Related
I know that you can use ' (aka quote) to create a list, and I use this all the time, like this:
> (car '(1 2 3))
1
But it doesn’t always work like I’d expect. For example, I tried to create a list of functions, like this, but it didn’t work:
> (define math-fns '(+ - * /))
> (map (lambda (fn) (fn 1)) math-fns)
application: not a procedure;
expected a procedure that can be applied to arguments
given: '+
When I use list, it works:
> (define math-fns (list + - * /))
> (map (lambda (fn) (fn 1)) math-fns)
'(1 -1 1 1)
Why? I thought ' was just a convenient shorthand, so why is the behavior different?
TL;DR: They are different; use list when in doubt.
A rule of thumb: use list whenever you want the arguments to be evaluated; quote “distributes” over its arguments, so '(+ 1 2) is like (list '+ '1 '2). You’ll end up with a symbol in your list, not a function.
An in-depth look at list and quote
In Scheme and Racket, quote and list are entirely different things, but since both of them can be used to produce lists, confusion is common and understandable. There is an incredibly important difference between them: list is a plain old function, while quote (even without the special ' syntax) is a special form. That is, list can be implemented in plain Scheme, but quote cannot be.
The list function
The list function is actually by far the simpler of the two, so let’s start there. It is a function that takes any number of arguments, and it collects the arguments into a list.
> (list 1 2 3)
(1 2 3)
This above example can be confusing because the result is printed as a quoteable s-expression, and it’s true, in this case, the two syntaxes are equivalent. But if we get slightly more complicated, you’ll see that it is different:
> (list 1 (+ 1 1) (+ 1 1 1))
(1 2 3)
> '(1 (+ 1 1) (+ 1 1 1))
(1 (+ 1 1) (+ 1 1 1))
What’s going on in the quote example? Well, we’ll discuss that in a moment, but first, take a look at list. It’s just an ordinary function, so it follows standard Scheme evaluation semantics: it evaluates each of its arguments before they get passed to the function. This means that expressions like (+ 1 1) will be reduced to 2 before they get collected into the list.
This behavior is also visible when supplying variables to the list function:
> (define x 42)
> (list x)
(42)
> '(x)
(x)
With list, the x gets evaluated before getting passed to list. With quote, things are more complicated.
Finally, because list is just a function, it can be used just like any other function, including in higher-order ways. For example, it can be passed to the map function, and it will work appropriately:
> (map list '(1 2 3) '(4 5 6))
((1 4) (2 5) (3 6))
The quote form
Quotation, unlike list, is a special part of Lisps. The quote form is special in part because it gets a special reader abbreviation, ', but it’s also special even without that. Unlike list, quote is not a function, and therefore it does not need to behave like one—it has rules of its own.
A brief discussion of Lisp source code
In Lisp, of which Scheme and Racket are derivatives, all code is actually made up of ordinary data structures. For example, consider the following expression:
(+ 1 2)
That expression is actually a list, and it has three elements:
the + symbol
the number 1
the number 2
All of these values are normal values that can be created by the programmer. It’s really easy to create the 1 value because it evaluates to itself: you just type 1. But symbols and lists are harder: by default, a symbol in the source code does a variable lookup! That is, symbols are not self-evaluating:
> 1
1
> a
a: undefined
cannot reference undefined identifier
As it turns out, though, symbols are basically just strings, and in fact we can convert between them:
> (string->symbol "a")
a
Lists do even more than symbols, because by default, a list in the source code calls a function! Doing (+ 1 2) looks at the first element in the list, the + symbol, looks up the function associated with it, and invokes it with the rest of the elements in the list.
Sometimes, though, you might want to disable this “special” behavior. You might want to just get the list or get the symbol without it being evaluated. To do this, you can use quote.
The meaning of quotation
With all this in mind, it’s pretty obvious what quote does: it just “turns off” the special evaluation behavior for the expression that it wraps. For example, consider quoteing a symbol:
> (quote a)
a
Similarly, consider quoteing a list:
> (quote (a b c))
(a b c)
No matter what you give quote, it will always, always spit it back out at you. No more, no less. That means if you give it a list, none of the subexpressions will be evaluated—do not expect them to be! If you need evaluation of any kind, use list.
Now, one might ask: what happens if you quote something other than a symbol or a list? Well, the answer is... nothing! You just get it back.
