I was a bit surprised by this racket code printing nay when I expected yeah:
(define five 5)
(case 5
[(five) "yeah"]
[else "nay"])
Looking at the racket documentation for case makes it clearer:
The selected clause is the first one with a datum whose quoted form is equal? to the result of val-expr.
So it's about quotation. I'm pretty sure that I did not yet fully grasp what quotation in lisps can buy me. I understand it in the viewpoint of macros and AST transformation. However I'm confused why is it helpful in the case of case for instance..?
I'm also curious, with this specification of case, can I use it to achieve what I wanted to (compare the actual values, not the quoted value), or should I use another construct for that? (cond, while strictly more powerful, is more verbose for simple cases, since you must repeat the predicate at each condition).
The problem is that case introduces implicit quote forms, which cause your example to work for 'five (whose value is 'five), instead of five (whose value is 5).
I almost never use case because of exactly this problem. Instead I use racket's match form with the == pattern:
(define five 5)
(define (f x)
(match x
[(== five) "yeah"]
[_ "nay"]))
(f 5) ; "yeah"
(f 6) ; "nay"
This produces "yeah" on only the value 5, just like you expected. If you wanted it to return "yeah" when it's equal to either five or six, you can use an or pattern:
(define five 5)
(define six 6)
(define (f x)
(match x
[(or (== five) (== six)) "yeah"]
[_ "nay"]))
(f 5) ; "yeah"
(f 6) ; "yeah"
(f 7) ; "nay"
And if you really want to match against quoted datums, you can do that by writing an explicit quote form.
(define (f x)
(match x
[(or 'five 'six) "yeah"]
[_ "nay"]))
(f 5) ; "nay"
(f 6) ; "nay"
(f 7) ; "nay"
(f 'five) ; "yeah"
(f 'six) ; "yeah"
These quote forms are implicit and invisible when you use case, lurking there waiting to cause confusion.
The Racket documentation gives this grammar:
(case val-expr case-clause ...)
where
case-clause = [(datum ...) then-body ...+]
| [else then-body ...+]
Let's compare to your example:
(define five 5)
(case 5 ; (case val-expr
[(five) "yeah"] ; [(datum) then-body1]
[else "nay"]) ; [else then-body2])
We see that (five) is interpreted as (datum). This means that five is
a piece of data (here a symbol), not an expression (later to be evaluated).
Your example of case is evaluated like this:
First the expression 5 is evaluated. The result is the value 5.
Now we look at a clause at a time. The first clause is [(five) "yeah"].
Is the value 5 equal (in the sense of equal?) to one of the datums in (five)? No, so we look at the next clause: [else "nay"]. It is an else-clause so the expression "nay" is evaluated and the result is the value "nay".
The result of the case-expression is thus the value "nay".
Note 1: The left-hand sides of case-clauses are datums (think: they are implicitly quoted).
Note 2: The result of val-expr is compared to the clause datums using equal?. (This is in contrast to Scheme, which uses eqv?.
UPDATE
Why include case? Let's see how one can write the example using cond:
(define five 5)
(let ([val five])
(cond
[(member val '(five)) "yeah"]
[(member val '(six seven)) "yeah"] ; added
[else "nay"])
This shows that one could do without case and just use cond.
However - which version is easier to read?
For a case expression it is easy to see which datums the value is compared to.
Here one must look closely to find the datums. Also in the example we know beforehand that we are trying to find the value among a few list of datums. In general we need to examine a cond-expression more closely to see that's what's happening.
In short: having a case-expression increases readability of your code.
For the historically interested: https://groups.csail.mit.edu/mac/ftpdir/scheme-mail/HTML/rrrs-1986/msg00080.html disussed whether to use eqv? or equal? for case.
UPDATE 2
I'll attempt to given an answer to:
I'm still not clear on the quotation vs working simply on the values though.
I'm wondering specifically why doing the quotation, why working on datum instead
of working on values. Didn't get that bit yet.
Both approaches make sense.
Let's for the sake of argument look at the case where case uses expressions rather than datums in the left hand side of a clause. Also following the Scheme tradition, let's assume eqv? is used for the comparison. Let's call such a
case-expression for ecase (short for expression-case).
The grammar becomes:
(ecase val-expr ecase-clause ...)
where
ecase-clause = [(expr ...) then-body ...+]
| [else then-body ...+]
Your example now becomes:
(define five 5)
(ecase five
[('five) "yeah"]
[else "nay")
This doesn't look too bad and the result is what we are used to.
However consider this example:
(ecase '(3 4)
[('five (list 3 4) "yeah"]
[else "nay")
The result of this would be "nay". The two lists resulting from evaluating the expressions '(3 4) and (list 3 4) are not equal in the sense of eqv?.
This shows that if one chooses to use eqv? for comparisions, having expressions available on the left hand side won't be helpful. The only values that work with eqv? atomic values - and therefore one could just as well use implicit quotations and restrict the left hand side to datums.
Now if equal? was used it would make much more sense to use expressions on the left hand side. The original Racket version of case was the same as the one in Scheme (i.e. it used eq?) later on it was changed to used equal?. If case was designed from scratch, I think, expressions would be allowed rather than datums.
