The endless droning

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Someone asked about better Lisp IDEs on reddit. Such things would obviously be desirable. But the comments are entirely full the usual sad endless droning from people who need there always to be something preventing them from doing what they pretend to want to do, and are happy to invent such barriers where none really exist. comp.lang.lisp lives on in spirit if not in fact.

The proper use of macros in Lisp

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People learning Lisp often try to learn how to write macros by taking an existing function they have written and turning it into a macro. This is a mistake: macros and functions serve different purposes and it is almost never useful to turn functions into macros, or macros into functions.

Let’s say you are learning Common Lisp1, and you have written a fairly obvious factorial function based on the natural mathematical definition: if $$n \in \mathbb{N}$$, then

$n! = \begin{cases} 1 &n \le 1\\ n \times (n - 1)! &n > 1 \end{cases}$

So this gives you a fairly obvious recursive definition of factorial:

(defun factorial (n)
(if (<= n 1)
1
(* n (factorial (1- n )))))

And so, you think you want to learn about macros so can you write factorial as a macro? And you might end up with something like this:

(defmacro factorial (n)
(if (<= ,n 1)
1
(* ,n (factorial ,(1- n )))))

And this superficially seems as if it works:

> (factorial 10)
3628800

But it doesn’t, in fact, work:

> (let ((x 3))
(factorial x))

Error: In 1- of (x) arguments should be of type number.

Why doesn’t this work and can it be fixed so it does? If it can’t what has gone wrong and how are macros meant to work and what are they useful for?

It can’t be fixed so that it works. trying to rewrite functions as macros is a bad idea, and if you want to learn what is interesting about macros you should not start there.

To understand why this is true you need to understand what macros actually are in Lisp.

What macros are: a first look

A macro is a function whose domain and range is syntax.

Macros are functions (quite explicitly so in CL: you can get at the function of a macro with macro-function, and this is something you can happily call the way you would call any other function), but they are functions whose domain and range is syntax. A macro is a function whose argument is a language whose syntax includes the macro and whose value, when called on an instance of that language, is a language whose syntax doesn’t include the macro. It may work recursively: its value may be a language which includes the same macro but in some simpler way, such that the process will terminate at some point.

So the job of macros is to provide a family of extended languages built on some core Lisp which has no remaining macros, only functions and function application, special operators & special forms involving them and literals. One of those languages is the language we call Common Lisp, but the macros written by people serve to extend this language into a multitude of variants.

As an example of this I often write in a language which is like CL, but is extended by the presence of a number of extra constructs, one of which is called ITERATE (but it predates the well-known one and is not at all the same):

(iterate next ((x 1))
(if (< x 10)
(next (1+ x))
x)

is equivalent to

(labels ((next (x)
(if (< x 10)
(next (1+ x))
x)))
(next 1))

Once upon a time when I first wrote iterate, it used to manually optimize the recursive calls to jumps in some cases, because the Symbolics I wrote it on didn’t have tail-call elimination. That’s a non-problem in LispWorks2. Anyone familiar with Scheme will recognise iterate as named let, which is where it came from (once, I think, it was known as nlet).

iterate is implemented by a function which maps from the language which includes it to a language which doesn’t include it, by mapping the syntax as above.

So compare this with a factorial function: factorial is a function whose domain is natural numbers and whose range is also natural numbers, and it has an obvious recursive definition. Well, natural numbers are part of the syntax of Lisp, but they’re a tiny part of it. So implementing factorial as a macro is, really, a hopeless task. What should

(factorial (+ x y (f z)))

Actually do when considered as a mapping between languages? Assuming you are using the recursive definition of the factorial function then the answer is it can’t map to anything useful at all: a function which implements that recursive definition simply has to be called at run time. The very best you could do would seem to be this:

(defun fact (n)
(if (< n 3)
n
(* n (fact (1- n)))))

(defmacro factorial (expression)
(fact ,expression))

And that’s not a useful macro (but see below).

So the answer is, again, that macros are functions which map between languages and they are useful where you want a new language: not just the same language with extra functions in it, but a language with new control constructs or something like that. If you are writing functions whose range is something which is not the syntax of a language built on Common Lisp, don’t write macros.

What macros are: a second look

Macroexpansion is compilation.

A function whose domain is one language and whose range is another is a compiler for the language of the domain, especially when that language is somehow richer than the language of the range, which is the case for macros.

But it’s a simplification to say that macros are this function: they’re not, they’re only part of it. The actual function which maps between the two languages is made up of macros and the macroexpander provided by CL itself. The macroexpander is what arranges for the functions defined by macros to be called in the right places, and also it is the thing which arranges for various recursive macros to actually make up a recurscive function. So it’s important to understand that the macroexpander is a critical part of the process: macros on their own only provide part of it.