> (quote 1)
1
> (quote "abcd")
"abcd"
This makes sense, since quote still just spits out exactly what you give it. This is why “literals” like numbers and strings are sometimes called “self-quoting” in Lisp parlance.
One more thing: what happens if you quote an expression containing quote? That is, what if you “double quote”?
> (quote (quote 3))
'3
What happened there? Well, remember that ' is actually just a direct abbreviation for quote, so nothing special happened at all! In fact, if your Scheme has a way to disable the abbreviations when printing, it will look like this:
> (quote (quote 3))
(quote 3)
Don’t be fooled by quote being special: just like (quote (+ 1)), the result here is just a plain old list. In fact, we can get the first element out of the list: can you guess what it will be?
> (car (quote (quote 3)))
quote
If you guessed 3, you are wrong. Remember, quote disables all evaluation, and an expression containing a quote symbol is still just a plain list. Play with this in the REPL until you are comfortable with it.
> (quote (quote (quote 3)))
''3
(quote (1 2 (quote 3)))
(1 2 '3)
Quotation is incredibly simple, but it can come off as very complex because of how it tends to defy our understanding of the traditional evaluation model. In fact, it is confusing because of how simple it is: there are no special cases, there are no rules. It just returns exactly what you give it, precisely as stated (hence the name “quotation”).
Appendix A: Quasiquotation
So if quotation completely disables evaluation, what is it good for? Well, aside from making lists of strings, symbols, or numbers that are all known ahead of time, not much. Fortunately, the concept of quasiquotation provides a way to break out of the quotation and go back into ordinary evaluation.
The basics are super simple: instead of using quote, use quasiquote. Normally, this works exactly like quote in every way:
> (quasiquote 3)
3
> (quasiquote x)
x
> (quasiquote ((a b) (c d)))
((a b) (c d))
What makes quasiquote special is that is recognizes a special symbol, unquote. Wherever unquote appears in the list, then it is replaced by the arbitrary expression it contains:
> (quasiquote (1 2 (+ 1 2)))
(1 2 (+ 1 2))
> (quasiquote (1 2 (unquote (+ 1 2))))
(1 2 3)
This lets you use quasiquote to construct templates of sorts that have “holes” to be filled in with unquote. This means it’s possible to actually include the values of variables inside of quoted lists:
> (define x 42)
> (quasiquote (x is: (unquote x)))
(x is: 42)
Of course, using quasiquote and unquote is rather verbose, so they have abbreviations of their own, just like '. Specifically, quasiquote is ` (backtick) and unquote is , (comma). With those abbreviations, the above example is much more palatable.
> `(x is: ,x)
(x is: 42)
One final point: quasiquote actually can be implemented in Racket using a rather hairy macro, and it is. It expands to usages of list, cons, and of course, quote.
Appendix B: Implementing list and quote in Scheme
Implementing list is super simple because of how “rest argument” syntax works. This is all you need:
(define (list . args)
args)
That’s it!
In contrast, quote is a lot harder—in fact, it’s impossible! It would seem totally feasible, since the idea of disabling evaluation sounds a lot like macros. Yet a naïve attempt reveals the trouble:
(define fake-quote
(syntax-rules ()
((_ arg) arg)))
We just take arg and spit it back out... but this doesn’t work. Why not? Well, the result of our macro will be evaluated, so all is for naught. We might be able to expand to something sort of like quote by expanding to (list ...) and recursively quoting the elements, like this:
(define impostor-quote
(syntax-rules ()
((_ (a . b)) (cons (impostor-quote a) (impostor-quote b)))
((_ (e ...)) (list (impostor-quote e) ...))
((_ x) x)))
Unfortunately, though, without procedural macros, we can’t handle symbols without quote. We could get closer using syntax-case, but even then, we would only be emulating quote’s behavior, not replicating it.
Appendix C: Racket printing conventions
When trying the examples in this answer in Racket, you may find that they do not print as one would expect. Often, they may print with a leading ', such as in this example:
> (list 1 2 3)
'(1 2 3)
This is because Racket, by default, prints results as expressions when possible. That is, you should be able to type the result into the REPL and get the same value back. I personally find this behavior nice, but it can be confusing when trying to understand quotation, so if you want to turn it off, call (print-as-expression #f), or change the printing style to “write” in the DrRacket language menu.
The behavior you are seeing is a consequence of Scheme not treating symbols as functions.
