A dynamic definite assignment puzzle, part 2

Sorry for that unexpected interlude into customer service horrors; thanks for your patience and let's get right back to coding horrors!

So as not to bury the lede: the C# compiler is correct to flag the program from the last episode as having a definite assignment problem. Thanks to everyone who participated in the discussion in the comments.

Particular mentions go out to Jeffrey L. Whitledge who sketched the solution mere moments after I posted the problem, C# architect Neal Gafter who posted a concise demonstration, and former C# compiler dev Chris Burrows who slyly pointed out on Twitter that he "solved the puzzle". Indeed, he solved it nine years ago when he designed and implemented this logic in the compiler.

The reasoning the compiler has to go through here is rather arcane, so let's dig into it.

First off, let's remind ourselves of the fundamental rule of dynamic in C#. We identify all the places in the program where the compiler deduces that the type of a thing is dynamic. Imagine that you then replaced that type judgment with the actual runtime type of the thing. The behaviour of the program with dynamic should be the same as the behaviour of the imaginary program where the compiler knew the runtime type.

This means that the compiler has to assume that the control flow of a program containing dynamic could be any control flow possible for any runtime type, including types that are legal but extremely weird. So let's make an extremely weird type or two and see what happens.

I've already discussed in this blog how operator overloading works for the && operator. Briefly, the rules in C# are:

• if left is false then left && right does not evaluate right, and has the same value as left
• otherwise, left && right evaluates both operands and has the same value as left & right

Obviously those rules work for bool; to overload &&, we therefore need a way to evaluate whether left is false or not, and a way to determine the value of left & right. The "operator true" and "operator false" operators tell you whether a given value is logically true or logically false, and both must be provided if you provide either. Let's look at an example, but we'll make it a crazy example!

class Crazy
{
public static bool operator true(Crazy c)
{
Console.WriteLine("operator true");
return true; // Um"
}
public static bool operator false(Crazy c)
{
Console.WriteLine("operator false");
return true; // Oh dear"
}
public static Crazy operator &(Crazy a, Crazy b)
{
Console.WriteLine("operator &");
return a;
}
}

This is obviously weird and wrong; this program says that all instances of Crazy are logically true, and also all of them are logically false (because "is this false?" returns true) and also if we & two of them together, we get the first unconditionally. But, whatever, this is a legal program. If we have

Crazy c = GetCrazy() && GetAnother();

The code generated will be the moral equivalent of:

Crazy c1 = GetCrazy();
Crazy c = Crazy.operator_false(c1) ?
c1 :
Crazy.operator_&(c1, GetAnother());

But of course at runtime for this particular program we will always say that c1 is false, and always just take the first operand without calling GetAnother().

Already we've got a very strange program, but maybe it is not entirely clear how this produces a code path that has a definite assignment error. To get that, we need a second super-weird class. This one overloads equality - C# requires you to overload equality and inequality in pairs as well - and instead of returning a Boolean, it returns, you guessed it, something crazy:

class Weird
{
public static Crazy operator ==(Weird p1, Weird p2)
{
Console.WriteLine("==");
return new Crazy();
}
public static Crazy operator !=(Weird p1, Weird p2)
{
Console.WriteLine("!=");
return new Crazy();
}
public override bool Equals(object obj) => true;
public override int GetHashCode() => 0;
}

Maybe now you see where this is going, but we're not quite there yet. Suppose we have

Weird obj = new Weird();
string str;
if (obj != null && GetString(out str))

What's going to happen? Weird overrides !=, so this will call Weird.operator_!=, which returns Crazy, and now we have Crazy && bool, which is not legal, so the program will not compile. But what if we make Crazy convertible from bool? Then it will work! So add this to Crazy:

public static implicit operator Crazy(bool b)
{
Console.WriteLine("implicit conversion from bool " + b);
return new Crazy();
}

Now what does if (obj != null && GetString(out str)) " mean? Let's turn it into method calls. This thing all together is:

Crazy c1 = Weird.operator_!=(obj, null);
bool b1 = Crazy.operator_false(c1);
// it is true that it is false
Crazy c2 = b1 ?
c1 : // This is the branch we take
Crazy.operator_implicit(GetString(out str));
if (c2) "

But wait a minute, is if(c2) legal? Yes it is, because Crazy implements operator true, which is then called by the if statement, and it returns true!

We have just demonstrated that it is possible to write the program fragment

if (obj != null && GetString(out str))
Console.WriteLine(obj + str);

and have the consequence get executed even if GetString is never called, and therefore str would be read before it was written, and therefore this program must be illegal. And indeed, if you remove the definite assignment error and run it, you get the output

!=operator falseoperator true

In the case where obj is dynamic, the compiler must assume the worst possible case, no matter how unlikely. It must assume that the dynamic value is one of these weird objects that overloads equality to return something crazy, and that the crazy thing can force the compiler to skip evaluating the right side but still result in a true value.

What is the moral of this story? There are several. The most practical moral is: do not compare any dynamic value to null! The correct way to write this code is

if ((object)obj != null && GetString(out str))

And now the problem goes away, because dynamic to object is always an identity conversion, and now the compiler can generate a null check normally, rather than going through all this rigamarole of having to see if the object overloads equality.

The moral for the programming language designer is: operator overloading is a horrible horrible feature that makes your life harder.

The moral for the compiler writer is: get good at coming up with weird and unlikely possible programs, because you're going to have to do that a lot when you write test cases.

Thanks to Stack Overflow contributor "NH." for the question which prompted this puzzle.