As i was saying last time, the nice thing about “no backtracking” is that it makes the language much easier to understand. Simple rules benefit both the compiler and the code reader; both are attempting to read the code to make sense of it. It is not always a good idea to take advantage of the compiler’s ability to search a large space if that makes it harder for a human to understand the code.

Suppose you have something like this mess: (*)

namespace XYZ.DEF
{ 
    public class GHI {} 
} 
namespace QRS.DEF.GHI 
{ 
    public class JKL { } 
} 

… in another file …

using QRS; 
namespace TUV  
{
    using XYZ;
    namespace ABC
    {
        namespace DEF
        {
            class GHI { }
            class MNO : DEF.GHI.JKL { }
        }
    }
} 

And now we must work out the base type of MNO.

With no backtracking we say “The nearest container of MNO that has a member DEF is ABC, therefore DEF means ABC.DEF”. Therefore GHI is ABC.DEF.GHI. Therefore JKL is ABC.DEF.GHI.JKL, which does not exist, therefore, give an error. The developer must fix the error by giving a type name that lets the compiler identify which DEF you meant.

If we had backtracking, what would we have to do? We’d get that error, and then we’d backtrack. Does XYZ contain a DEF? Yes. Does it contain a GHI? Yes. Does it contain a JKL? No. Backtrack again. Does QRS contain an DEF.GHI.JKL? Yes.

That works, but can we logically conclude from the fact that it works that it is the one the user meant?

Who the heck knows in this crazy situation? We got all kinds of good bindings in there that then went bad very late in the game. The idea that we stumbled upon the desired answer after going down so many blind alleys seems highly suspect. Maybe there is yet another choice in there that is the one the user meant. We cannot know that unless we try all of them, and again, that could involve a lot of searching.

The correct thing to do here is not to backtrack multiple times and try out all kinds of worse bindings for every stage of the lookup. The correct thing to do is to say “buddy, the best possible match for this lookup gives nonsensical results; give me something less ambiguous to work with here please.”

An unfortunate fact about writing a language where the compiler by design complains loudly if the best match is something that doesn’t work, is that developers frequently say “well, sure, in general I want the compiler to point out all my mistakes — or, rather, all my coworker’s mistakes. But for this specific case, I know what I am doing, so please, compiler, do what I mean, not what I say.”

Trouble is, you can’t have it both ways. You can’t have both a compiler that both enforces rigid rules that make it highly likely that suspicious code will be aggressively identified as erroneous and allow crazy code via compiler heuristics that figure out “what I really meant” when you write something that the compiler quite rightly sees as ambiguous or wrong.

I could discuss many more places where we could do backtracking but do not. Method type inference, for example, always either makes progress or fails; it never backtracks in C# (**). But I think I will leave it at that. Except to say that there is one place where we do use backtracking, and that’s analysis of overload resolution in nested lambdas. I wrote about that here.

(*) When reading this remember that in C#, “namespace XYZ.DEF { }” is a short form for “namespace XYZ{ namespace DEF { } }”

(**) Unlike in, say, F#.

Commentary from 2021:

There were a number of good comments to the original post, which pointed out:

• Error analysis often has to perform some sort of backtracking to give a good error. In the examples I gave, for instance, we could run a backtracking analysis and then give a “did you mean…?” style error message if backtracking finds a solution, and a different error if no solution can be found. That’s a great idea in principle, but in practice we often end up in situations where backtracking is expensive, so we’d want to put a box around the amount of time and resources the error reporting system uses.
• This is even more the case when IntelliSense is involved. IntelliSense by definition is typically operating on code that is wrong; if the code was right, why are you editing it? IntelliSense needs to give good options, but it needs to do so in under 30 milliseconds, so it uses lots of tricks to meet its time budget.
• There was also some good discussion of the deeper language design point here. There are times when we want to use backtracking algorithms, and times we do not; what motivates that choice? There is not just one thing; like all design and implementation problems it comes down to resolving conflicts between incompatible principles. One of the nice things about “no backtracking” algorithms is that they often allow “local” analysis; if we see ABC.DEF.GHI and immediately after, ABC.DEF.XYZ, we know that no matter what, the ABC.DEF means the same thing; the meanings of ABC and ABC.DEF do not change depending on whether GHI and XYZ produce errors or not. That’s a nice property to have.