The expression '(+ - * /) produces a value which is a list of symbols. That's simply because (+ - * /) is a list of symbols, and we are just quoting it to suppress evaluation in order to get that object literally as a value.
The expression (list + - * /) produces a list of functions. This is because it is a function call. The symbolic expressions list, +, -, * and / are evaluated. They are all variables which denote functions, and so are reduced to those functions. The list function is then called, and returns a list of those remaining four functions.
In ANSI Common Lisp, calling symbols as functions works:
[1]> (mapcar (lambda (f) (funcall f 1)) '(+ - * /))
(1 -1 1 1)
When a symbol is used where a function is expected, the top-level function binding of the symbol is substituted, if it has one, and everything is cool. In effect, symbols are function-callable objects in Common Lisp.
If you want to use list to produce a list of symbols, just like '(+ - * /), you have to quote them individually to suppress their evaluation:
(list '+ '- '* '/)
Back in the Scheme world, you will see that if you map over this, it will fail in the same way as the original quoted list. The reason is the same: trying to use a symbol objects as a functions.
The error message you are being shown is misleading:
expected a procedure that can be applied to arguments
given: '+
This '+ being shown here is (quote +). But that's not what the application was given; it was given just +, the issue being that the symbol object + isn't usable as a function in that dialect.
What's going on here is that the diagnostic message is printing the + symbol in "print as expression" mode, a feature of Racket, which is what I guess you're using.
In "print as expression" mode, objects are printed using a syntax which must be read and evaluated to produce a similar object. See the StackOverflow question "Why does the Racket interpreter write lists with an apostroph before?"
i am currently learning common lisp and stumbled upon a question i could not
answer my self:
(defun find-all (item seq &rest keyword-args &key (test #'eql)
test-not &allow-other-keys)
(if test-not
(apply #'remove item seq
:test-not (complement test-not) keyword-args)
(apply #'remove item seq
:test (complement test) keyword-args)))
The function is used to find every element in seq matching item according
to a test function.
Sadly I dont understand why the function 'apply' was used here.
Shouldn't it be possible to just invoke remove without apply?
A warnings says: "The function has an odd number of arguments in the keyword portion" if I invoke remove without apply.
I hope you can help me,
Thanks in advance!
REMOVE and FIND-ALL
REMOVE takes a bunch of keyword arguments: from-end, test, test-not, start, end, count and key.
Now in the function FIND-ALL you want to modify just one: either test or test-not and then call REMOVE.
How to write the parameter list of FIND-ALL
Now you have basically three options to write the parameter list of FIND-ALL, since it is basically the same as REMOVE with only one argument changed.
list every keyword argument with their defaults and later pass them on with the necessary changed to REMOVE.
list only a rest arguments list, manipulate this argument list and pass the new one via APPLY to REMOVE.
mix of 1. and 2. like in the example above. List only the required arguments and the keyword arguments to modify plus a rest list of the other keyword arguments provided at call time. Call REMOVE via APPLY.
How good are the three possibilities?
Now 1. has the advantage that you get to see a full parameter list for FIND-ALL and that it does not need to cons an argument list. Lisp can check some of that. But you really need to copy all arguments in your parameter list and later in the calls to REMOVE. Possible but not that great.
Then 2. has the disadvantage of not having a visible parameter list for FIND-ALL, but it might be easier to write with a few functions to manipulate argument lists. 3. is relatively easy to write, but also lacks the full parameter list.
Your example
Thus in your example it is version 3 of above:
the required arguments are passed in first
next are the modified arguments
last is the list of all keyword arguments
If you want to pass an existing list of arguments as part of the arguments to the function, then you need APPLY. That's why it is used there. APPLY calls a function with the arguments taken from a list.
CL-USER 1 > (apply #'+ 1 2 '(3 4 5))
15
CL-USER 2 > (let ((numbers '(3 4 5)))
(apply #'+ 1 2 numbers))
15
CL-USER 3 > (+ 1 2 3 4 5)
15
Let's take for example the signature of REMOVE:
remove item sequence &key from-end test test-not start end count key
=> result-sequence
The above means the function accepts 2 mandatory arguments, item and sequence, and additional keywords arguments that should be given with the :key value syntax. If you only give the key or the value, like:
(remove x list :count)
Then it is invalid. A simple test is to look at the number of parameters in keyword position and check that the number of given arguments is even. When it is odd, you know something is wrong.
If you call:
(remove item seq args)
You are in the same case. It does not matter that args is a list in your specific case.