An example: two versions of a recursive macro

People often say that you should not write recursive macros, but this prohibition on recursive macros is pretty specious: they’re just fine. Consider a language which only has lambda and doesn’t have let. Well, we can write a simple version of let, which I’ll call bind as a macro: a function which takes this new language and turns it into the more basic one. Here’s that macro:

(defmacro bind ((&rest bindings) &body forms)
((lambda ,(mapcar #'first bindings) ,@forms)
,@(mapcar #'second bindings)))

And now

> (bind ((x 1) (y 2))
(+ x y))
(bind ((x 1) (y 2)) (+ x y))
-> ((lambda (x y) (+ x y)) 1 2)
3

(These example expansions come via use of my trace-macroexpand package, available in a good Lisp near you: see appendix for configuration).

So now we have a language with a binding form which is more convenient than lambda. But maybe we want to be able to bind sequentially? Well, we can write a let* version, called bind*, which looks like this

(defmacro bind* ((&rest bindings) &body forms)
(if (null (rest bindings))
(bind ,bindings ,@forms)
(bind (,(first bindings))
(bind* ,(rest bindings) ,@forms))))

And you can see how this works: it checks if there’s just one binding in which case it’s just bind, and if there’s more than one it peels off the first and then expands into a bind* form for the rest. And you can see this working (here both bind and bind* are being traced):

> (bind* ((x 1) (y (+ x 2)))
(+ x y))
(bind* ((x 1) (y (+ x 2))) (+ x y))
-> (bind ((x 1)) (bind* ((y (+ x 2))) (+ x y)))
(bind ((x 1)) (bind* ((y (+ x 2))) (+ x y)))
-> ((lambda (x) (bind* ((y (+ x 2))) (+ x y))) 1)
(bind* ((y (+ x 2))) (+ x y))
-> (bind ((y (+ x 2))) (+ x y))
(bind ((y (+ x 2))) (+ x y))
-> ((lambda (y) (+ x y)) (+ x 2))
(bind* ((y (+ x 2))) (+ x y))
-> (bind ((y (+ x 2))) (+ x y))
(bind ((y (+ x 2))) (+ x y))
-> ((lambda (y) (+ x y)) (+ x 2))
4

You can see that, in this implementation, which is LW again, some of the forms are expanded more than once: that’s not uncommon in interpreted code: since macros should generally be functions (so, not have side-effects) it does not matter that they may be expanded multiple times. Compilation will expand macros and then compile the result, so all the overhead of macroexpansion happend ahead of run-time:

 (defun foo (x)
(bind* ((y (1+ x)) (z (1+ y)))
(+ y z)))
foo

> (compile *)
(bind* ((y (1+ x)) (z (1+ y))) (+ y z))
-> (bind ((y (1+ x))) (bind* ((z (1+ y))) (+ y z)))
(bind ((y (1+ x))) (bind* ((z (1+ y))) (+ y z)))
-> ((lambda (y) (bind* ((z (1+ y))) (+ y z))) (1+ x))
(bind* ((z (1+ y))) (+ y z))
-> (bind ((z (1+ y))) (+ y z))
(bind ((z (1+ y))) (+ y z))
-> ((lambda (z) (+ y z)) (1+ y))
foo
nil
nil

> (foo 3)
9

There’s nothing wrong with macros like this, which expand into simpler versions of themselves. You just have to make sure that the recursive expansion process is producing successively simpler bits of syntax and has a well-defined termination condition.

Macros like this are often called ‘recursive’ but they’re actually not: the function associated with bind* does not call itself. What is recursive is the function implicitly defined by the combination of the macro function and the macroexpander: the bind* function simply expands into a bit of syntax which it knows will cause the macroexpander to call it again.

It is possible to write bind* such that the macro function itself is recursive:

(defmacro bind* ((&rest bindings) &body forms)
(labels ((expand-bind (btail)
(if (null (rest btail))
(bind ,btail
,@forms)
(bind (,(first btail))
,(expand-bind (rest btail))))))
(expand-bind bindings)))

And now compiling foo again results in this output from tracing macroexpansion:

(bind* ((y (1+ x)) (z (1+ y))) (+ y z))
-> (bind ((y (1+ x))) (bind ((z (1+ y))) (+ y z)))
(bind ((y (1+ x))) (bind ((z (1+ y))) (+ y z)))
-> ((lambda (y) (bind ((z (1+ y))) (+ y z))) (1+ x))
(bind ((z (1+ y))) (+ y z))
-> ((lambda (z) (+ y z)) (1+ y))

You can see that now all the recursion happens within the macro function for bind* itself: the macroexpander calls bind*’s macro function just once.

