example.... but I fear I am too much of a C# newbie and have missed the point of this question completely!

Answer 5

Actually, I'll call this a bug now, with the clearer example. This still holds, but the double-evaluation is certainly not good.

It seems as though A ?? B is implemented as A.HasValue ? A : B. In this case, there's a lot of casting too (following the regular casting for the ternary ?: operator). But if you ignore all that, then this makes sense based on how it's implemented:

1. A ?? B expands to A.HasValue ? A : B
2. A is our x ?? y. Expand to x.HasValue : x ? y
3. replace all occurrences of A -> (x.HasValue : x ? y).HasValue ? (x.HasValue : x ? y) : B

Here you can see that x.HasValue is checked twice, and if x ?? y requires casting, x will be cast twice.

I'd put it down simply as an artifact of how ?? is implemented, rather than a compiler bug. Take-Away: Don't create implicit casting operators with side effects.

It seems to be a compiler bug revolving around how ?? is implemented. Take-away: don't nest coalescing expressions with side-effects.