Apply
APPLY is a more generic way of calling a function: its last argument is a list of additional arguments.
Suppose args is bound to (3 4), then:
(apply #'+ 1 args)
is equivalent to:
(+ 1 3 4)
This works with keyword arguments too; if args is the list (:count 1), then:
(apply #'remove item seq args)
is equivalent to:
(remove item seq :count 1)
And here, the number of arguments in keyword position is even.
I know that you can use ' (aka quote) to create a list, and I use this all the time, like this:
> (car '(1 2 3))
1
But it doesn’t always work like I’d expect. For example, I tried to create a list of functions, like this, but it didn’t work:
> (define math-fns '(+ - * /))
> (map (lambda (fn) (fn 1)) math-fns)
application: not a procedure;
expected a procedure that can be applied to arguments
given: '+
When I use list, it works:
> (define math-fns (list + - * /))
> (map (lambda (fn) (fn 1)) math-fns)
'(1 -1 1 1)
Why? I thought ' was just a convenient shorthand, so why is the behavior different?
TL;DR: They are different; use list when in doubt.
A rule of thumb: use list whenever you want the arguments to be evaluated; quote “distributes” over its arguments, so '(+ 1 2) is like (list '+ '1 '2). You’ll end up with a symbol in your list, not a function.
An in-depth look at list and quote
In Scheme and Racket, quote and list are entirely different things, but since both of them can be used to produce lists, confusion is common and understandable. There is an incredibly important difference between them: list is a plain old function, while quote (even without the special ' syntax) is a special form. That is, list can be implemented in plain Scheme, but quote cannot be.
The list function
The list function is actually by far the simpler of the two, so let’s start there. It is a function that takes any number of arguments, and it collects the arguments into a list.
> (list 1 2 3)
(1 2 3)
This above example can be confusing because the result is printed as a quoteable s-expression, and it’s true, in this case, the two syntaxes are equivalent. But if we get slightly more complicated, you’ll see that it is different:
> (list 1 (+ 1 1) (+ 1 1 1))
(1 2 3)
> '(1 (+ 1 1) (+ 1 1 1))
(1 (+ 1 1) (+ 1 1 1))
What’s going on in the quote example? Well, we’ll discuss that in a moment, but first, take a look at list. It’s just an ordinary function, so it follows standard Scheme evaluation semantics: it evaluates each of its arguments before they get passed to the function. This means that expressions like (+ 1 1) will be reduced to 2 before they get collected into the list.
This behavior is also visible when supplying variables to the list function:
> (define x 42)
> (list x)
(42)
> '(x)
(x)
With list, the x gets evaluated before getting passed to list. With quote, things are more complicated.
Finally, because list is just a function, it can be used just like any other function, including in higher-order ways. For example, it can be passed to the map function, and it will work appropriately:
> (map list '(1 2 3) '(4 5 6))
((1 4) (2 5) (3 6))
The quote form
Quotation, unlike list, is a special part of Lisps. The quote form is special in part because it gets a special reader abbreviation, ', but it’s also special even without that. Unlike list, quote is not a function, and therefore it does not need to behave like one—it has rules of its own.
A brief discussion of Lisp source code
In Lisp, of which Scheme and Racket are derivatives, all code is actually made up of ordinary data structures. For example, consider the following expression:
(+ 1 2)
That expression is actually a list, and it has three elements:
the + symbol
the number 1
the number 2
All of these values are normal values that can be created by the programmer. It’s really easy to create the 1 value because it evaluates to itself: you just type 1. But symbols and lists are harder: by default, a symbol in the source code does a variable lookup! That is, symbols are not self-evaluating:
> 1
1
> a
a: undefined
cannot reference undefined identifier
As it turns out, though, symbols are basically just strings, and in fact we can convert between them:
> (string->symbol "a")
a
Lists do even more than symbols, because by default, a list in the source code calls a function! Doing (+ 1 2) looks at the first element in the list, the + symbol, looks up the function associated with it, and invokes it with the rest of the elements in the list.
Sometimes, though, you might want to disable this “special” behavior. You might want to just get the list or get the symbol without it being evaluated. To do this, you can use quote.
The meaning of quotation
With all this in mind, it’s pretty obvious what quote does: it just “turns off” the special evaluation behavior for the expression that it wraps. For example, consider quoteing a symbol:
> (quote a)
a
Similarly, consider quoteing a list:
> (quote (a b c))
(a b c)
No matter what you give quote, it will always, always spit it back out at you. No more, no less. That means if you give it a list, none of the subexpressions will be evaluated—do not expect them to be! If you need evaluation of any kind, use list.