While it’s possible to write macros like this second version of bind*, it is normally easier to write the first version and to allow the combination of the macroexpander and the macro function to implement the recursive expansion.

Two historical uses for macros

There are two uses for macros — both now historical — where they were used where functions would be more natural.

The first of these is function inlining, where you want to avoid the overhead of calling a small function many times. This overhead was a lot on computers made of cardboard, as all computers were, and also if the stack got too deep the cardboard would tear and this was bad. It makes no real sense to inline a recursive function such as the above factorial: how would the inlining process terminate? But you could rewrite a factorial function to be explicitly iterative:

(defun factorial (n)
(do* ((k 1 (1+ k))
(f k (* f k)))
((>= k n) f)))

And now, if you have very many calls to factorial, you wanted to optimise the function call overhead away, and it was 1975, you might write this:

(defmacro factorial (n)
(let ((nv ,n))
(do* ((k 1 (1+ k))
(f k (* f k)))
((>= k nv) f))))

And this has the effect of replacing (factorial n) by an expression which will compute the factorial of n. The cost of that is that (funcall #'factorial n) is not going to work, and (funcall (macro-function 'factorial) ...) is never what you want.

Well, that’s what you did in 1975, because Lisp compilers were made out of the things people found down the sides of sofas. Now it’s no longer 1975 and you just tell the compiler that you want it to inline the function, please:

(declaim (inline factorial))
(defun factorial (n) ...)

and it will do that for you. So this use of macros is now purely historicl.

The second reason for macros where you really want functions is computing things at compile time. Let’s say you have lots of expressions like (factorial 32) in your code. Well, you could do this:

(defmacro factorial (expression)
(typecase expression
((integer 0)
(factorial/fn expression))
(number
(error "factorial of non-natural literal ~S" expression))
(t
(factorial/fn ,expression))))

So the factorial macro checks to see if its argument is a literal natural number and will compute the factorial of it at macroexpansion time (so, at compile time or just before compile time). So a function like

(defun foo ()
(factorial 32))

will now compile to simply return 263130836933693530167218012160000000. And, even better, there’s some compile-time error checking: code which is, say, (factorial 12.3) will cause a compile-time error.

Well, again, this is what you would do if it was 1975. It’s not 1975 any more, and CL has a special tool for dealing with just this problem: compiler macros.

(defun factorial (n)
(do* ((k 1 (1+ k))
(f k (* f k)))
((>= k n) f)))

(define-compiler-macro factorial (&whole form n)
(typecase n
((integer 0)
(factorial n))
(number
(error "literal number is not a natural: ~S" n))
(t form)))

Now factorial is a function and works the way you expect — (funcall #'factoial ...) will work fine. But the compiler knows that if it comes across (factorial ...) then it should give the compiler macro for factorial a chance to say what this expression should actually be. And the compiler macro does an explicit check for the argument being a literal natural number, and if it is computes the factorial at compile time, and the same check for a literal number which is not a natural, and finally just says ’I don’t know, call the function’. Note that the compiler macro itself calls factorial, but since the argument isn’t a literal there’s no recursive doom.

So this takes care of the other antique use of macros where you would expect functions. And of course you can combine this with inlining and it will all work fine: you can write functions which will handle special cases via compiler macros and will otherwise be inlined.

That leaves macros serving the purpose they are actually useful for: building languages.

Appendix: setting up trace-macroexpand

(use-package :org.tfeb.hax.trace-macroexpand)

;;; Don't restrict print length or level when tracing
(setf *trace-macroexpand-print-level* nil
*trace-macroexpand-print-length* nil)

;;; Enable tracing
(trace-macroexpand)

;;; Trace the macros you want to look at ...
(trace-macro ...)

;;; ... and ntrace them
(untrace-macro ...)

1. All the examples in this article are in Common Lisp except where otherwise specified. Other Lisps have similar considerations, although macros in Scheme are not explicitly functions in the way they are in CL.

2. This article originated as a message on the lisp-hug` mailing list for LispWorks users. References to ‘LW’ mean LispWorks, although everything here should apply to any modern CL. (In terms of tail call elimination I would define a CL which does not eliminate tail self-calls in almost all cases under reasonable optimization settings as pre-modern: I don’t use such implementations.)

A letter to my MP

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On the occasion of the johnsonites’ rewriting the rules on political corruption to suit themselves.