A number of the articles I’ve published over the years involve “backtracking” algorithms; most recently my series on how to solve Sudoku puzzles (and more generally, all graph colouring problems) by backtracking. The idea of a backtracking algorithm is really simple and powerful: when faced with a choice, try every possibility. If all of them go wrong, backtrack to the previous choice and try a different possibility. Backtracking algorithms are essentially a depth-first search of the space of possible solutions.

But what if the solution space is large? There are more ways to fill in nine digits in 81 places than there are protons in the entire Universe; we can’t check them all. The reason that backtracking works so rapidly for Sudoku puzzles is because typically every guess eliminates many of the branches of the solution tree. If you guess that the first square is 1 then you do not need to explore the possibility that the second square is 1; you can just prune that branch and never explore it. Since that branch itself is vast, that’s a good strategy! (There are pathological cases for this algorithm, but in practice it is fast.)

Backtracking algorithms work well for many situations, but we have consciously eschewed them in the design of C#. (With a few exceptions). For example, there is no backtracking that ever crosses “phases” of compilation in C#.(*) When we are presented with some source code the first thing we do is “lex” it; break it up into “tokens”. This is akin to finding the words in a sentence by looking at the spacing and punctuation. Then we do a grammatical analysis of the token stream, and then we do a semantic analysis of the grammatical analysis. When one of those phases finds an error, we never backtrack to a previous phases and ask for more analysis. Suppose for example you have this silly code:

x = j+++++1;

That is lexically correct C# code. The lexer is a greedy lexer; it tries to make the largest token it can at any given time. So it tokenizes this as:

x = j ++ ++ + 1 ;

Now the grammar analyzer takes that token stream and says “this is an expression involving two identifiers, one constant and four operators; how should I parenthesize that?” The grammar analyzer determines using the rules of precedence and associativity that this means

x=(((j++)++)+1);

So this means “evaluate x, then increment j, then increment whatever the previous result was, then add 1 to the previous result, then assign the previous result to x, done”

Then the semantic analyzer checks that to make sure that all those operations make sense. They do not, because the result of j++ is a value, not a variable, but the second ++ requires a variable. The semantic analyzer gives an error, and the program does not compile.

If we had a backtracking compiler, the semantic analyzer could tell the parser “no, you did something wrong, give me a different parse tree if there is one”.

Suppose the parser does that. It could say well, maybe that last + was not a binary operator but rather the unary operator + applied to 1, oh, but if we do that then we have two expressions next to each other with no intervening operator. Same thing if the second ++ applies to the +1 instead of the j++. So no, there is no other parse of this. So push back on the lexer and ask it to backtrack.

The lexer could then say, oh, sure, if you don’t like that then maybe my guess that the second ++ was ++ was wrong. Try this one:

x = j ++ + ++ 1 ;

and then the parser says OK, if that’s where the token breaks are then the grammatical analysis is:

x=((j++)+(++1));

and the semantic analyzer determines that’s wrong too, this time because you can’t have a ++ on a constant. So the semantic analyzer tells the parser, no, that’s not it either, and the parser tells the lexer, no, that’s not it either. And the lexer says “well, maybe that wasn’t a ++ at all the second time, maybe that should be

x = j ++ + + + 1 ;

And now the parser says “oh, I see what that is, it is two unary plus operators applied to 1 and an addition operator:

x=((j++)+(+(+1)));

and the semantic analyzer says “finally! something I can work with.”

That’s not what we do. This is a silly example, but perhaps you can see that there are two major problems with this strategy.

First, though most of the time lexing is unambiguous, in some pathological cases there are potentially enormously many ways that simple programs could be lexed. Just determining where the breaks could go between five plus signs gives eight possibilities; it grows exponentially from there. Since most of the time programs lex unambiguously, and when they do lex ambiguously, the space to explore could be really big, it’s better to always lex greedily and not backtrack ever. If a program fails grammatical analysis because of an unintended lexical structure then the best thing to do is tell the user and they’ll fix it.

The second major problem is that when backtracking, how do you know which backtracking is better? Of those eight possibilities, two give program fragments that pass semantic analysis: the one we just found, and x=(j+(+(+(+(+1))))); Is it at all clear which of the two possibilities is the one the user meant when they wrote this ridiculous statement? Remember, we’re not trying to solve a Sudoku puzzle here, where any given combination of numbers is as meaningless as the next; we are attempting to output IL that has the same meaning as the meaning intended by the user! One of these possibilities mutates j and the other does not, so it makes a difference. C# is not a “guess what you meant” language, it’s a “do what you said” language(**). If what you said was ambiguous, we should tell you about it.