Now, one might ask: what happens if you quote something other than a symbol or a list? Well, the answer is... nothing! You just get it back.
> (quote 1)
1
> (quote "abcd")
"abcd"
This makes sense, since quote still just spits out exactly what you give it. This is why “literals” like numbers and strings are sometimes called “self-quoting” in Lisp parlance.
One more thing: what happens if you quote an expression containing quote? That is, what if you “double quote”?
> (quote (quote 3))
'3
What happened there? Well, remember that ' is actually just a direct abbreviation for quote, so nothing special happened at all! In fact, if your Scheme has a way to disable the abbreviations when printing, it will look like this:
> (quote (quote 3))
(quote 3)
Don’t be fooled by quote being special: just like (quote (+ 1)), the result here is just a plain old list. In fact, we can get the first element out of the list: can you guess what it will be?
> (car (quote (quote 3)))
quote
If you guessed 3, you are wrong. Remember, quote disables all evaluation, and an expression containing a quote symbol is still just a plain list. Play with this in the REPL until you are comfortable with it.
> (quote (quote (quote 3)))
''3
(quote (1 2 (quote 3)))
(1 2 '3)
Quotation is incredibly simple, but it can come off as very complex because of how it tends to defy our understanding of the traditional evaluation model. In fact, it is confusing because of how simple it is: there are no special cases, there are no rules. It just returns exactly what you give it, precisely as stated (hence the name “quotation”).
Appendix A: Quasiquotation
So if quotation completely disables evaluation, what is it good for? Well, aside from making lists of strings, symbols, or numbers that are all known ahead of time, not much. Fortunately, the concept of quasiquotation provides a way to break out of the quotation and go back into ordinary evaluation.
The basics are super simple: instead of using quote, use quasiquote. Normally, this works exactly like quote in every way:
> (quasiquote 3)
3
> (quasiquote x)
x
> (quasiquote ((a b) (c d)))
((a b) (c d))
What makes quasiquote special is that is recognizes a special symbol, unquote. Wherever unquote appears in the list, then it is replaced by the arbitrary expression it contains:
> (quasiquote (1 2 (+ 1 2)))
(1 2 (+ 1 2))
> (quasiquote (1 2 (unquote (+ 1 2))))
(1 2 3)
This lets you use quasiquote to construct templates of sorts that have “holes” to be filled in with unquote. This means it’s possible to actually include the values of variables inside of quoted lists:
> (define x 42)
> (quasiquote (x is: (unquote x)))
(x is: 42)
Of course, using quasiquote and unquote is rather verbose, so they have abbreviations of their own, just like '. Specifically, quasiquote is ` (backtick) and unquote is , (comma). With those abbreviations, the above example is much more palatable.
> `(x is: ,x)
(x is: 42)
One final point: quasiquote actually can be implemented in Racket using a rather hairy macro, and it is. It expands to usages of list, cons, and of course, quote.
Appendix B: Implementing list and quote in Scheme
Implementing list is super simple because of how “rest argument” syntax works. This is all you need:
(define (list . args)
args)
That’s it!
In contrast, quote is a lot harder—in fact, it’s impossible! It would seem totally feasible, since the idea of disabling evaluation sounds a lot like macros. Yet a naïve attempt reveals the trouble:
(define fake-quote
(syntax-rules ()
((_ arg) arg)))
We just take arg and spit it back out... but this doesn’t work. Why not? Well, the result of our macro will be evaluated, so all is for naught. We might be able to expand to something sort of like quote by expanding to (list ...) and recursively quoting the elements, like this:
(define impostor-quote
(syntax-rules ()
((_ (a . b)) (cons (impostor-quote a) (impostor-quote b)))
((_ (e ...)) (list (impostor-quote e) ...))
((_ x) x)))
Unfortunately, though, without procedural macros, we can’t handle symbols without quote. We could get closer using syntax-case, but even then, we would only be emulating quote’s behavior, not replicating it.
Appendix C: Racket printing conventions
When trying the examples in this answer in Racket, you may find that they do not print as one would expect. Often, they may print with a leading ', such as in this example:
> (list 1 2 3)
'(1 2 3)
This is because Racket, by default, prints results as expressions when possible. That is, you should be able to type the result into the REPL and get the same value back. I personally find this behavior nice, but it can be confusing when trying to understand quotation, so if you want to turn it off, call (print-as-expression #f), or change the printing style to “write” in the DrRacket language menu.
