Skip to content

Creating Quoted Lists and Pairs


As we have seen earlier it is not possible to directly write a list object because Scheme will try to apply the first element as a procedure to the remaining elements:

guile> (1 2 3)
standard input:98:1: In expression (1 2 3):
standard input:98:1: Wrong type to apply: 1
ABORT: (misc-error)

Scheme tries to read 1 as a procedure and apply it to 2 and 3, which obviously fails. The “wrong type” in the error message refers to 1 being a number and not a procedure. The regular syntax to create that list is to use the list procedure in the first place:

guile> (list 1 2 3)
(1 2 3)

Numbers are self-evaluating, but if we pass symbols to list they are evaluated before being added to the list:

guile> (list red green blue)
((1.0 0.0 0.0) (0.0 1.0 0.0) (0.0 0.0 1.0))

But what if we want to store the symbols as names, that is we want the result to evaluate to (red green blue)? Of course we can quote the symbols, using either quote or the shorthand notation:

guile> (list (quote red) (quote green) (quote blue))
(red green blue)

guile> (list 'red 'green 'blue)
(red green blue)

This may become quite unwieldy in practice, therefore Scheme provides a shortcut and allows the quoting of the expression as a whole:

guile> (quote (red green blue))
(red green blue)

Please note that the three elements are surrounded by nested parens: quote quotes one single object, and that is the whole list here. It is also possible to use the shorthand notation, and this is what you will likely see and use most:

guile> '(red green blue)
(red green blue)

Or in LilyPond syntax:

myList = #'(red green blue)


The same is true for pairs. You can't enter them literally but you have to either use the cons procedure or quote them through quote or ':

guile> (1 . 2)
standard input:82:1: In expression (1 . 2):
standard input:82:1: Wrong number of arguments to 1
ABORT: (wrong-number-of-args)

guile> (cons 1 2)
(1 . 2)

guile> (quote (1 . 2))
(1 . 2)

guile> '(1 . 2)
(1 . 2)

(Please don't ask about the error message in the first example. Obviously this expression is so fishy that Scheme isn't even able to produce a meaningful error ...)

And as with lists and the list constructor elements that are not quoted will be evaluated:

guile> (cons red random)
((1.0 0.0 0.0) . #<primitive-procedure random>)

guile> (cons red blue)
((1.0 0.0 0.0) 0.0 0.0 1.0)


But wait a minute, this last one does not look like a pair, isn't it?

Well, this is one of these moments where Scheme's syntactical structures can drive you crazy when you haven't really dug into the core. Although it isn't the topic of this chapter it seems appropriate to take the opportunity of a practical recap of what we have seen in the chapter about the internal structure of lists.

We can properly retrieve the car and cdr of this expression:

guile> (car (cons red blue))
(1.0 0.0 0.0)
guile> (cdr (cons red blue))
(0.0 0.0 1.0)

But should that expression not evaluate to something that looks like

((1.0 0.0 0.0) . (0.0 0.0 1.0))


OK, let's dissect it: we have a pair where both the car and the cdr are lists. And earlier we have seen a definition of just that: a “pair whose cdr is a list”. This definition describes - a list. So our pair is just a special case: when the cdr of a pair happens to be a list the whole expression becomes a list as well. Sounds somewhat strange but obviously doesn't do any harm (if you don't take countless hours of scratching newbies' heads into account ...).

We can verify that assumption quite easily. Usually a list is also a pair but a pair is not a list:

guile> (pair? '(red blue))

guile> (list? '(red . blue))

But here we can see that our pair is at the same time a list:

guile> (list? (cons red random))
guile> (list? (cons red blue))

But there's one more thing to it. If we explicitly create a list from red and blue it takes yet another form:

guile> (list red blue)
((1.0 0.0 0.0) (0.0 0.0 1.0))

What is that? Well, a proper list is a list whose last element is a pair with an empty list as its cdr. Let's see what the cdr of our last (second) list element is:

guile> (cdr (list red blue))
((0.0 0.0 1.0))

So why are we getting two nested paren levels? It gets more and more confusing ... Well, blue is a list, so the cdr of our initial expression should be a list, isn't it? And as we have seen earlier the last element in a list is not the element itself but a pair with the element and an empty list as its elements. So we can check the car and the cdr of that last expression:

guile> (car (cdr (list red blue)))
(0.0 0.0 1.0)

So this is the “blue” list we'd expect.

guile> (cdr (cdr (list red blue)))

And this is the empty list that makes it a “proper list”.

So if we try to wrap that up we can say: (list red blue) creates a proper list with two elements that are both proper lists as well. (cons red blue) on the other hand creates a list whose first element is a list (“red”) while “blue” is represented as three individual elements.

Being confronted with this kind of stuff can be pretty confusing, not only for the new user. But everything can be stripped down to some very basic fundamental concepts, so don't be frightened but try to dissect things one by one - and ask on the mailing lists for clarifications, and don't hesitate to keep asking until you have fully understood the case.

Last update: November 3, 2022