Next time on FAIC: we’ll return to this point by looking at how the name resolution algorithm also does not backtrack.

(*) Unlike JavaScript, interestingly enough. The ECMAScript specification notes that there are lexical ambiguities involving the / character; it could be introducing a comment as // or as /*, or closing a comment as */, or could be a normal division operator, or a /= operator, or opening or closing a regular expression. Some of those overlap; The spec actually says that the correct lex is the one that produces a grammatically sensible program!

(**) Again, unlike JavaScript.

Commentary from 2021:

There were a lot of good comments (and some needlessly pedantic comments, imagine that) on the original article that went down a few ratholes mentioned above:

• There were some points made about the difficulty of having lexical ambiguities involving the / symbol in JS that are exacerbated by automatic semicolon insertion
• Some readers pointed out that sudoku puzzles are required by the rules of the puzzle to have a unique solution; though that is true, I was just using it as an example of a problem amenable to a backtracking algorithm; I was not intending to draw a deeper analogy between solving sudoku puzzles and building a compiler. (Fun fact though: it is possible to encode a sudoku puzzle into a C# program such that the compiler is required to find a solution or produce an error if the puzzle is ambiguous or unsolvable.)
• Some readers asked about scenarios that involve a kind of backtracking in the parser, but not “across phases”. For example, consider: (T)-1 — does that parse as a subtraction or as a cast of -1? Or my favourite: A(B<C,D>(E)) which parses as two arguments to A in C# 1, and one in C# 2. In both these situations the parser is required to “look ahead” arbitrarily far in the token stream to determine how the parse works; most of the time the parser only has to look ahead one token to determine what the parse should be. A “look ahead” and a “backtrack” are kinda sorta the same thing. Similarly there are scenarios where it is not clear whether >> is a shift operator or the end of a nested generic type.

In 2016 I worked on rearchitecting the Hack parser; Hack is a gradually-typed variant of PHP used inside Facebook. One of the major difficulties I faced in writing a purely functional, full-fidelity Hack parser is that Hack has XML literals in the language, much like VB. The lexical and grammatical rules of the language are completely different inside these XML regions and I did end up having to build some cross-phase backtracking into the parser, which was quite vexing.

As you know if you’ve read this blog for a while, I’m disturbed by the myth that “value types go on the stack”. Unfortunately, there are plenty of examples in our own documentation and in many books that reinforce this myth, either subtly or overtly. I’m opposed to it because:

1. It is usually stated incorrectly: the statement should be “value types can be stored on the stack”, instead of the more common “value types are always stored on the stack”.
2. It is almost always irrelevant. We’ve worked hard to make a managed environment where the distinctions between different kinds of storage are hidden from the user. Unlike some languages, in which you must know whether a particular storage is on the stack or the heap for correctness reasons.
3. It is incomplete. What about references? References are neither value types nor instances of reference types, but they are values. They’ve got to be stored somewhere. Do they go on the stack or the heap? Why does no one ever talk about them? Just because they don’t have a type in the C# type system is no reason to ignore them.

The way in the past I’ve usually pushed back on this myth is to say that the real statement should be “in the Microsoft implementation of C# on the desktop CLR, value types are stored on the stack when the value is a local variable or temporary that is not a closed-over local variable of a lambda or anonymous method, and the method body is not an iterator block, and the jitter chooses to not enregister the value.”

The sheer number of weasel words in there is astounding, but they’re all necessary:

• Versions of C# provided by other vendors may choose other allocation strategies for their temporary variables; there is no language requirement that a data structure called “the stack” be used to store locals of value type.
• We have many versions of the CLI that run on embedded systems, in web browsers, and so on. Some may run on exotic hardware. I have no idea what the memory allocation strategies of those versions of the CLI are. The hardware might not even have the concept of “the stack” for all I know. Or there could be multiple stacks per thread. Or everything could go on the heap.
• Lambdas and anonymous methods hoist local variables to become heap-allocated fields; those are not on the stack anymore.
• Iterator blocks in today’s implementation of C# on the desktop CLR also hoist locals to become heap-allocated fields. They do not have to! We could have chosen to implement iterator blocks as coroutines running on a fiber with a dedicated stack. In that case, the locals of value type could go on the stack of the fiber.
• People always seem to forget that there is more to memory management than “the stack” and “the heap”. Registers are neither on the stack or the heap, and it is perfectly legal for a value type to go in a register if there is one of the right size. If if is important to know when something goes on the stack, then why isn’t it important to know when it goes in a register? Conversely, if the register scheduling algorithm of the jit compiler is unimportant for most users to understand, then why isn’t the stack allocation strategy also unimportant?