The behavior you are seeing is a consequence of Scheme not treating symbols as functions.
The expression '(+ - * /) produces a value which is a list of symbols. That's simply because (+ - * /) is a list of symbols, and we are just quoting it to suppress evaluation in order to get that object literally as a value.
The expression (list + - * /) produces a list of functions. This is because it is a function call. The symbolic expressions list, +, -, * and / are evaluated. They are all variables which denote functions, and so are reduced to those functions. The list function is then called, and returns a list of those remaining four functions.
In ANSI Common Lisp, calling symbols as functions works:
[1]> (mapcar (lambda (f) (funcall f 1)) '(+ - * /))
(1 -1 1 1)
When a symbol is used where a function is expected, the top-level function binding of the symbol is substituted, if it has one, and everything is cool. In effect, symbols are function-callable objects in Common Lisp.
If you want to use list to produce a list of symbols, just like '(+ - * /), you have to quote them individually to suppress their evaluation:
(list '+ '- '* '/)
Back in the Scheme world, you will see that if you map over this, it will fail in the same way as the original quoted list. The reason is the same: trying to use a symbol objects as a functions.
The error message you are being shown is misleading:
expected a procedure that can be applied to arguments
given: '+
This '+ being shown here is (quote +). But that's not what the application was given; it was given just +, the issue being that the symbol object + isn't usable as a function in that dialect.
What's going on here is that the diagnostic message is printing the + symbol in "print as expression" mode, a feature of Racket, which is what I guess you're using.
In "print as expression" mode, objects are printed using a syntax which must be read and evaluated to produce a similar object. See the StackOverflow question "Why does the Racket interpreter write lists with an apostroph before?"
I know that you can use ' (aka quote) to create a list, and I use this all the time, like this:
> (car '(1 2 3))
1
But it doesn’t always work like I’d expect. For example, I tried to create a list of functions, like this, but it didn’t work:
> (define math-fns '(+ - * /))
> (map (lambda (fn) (fn 1)) math-fns)
application: not a procedure;
expected a procedure that can be applied to arguments
given: '+
When I use list, it works:
> (define math-fns (list + - * /))
> (map (lambda (fn) (fn 1)) math-fns)
'(1 -1 1 1)
Why? I thought ' was just a convenient shorthand, so why is the behavior different?
TL;DR: They are different; use list when in doubt.
A rule of thumb: use list whenever you want the arguments to be evaluated; quote “distributes” over its arguments, so '(+ 1 2) is like (list '+ '1 '2). You’ll end up with a symbol in your list, not a function.
An in-depth look at list and quote
In Scheme and Racket, quote and list are entirely different things, but since both of them can be used to produce lists, confusion is common and understandable. There is an incredibly important difference between them: list is a plain old function, while quote (even without the special ' syntax) is a special form. That is, list can be implemented in plain Scheme, but quote cannot be.
The list function
The list function is actually by far the simpler of the two, so let’s start there. It is a function that takes any number of arguments, and it collects the arguments into a list.
> (list 1 2 3)
(1 2 3)
This above example can be confusing because the result is printed as a quoteable s-expression, and it’s true, in this case, the two syntaxes are equivalent. But if we get slightly more complicated, you’ll see that it is different:
> (list 1 (+ 1 1) (+ 1 1 1))
(1 2 3)
> '(1 (+ 1 1) (+ 1 1 1))
(1 (+ 1 1) (+ 1 1 1))
What’s going on in the quote example? Well, we’ll discuss that in a moment, but first, take a look at list. It’s just an ordinary function, so it follows standard Scheme evaluation semantics: it evaluates each of its arguments before they get passed to the function. This means that expressions like (+ 1 1) will be reduced to 2 before they get collected into the list.
This behavior is also visible when supplying variables to the list function:
> (define x 42)
> (list x)
(42)
> '(x)
(x)
With list, the x gets evaluated before getting passed to list. With quote, things are more complicated.
Finally, because list is just a function, it can be used just like any other function, including in higher-order ways. For example, it can be passed to the map function, and it will work appropriately:
> (map list '(1 2 3) '(4 5 6))
((1 4) (2 5) (3 6))
The quote form
Quotation, unlike list, is a special part of Lisps. The quote form is special in part because it gets a special reader abbreviation, ', but it’s also special even without that. Unlike list, quote is not a function, and therefore it does not need to behave like one—it has rules of its own.