The best Lisp

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People sometimes ask which is the best Lisp dialect? That’s a category error, and here’s why.

Programming in Lisp — any Lisp — is about building languages: in Lisp the way you solve a problem is by building a language — a jargon, or a dialect if you like — to talk about the problem and then solving the problem in that language. Lisps are, quite explicitly, language-building languages.

This is, in fact, how people solve large problems in all programming languages: Greenspun’s tenth rule isn’t really a statement about Common Lisp, it’s a statement that all sufficiently large software systems end up having some hacked-together, informally-specified, half-working language in which the problem is actually solved. Often people won’t understand that the thing they’ve built is in fact a language, but that’s what it is. Everyone who has worked on large-scale software will have come across these things: often they are very horrible, and involve much use of language-in-a-string1.

The Lisp difference is two things: when you start solving a problem in Lisp, you know, quite explicitly, that this is what you are going to do; and the language has wonderful tools which let you incrementally build a series of lightweight languages, ending up with one or more languages in which to solve the problem.

So, after that preface, why is this question the wrong one to ask? Well, if you are going to program in Lisp you are going to be building languages, and you want those languages not to be awful. Lisp makes it it far easier to build languages which are not awful, but it doesn’t prevent you doing so if you want to. And again, anyone who has dealt with enough languages built on Lisps will have come across some which are, in fact, awful.

If you are going to build languages then you need to understand how languages work — what makes a language habitable to its human users (the computer does not care with very few exceptions). That means you will need to be a linguist. So the question then is: how do you become a linguist? Well, we know the answer to that, because there are lots of linguists and lots of courses on linguistics. You might say that, well, those people study natural languages, but that’s irrelevant: natural languages have been under evolutionary pressure for a very long time and they’re really good for what they’re designed for (which is not the same as what programming languages are designed for, but the users — humans — are the same).

So, do you become a linguist by learning French? Or German? Or Latin? Or Cuzco Quechua? No, you don’t. You become a linguist by learning enough about enough languages that you can understand how languages work. A linguist isn’t someone who speaks French really well: they’re someone who understands that French is a Romance language, that German isn’t but has many Romance loan words, that English is closer to German than it is French but got a vast injection of Norman French, which in turn wasn’t that close to modern French, that Swiss German has cross-serial dependencies but Hochdeutsch does not and what that means, and so on. A linguist is someone who understands things about the structure of languages: what do you see, what do you never see, how do different languages do equivalent things? And so on.

The way you become a linguist is not by picking a language and learning it: it’s by looking at lots of languages enough to understand how they work.

If you want to learn to program in Lisp, you will need to become a linguist. The very best way to ensure you fail at that is to pick a ‘best’ Lisp and learn that. There is no best Lisp, and in order to program well in any Lisp you must be exposed to as many Lisps and as many other languages as possible.

If you think there’s a distinction between a ‘dialect’, a ‘jargon’ and a ‘language’ then I have news for you: there is. A language is a dialect with a standards committee. (This is stolen from a quote due to Max Weinrich that all linguists know:

אַ שפּראַך איז אַ דיאַלעקט מיט אַן אַרמיי און פֿלאָט

a shprakh iz a dyalekt mit an armey un flot.)

1. ‘Language-in-a-string’ is where a programming language has another programming language embedded in strings in the outer language. Sometimes programs in that inner programming language will be made up by string concatenation in the outer language. Sometimes that inner language will, in turn, have languages embedded in its strings. It’s a terrible, terrible thing.

An age of optimism

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On the occasion of Rishi Sunak’s budget.

Computer insecurity

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Making computer systems secure is very difficult. The consequences of insecure systems are already extremely serious and will be catastrophic in future if they are not already. Malignant people, often sponsored by malignant states, are actively attacking computer systems and have had considerable success doing so.

So it is surprising that companies whose stated aims are to increase security are effectively working to make their customers’ systems less secure.

Neurodivergent

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Recently I wrote two articles about Richard Stallman (RMS) and the Free Software Foundation (FSF). Someone who is autistic wrote to me and pointed out some unfortunate implications of what I wrote, which were both wrong and offensive to neurodivergent people: I am sorry for that. The remainder of this article is an attempt to correct those mistakes.

The lost cause of the Free Software Foundation

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The Free Software Foundation has reelected Richard Stallman to its board. At first glance this looks like a wilful act of self-harm by the FSF: RMS has expressed opinions which are abhorrent and has behaved appallingly towards women, at least. This is to misunderstand both what the cause of the FSF really is and what their options for that cause now are.