Having made these points many times in the last few years, I’ve realized that the fundamental problem is in the mistaken belief that the type system has anything whatsoever to do with the storage allocation strategy. It is simply false that the choice of whether to use the stack or the heap has anything fundamentally to do with the type of the thing being stored. The truth is: the choice of allocation mechanism has to do only with the known required lifetime of the storage.

Once you look at it that way then everything suddenly starts making much more sense. Let’s break it down into some simple declarative sentences.

• There are three kinds of values: (1) instances of value types, (2) instances of reference types, and (3) references. (Code in C# cannot manipulate instances of reference types directly; it always does so via a reference. In unsafe code, pointer types are treated like value types for the purposes of determining the storage requirements of their values.)
• There exist “storage locations” which can store values.
• Every value manipulated by a program is stored in some storage location.
• Every reference (except the null reference) refers to a storage location.
• Every storage location has a “lifetime”. That is, a period of time in which the storage location’s contents are valid.
• The time between a start of execution of a particular method and the method returning normally or throwing an exception is the “activation period” of that method execution.
• Code in a method can require the use of a storage location. If the required lifetime of the storage location is longer than the activation period of the current method execution then the storage is said to be “long lived”. Otherwise it is “short lived”. (Note that when method M calls method N, the use of the storage locations for the parameters passed to N and the value returned by N is required by M.)

Now we come to implementation details. In the Microsoft implementation of C# on the CLR:

• There are three kinds of storage locations: stack locations, heap locations, and registers.
• Long-lived storage locations are always heap locations.
• Short-lived storage locations are always stack locations or registers.
• There are some situations in which it is difficult for the compiler or runtime to determine whether a particular storage location is short-lived or long-lived. In those cases, the prudent decision is to treat them as long-lived. In particular, the storage locations of instances of reference types are always treated as though they are long-lived, even if they are provably short-lived. Therefore they always go on the heap.

And now things follow very naturally:

• We see that references and instances of value types are essentially the same thing as far as their storage is concerned; they go on either the stack, in registers, or the heap depending on whether the storage of the value needs to be short-lived or long-lived.
• It is frequently the case that array elements, fields of reference types, locals in an iterator block and closed-over locals of a lambda or anonymous method must live longer than the activation period of the method that first required the use of their storage. And even in the rare cases where their lifetimes are shorter than that of the activation of the method, it is difficult or impossible to write a compiler that knows that. Therefore we must be conservative: all of these storage locations go on the heap.
• It is frequently the case that local variables and temporary values can be shown via compile-time analysis to be unused after the activation period ends, and therefore can be treated short-lived, and therefore can go onto the stack or put into registers.

Once you abandon entirely the crazy idea that the type of a value has anything whatsoever to do with the storage, it becomes much easier to reason about it. Of course, my point above stands: you don’t need to reason about it unless you are writing unsafe code or doing some sort of heavy interoperating with unmanaged code. Let the compiler and the runtime manage the lifetime of your storage locations; that’s what its good at.

Today a question about the is and as operators: is the is operator implemented as a syntactic sugar for the as operator, or is the as operator implemented as a syntactic sugar for the is operator? More briefly, is is as or is as is?

Perhaps some sample code would be more clear. This code

bool b = x is Foo;

could be considered as a syntactic sugar for

bool b = (x as Foo) != null;

in which case is is a syntactic sugar for as. Similarly,

Foo f = x as Foo;

could be considered to be a syntactic sugar for

var temp = x;
Foo f = (temp is Foo) ? (Foo)temp : (Foo)null;

in which case as is a syntactic sugar for is. Clearly we cannot have both of these be sugars because then we have an infinite regress!

The specification is clear on this point: as (in the non-dynamic case) is defined as a syntactic sugar for is.