A brief discussion of Lisp source code
In Lisp, of which Scheme and Racket are derivatives, all code is actually made up of ordinary data structures. For example, consider the following expression:
(+ 1 2)
That expression is actually a list, and it has three elements:
the + symbol
the number 1
the number 2
All of these values are normal values that can be created by the programmer. It’s really easy to create the 1 value because it evaluates to itself: you just type 1. But symbols and lists are harder: by default, a symbol in the source code does a variable lookup! That is, symbols are not self-evaluating:
> 1
1
> a
a: undefined
cannot reference undefined identifier
As it turns out, though, symbols are basically just strings, and in fact we can convert between them:
> (string->symbol "a")
a
Lists do even more than symbols, because by default, a list in the source code calls a function! Doing (+ 1 2) looks at the first element in the list, the + symbol, looks up the function associated with it, and invokes it with the rest of the elements in the list.
Sometimes, though, you might want to disable this “special” behavior. You might want to just get the list or get the symbol without it being evaluated. To do this, you can use quote.
The meaning of quotation
With all this in mind, it’s pretty obvious what quote does: it just “turns off” the special evaluation behavior for the expression that it wraps. For example, consider quoteing a symbol:
> (quote a)
a
Similarly, consider quoteing a list:
> (quote (a b c))
(a b c)
No matter what you give quote, it will always, always spit it back out at you. No more, no less. That means if you give it a list, none of the subexpressions will be evaluated—do not expect them to be! If you need evaluation of any kind, use list.
Now, one might ask: what happens if you quote something other than a symbol or a list? Well, the answer is... nothing! You just get it back.
> (quote 1)
1
> (quote "abcd")
"abcd"
This makes sense, since quote still just spits out exactly what you give it. This is why “literals” like numbers and strings are sometimes called “self-quoting” in Lisp parlance.
One more thing: what happens if you quote an expression containing quote? That is, what if you “double quote”?
> (quote (quote 3))
'3
What happened there? Well, remember that ' is actually just a direct abbreviation for quote, so nothing special happened at all! In fact, if your Scheme has a way to disable the abbreviations when printing, it will look like this:
> (quote (quote 3))
(quote 3)
Don’t be fooled by quote being special: just like (quote (+ 1)), the result here is just a plain old list. In fact, we can get the first element out of the list: can you guess what it will be?
> (car (quote (quote 3)))
quote
If you guessed 3, you are wrong. Remember, quote disables all evaluation, and an expression containing a quote symbol is still just a plain list. Play with this in the REPL until you are comfortable with it.
> (quote (quote (quote 3)))
''3
(quote (1 2 (quote 3)))
(1 2 '3)
Quotation is incredibly simple, but it can come off as very complex because of how it tends to defy our understanding of the traditional evaluation model. In fact, it is confusing because of how simple it is: there are no special cases, there are no rules. It just returns exactly what you give it, precisely as stated (hence the name “quotation”).
Appendix A: Quasiquotation
So if quotation completely disables evaluation, what is it good for? Well, aside from making lists of strings, symbols, or numbers that are all known ahead of time, not much. Fortunately, the concept of quasiquotation provides a way to break out of the quotation and go back into ordinary evaluation.
The basics are super simple: instead of using quote, use quasiquote. Normally, this works exactly like quote in every way:
> (quasiquote 3)
3
> (quasiquote x)
x
> (quasiquote ((a b) (c d)))
((a b) (c d))
What makes quasiquote special is that is recognizes a special symbol, unquote. Wherever unquote appears in the list, then it is replaced by the arbitrary expression it contains:
> (quasiquote (1 2 (+ 1 2)))
(1 2 (+ 1 2))
> (quasiquote (1 2 (unquote (+ 1 2))))
(1 2 3)
This lets you use quasiquote to construct templates of sorts that have “holes” to be filled in with unquote. This means it’s possible to actually include the values of variables inside of quoted lists:
> (define x 42)
> (quasiquote (x is: (unquote x)))
(x is: 42)
Of course, using quasiquote and unquote is rather verbose, so they have abbreviations of their own, just like '. Specifically, quasiquote is ` (backtick) and unquote is , (comma). With those abbreviations, the above example is much more palatable.
> `(x is: ,x)
(x is: 42)
One final point: quasiquote actually can be implemented in Racket using a rather hairy macro, and it is. It expands to usages of list, cons, and of course, quote.
Appendix B: Implementing list and quote in Scheme
Implementing list is super simple because of how “rest argument” syntax works. This is all you need:
(define (list . args)
args)
That’s it!
In contrast, quote is a lot harder—in fact, it’s impossible! It would seem totally feasible, since the idea of disabling evaluation sounds a lot like macros. Yet a naïve attempt reveals the trouble:
(define fake-quote
(syntax-rules ()
((_ arg) arg)))
We just take arg and spit it back out... but this doesn’t work. Why not? Well, the result of our macro will be evaluated, so all is for naught. We might be able to expand to something sort of like quote by expanding to (list ...) and recursively quoting the elements, like this:
(define impostor-quote
(syntax-rules ()
((_ (a . b)) (cons (impostor-quote a) (impostor-quote b)))
((_ (e ...)) (list (impostor-quote e) ...))
((_ x) x)))
Unfortunately, though, without procedural macros, we can’t handle symbols without quote. We could get closer using syntax-case, but even then, we would only be emulating quote’s behavior, not replicating it.
Appendix C: Racket printing conventions
When trying the examples in this answer in Racket, you may find that they do not print as one would expect. Often, they may print with a leading ', such as in this example:
> (list 1 2 3)
'(1 2 3)
This is because Racket, by default, prints results as expressions when possible. That is, you should be able to type the result into the REPL and get the same value back. I personally find this behavior nice, but it can be confusing when trying to understand quotation, so if you want to turn it off, call (print-as-expression #f), or change the printing style to “write” in the DrRacket language menu.
The behavior you are seeing is a consequence of Scheme not treating symbols as functions.
The expression '(+ - * /) produces a value which is a list of symbols. That's simply because (+ - * /) is a list of symbols, and we are just quoting it to suppress evaluation in order to get that object literally as a value.
The expression (list + - * /) produces a list of functions. This is because it is a function call. The symbolic expressions list, +, -, * and / are evaluated. They are all variables which denote functions, and so are reduced to those functions. The list function is then called, and returns a list of those remaining four functions.
In ANSI Common Lisp, calling symbols as functions works:
[1]> (mapcar (lambda (f) (funcall f 1)) '(+ - * /))
(1 -1 1 1)
When a symbol is used where a function is expected, the top-level function binding of the symbol is substituted, if it has one, and everything is cool. In effect, symbols are function-callable objects in Common Lisp.
If you want to use list to produce a list of symbols, just like '(+ - * /), you have to quote them individually to suppress their evaluation:
(list '+ '- '* '/)
Back in the Scheme world, you will see that if you map over this, it will fail in the same way as the original quoted list. The reason is the same: trying to use a symbol objects as a functions.
The error message you are being shown is misleading:
expected a procedure that can be applied to arguments
given: '+
This '+ being shown here is (quote +). But that's not what the application was given; it was given just +, the issue being that the symbol object + isn't usable as a function in that dialect.
What's going on here is that the diagnostic message is printing the + symbol in "print as expression" mode, a feature of Racket, which is what I guess you're using.
In "print as expression" mode, objects are printed using a syntax which must be read and evaluated to produce a similar object. See the StackOverflow question "Why does the Racket interpreter write lists with an apostroph before?"
Is it possible to dereference a variable contained inside a list, to obtain its value? For example:
(define one 1)
(define two 2)
(define list '(one two))
(display (list-ref list 0))
Here list-ref references to one, and display shows one in letters. Could instead one dereference to the value contained by the homonym variable?
Eval can definitely solve this problem... but like most situations where eval is applicable, it's a large and dangerous hammer.
Matthew Flatt's blog post on this topic has become the go-to explanation:
http://blog.racket-lang.org/2011/10/on-eval-in-dynamic-languages-generally.html
Here's how you might do it without eval in Racket. Stripping away the cruft, "dict-ref" can find a named
element (or elements) in an "association list".
#lang racket
(define data
'((one 1)
(two 2)))
(define wanted-list '(two one))
;; evaluates to '((2) (1)):
(for/list ([wanted wanted-list])
(dict-ref data wanted))
That's happening because the quoted list contains two symbols, one and two. Try the following:
(display (eval (list-ref list 0)))
To resolve that symbol.