However, in practice the CLR provides us instruction isinst, which ironically acts like as. We have an instruction which implements the semantics of as pretty well, from which we can build an implementation of is. In short, de jure is is is, and as is as is is, but de facto is is as and as is isinst.

I now invite you to leave the obvious jokes about President Clinton in the comments.

Commentary from 2020:

• This article was based on a serious question I had been asked, but of course I relished the opportunity to make silly sentences such as “is is as or is as is?”
• The bit at the end about President Clinton confused some readers, understandably. Read the explainer here for details. The TLDR is: President Clinton famously attempted to explain a lie with “it depends on what the meaning of ‘is’ is”.
• In the original comments a reader asked why anyone should care about this level of implementation detail. My response to that question gets straight to the philosophy of why I’ve written a blog for 17 years:

I reject the premise of the question; I’m not saying that you or anyone else should care about this detail. I do not necessarily write about things that are important for developers to understand; I write about things that are fabulous! And the person who gets to decide what is fabulous is me. If some of those fabulous things happen to be important, that’s super, but that’s not my aim. The name of the blog is not “Important Stuff You Should Know About Coding“.

I blog because I’ve learned a bunch of interesting stuff — interesting to me — and I like to share things that I find interesting.

• • Another good question in the comments was “does the compiler statically analyze is and as expressions?”  Yes! If you write something like if(x is string) and we already know that x is some class type Foo then the expression will always be false and you get a warning.However, there are subtle problems here. The rules of isinst and the rules of C# are slightly different. In particular, it is illegal in C# to convert int[] to uint[] via reference conversion, but legal in the CLR. You can therefore get into situations where the compiler will warn about “impossible” situations that are actually possible.
  • Several readers pointed out that there is a missing language feature: once we know that x is Foo is true, it should be legal to use the members of Foo from x. A form of this feature was added to C# 6; if(x is Foo f)

Today, another entertaining question from StackOverflow. Presented again as a dialogue, as is my wont.

The specification says “even though constants are considered static members, a constant-declaration neither requires nor allows a static modifier.” Why was the decision made to not force constants to use the static modifier if they are considered static?

Let’s grant that it is sensible that constants are considered to be “static” – that is, members associated with the type itself, rather than members of a particular instance. Let’s also grant that “const” has to be in there somewhere. There are three possible choices for how to notate a constant declaration:

1) Make static optional: “const int x…” or “static const int x…” are both legal.
2) Make static required: “const int x…” is illegal, “static const int x…” is legal
3) Make static illegal: “const int x…” is legal, “static const int x…” is illegal.

Does that pretty much sum it up?

Yes. Why did the design team choose option (3) instead of (1) or (2)?

The design notes from 1999 do not say. But we can deduce what was probably going through the language designer’s heads.

The problem with (1) is that you could read code that uses both “const int x…” and “static const int y…” and then you would naturally ask yourself “what’s the difference?” Since the default for non-constant fields and methods is “instance” unless marked “static”, the natural conclusion would be that some constants are per-instance and some are per-type, and that conclusion would be wrong. This is bad because it is misleading.

The problem with (2) is that first off, it is redundant. It’s just more typing without adding clarity or expressiveness to the language. And second, I don’t know about you, but I personally hate it when the compiler gives me the error “You forgot to say the magic word right here. I know you forgot to say the magic word, I am one hundred percent capable of figuring out that the magic word needs to go there, but I’m not going to let you get any work done until you say the magic word”. I once heard that tendency of compilers compared to the guy at the party who interrupts your conversation to point out that you said “a historic occasion” when clearly you meant to say “an historic occasion“. No one likes that guy.

The problem with (3) is that the developer is required to know that const logically implies static. However, once the developer learns this fact, they’ve learned it. It’s not like this is a complex idea that is hard to figure out.

The solution which presents the fewest problems and costs to the end user is (3).

That seems reasonable. Does the C# language apply this principle of eschewing redundancy consistently?

Nope! It is interesting to compare and contrast this with other places in the language where different decisions were made.

For example, overloaded operators are required to be both public and static. In this case, again we are faced with three options:

(1) make public static optional,
(2) make it required, or
(3) make it illegal.

For overloaded operators we chose (2). Since the “natural state” of a method is private/instance it seems bizarre and misleading to make something that looks like a method public/static invisibly, as (1) and (3) both require.

For another example, a virtual method with the same signature as a virtual method in a base class is supposed to have either “new” or “override” on it. Again, three choices.

(1) make it optional: you can say new, or override, or nothing at all, in which case we default to new.
(2) make it required: you have to say new or override, or
(3) make it illegal: you cannot say new at all, so if you don’t say override then it is automatically new.

In this case we chose (1) because that works best for the brittle base class situation of someone adds a virtual method to a base class that you don’t realize you are now overriding. This produces a warning, but not an error.

Each of these situations has to be considered on a case-by-case basis. The general guidance is not , say “always pick option 3”, but rather, to figure out what is least misleading to the typical user and go with that.

When you override a virtual method in C# you are required to ensure that the stated accessibility of the overridden method – that is, whether it is public, internal, protected or protected internal(*) – is exactly re-stated in the overriding method. Except in one case. I refer you to section 10.6.4 of the specification, which states:

an override declaration cannot change the accessibility of the virtual method. However, if the overridden base method is protected internal and it is declared in a different assembly than the assembly containing the override method then the override method’s declared accessibility must be protected.

What the heck is up with that? Surely if an overridden method is protected internal then it only makes sense that the overriding method should be exactly the same: protected internal.

I’ll explain why we have this rule, but first, a brief digression.

A certain island is inhabited by only knights and knaves. Knights make only true statements and only answer questions truthfully; knaves make only false statements and only answer questions untruthfully. If you walk up to an inhabitant of the (aptly-named) Island of Knights and Knaves you can rapidly ascertain whether a particular individual is a knight or a knave by asking a question you know the answer to. For example “does two plus two equal four?” A knight will answer “yes” (**), and a knave will answer “no”. Knaves are prone to saying things like “my mother is a male knight”, which is plainly false.

It might seem at first glance that there is no statement which could be made by both a knight and a knave. Since knights tell the truth and knaves lie, they cannot both make the same statement, right? But in fact there are many statements that can be made by both. Can you think of one?

.

.

.

.

.

.

.

Both a knight and a knave can say “I am a knight.”

How does that work? The reason this works is because the pronoun “I” refers to different people when uttered by different people. If Alice, a knight, makes the statement “I am a knight”, she is asserting the truth that “Alice is a knight”. If Bob, a knave, makes the statement “I am a knight”, he is not asserting the true statement “Alice is a knight” but rather the false statement “Bob is a knight”. Similarly, both Alice and Bob can assert the statement “My name is Alice,” for the same reason.

And that’s why overriding methods in a different assembly aren’t “protected internal”. The modifier “internal” is like a pronoun; it refers to the current assembly. When used in two different assemblies it means different things. The purpose of the rule is to ensure that a derived class does not make the accessibility domain of the virtual member any larger or smaller.

An analogy might help. Suppose a protected resource is a car, an assembly is a dwelling, a person is a class and descendent is a derived class.

• Alice has a Mazda and lives in House with her good friend Charlie.
• Charlie has a child, Diana, who lives in Apartment.
• Alice has a child, Elroy, who lives in Condo with his good friend Greg.
• Elroy has a child – Alice’s grandchild — Frank, who lives in Yurt.

Alice grants access to Mazda to anyone living in House and any descendent of Alice. The people who can access Mazda are Alice, Charlie, Elroy, and Frank.

Diana does not get access to Mazda because she is not Alice, not a descendent of Alice, and not a resident of House. That she is a child of Alice’s housemate is irrelevant.

Greg does not get access to Mazda for the same reason: he is not Alice, not a descendent of Alice, and not a resident of House. That he is a housemate of a descendent of Alice is irrelevant.

Now we come to the crux of the matter. Elroy is not allowed to extend his access to Mazda to Greg. Alice owns that Mazda and she said “myself, my descendents and my housemates”. Her children don’t have the right to extend the accessibility of Mazda beyond what she initially set up. Nor may Elroy deny access to Frank; as a descendent of Alice, Frank has a right to borrow the car and Greg cannot stop him by making it “private”.

When Elroy describes what access he has to Mazda he is only allowed to say “I grant access to this to myself and my descendents” because that is what Alice already allowed. He cannot say “I grant access to Mazda to myself, my descendents and to the other residents of Condo”.

(*) Private virtual methods are illegal in C#, which irks me to no end. I would totally use that feature if we had it.

(**) Assuming that they answer at all, of course; there could be mute knights and knaves.