The only remaining issue: Why did the authors of Scheme choose eqv? over equal? for comparisons? My intuition is that the reason were performance (which back in the day was more important than now). The linked to post from the rrrs-authors mailing list gives two options. If you dig a little further you might be able to find responses.
I can't find a reference right now, but case statements use literal, unevaluated data in their different clauses because it is both a frequent use-case and more easily subject to efficient compilation.
You could probably write your own version of Clojure's condp macro or a custom conditional operator to handle your use case.
Related
I need to complete an assignment for my college course using Scheme. I've never coded in Scheme before so I have no clue what I'm doing. Our assignment is to define an anonymous function that computes the discriminant of a quadratic function. I keep running into the error: "Invalid `define'. Any help would be appreciated.
(define roots (lambda(abc))(
(lambda(discriminant))(
list(/(+(-b) discriminant)(*2a))
(/(-(-b) discriminant)(*2a))
)
(sqrt - (*bb)(*4ac))
)
First, you should learn a bit about what Scheme code looks like; find some example code (in your textbook, or online, or in answers here on SO) and notice how parentheses and whitespace are used. Then emulate that. You can't arbitrarily place parentheses or arbitrarily remove whitespace in Scheme (or in any Lisp).
For example, in the posted code (-b) gets two things wrong. First, -b is treated as one symbol, not as the negation of the value of b. Further, placing the symbol in parentheses indicates a procedure call; given an s-expression (f x), f is either a syntactic keyword (in which case (f x) is interpreted as a macro call), or (f x) is interpreted as a procedure call. If it is a procedure call and f is not bound to a procedure, then an exception is raised. So (-b) attempts to call a procedure named -b, which does not exist (unless you have defined it), raising an exception. You can use (- b), with a space between the - procedure and the symbol b; this evaluates to the negation of the value of b.
Similarly, *2a is interpreted as a symbol, not an expression; placing the *2a between parentheses is interpreted as a procedure call. The interpreter (or compiler) is expecting that *2a is a procedure which takes no arguments. You need to add the spaces: (* 2 a); this is interpreted as a call to the procedure * with the arguments 2 and a.
(*bb) and (*4ac) have exactly the same problems. The second case is interesting because when it is correctly written it illustrates one of the advantages of prefix notation. Since * is associative, it does not matter what order multiple values are multiplied in. To express naively 4 * a * c in prefix notation you could write (* 4 (* a c)), explicitly ordering the multiplications. You could also write this as (* (* 4 a) c), multiplying in a different order. It does not matter what order you multiply in, so you might as well just write (* 4 a c), so long as your language supports this notation. It turns out that Scheme and other Lisps do support this notation.
Another problem with s-expression notation in the posted code (after fixing the problems noted above): (sqrt - (* b b) (* 4 a c)). This is attempting to call the sqrt procedure on the arguments -, (* b b), and (* 4 a c). But sqrt is not a higher-order procedure (i.e., it does not take procedures as arguments), and it in fact only takes one argument. It was meant to apply the - procedure to the arguments (* b b) and (* 4 a c), subtracting them before taking the square root: (sqrt (- (* b b) (* 4 a c))).
The first lambda expression has a formal parameter list containing only one parameter: abc. As before, this is a mistake. The intention was to define three parameters: don't skimp on spaces: (lambda (a b c)).
The other significant problem is that there are syntax errors in the lambda expressions: (lambda (a b c)) has no body, but a lambda expression must have at least one expression in its body. This was probably intended to wrap the lambda expression which follows. Similarly, the inner lambda expression is missing its body. It was probably intended to wrap the (list ;;...) form that follows.
With that done, the inner lambda expression is itself inside of a pair of parentheses, taking the expression (sqrt (- (* b b) (* 4 a c))) as its argument. This is the lambda form of a let binding. Thus, the inner lambda takes one argument, discriminant, and evaluates the list form that is its body. Since the inner lambda expression itself occurs in the first position of an s-expression, it is part of a procedure call, and this inner anonymous procedure is then called on its argument, binding discriminant to the value obtained by evaluating that argument, which is (sqrt (- (* b b) (* 4 a c))). This all occurs inside of the outer lambda, which takes the three arguments a, b, and c. So, root is a function taking three arguments, and returning a list of roots, after binding the result of the discriminant calculation to discriminant (as a way of both simplifying the expression of the roots and ensuring that the discriminant need only be calculated one time).
Here is the fixed-up code. Note that I only added some spaces and added or moved a few parentheses; nothing else was changed:
(define roots
(lambda (a b c)
((lambda (discriminant)
(list (/ (+ (- b) discriminant) (* 2 a))
(/ (- (- b) discriminant) (* 2 a))))
(sqrt (- (* b b) (* 4 a c))))))
Pay attention to what this looks like. In Lisps you should almost never leave parentheses hanging on lines by themselves, and you should always place a space between arguments. Remember that everything is a procedure call.
Here is a sample interaction. Notice that you can represent negative numbers as -1 instead of (- 1) (you can do either if you wish). You just can't express a negative value using a variable as -b.
> (roots 1 0 -1)
(1 -1)
> (roots 1 8 15)
(-3 -5)
I'm having a hard time understanding the syntax of let vs some of the other statements. For example, a "normal" statement has one parentheses:
(+ 2 2)
$2 = 4
Yet the let statement has two:
(let ((x 2)) (+ x 2))
$3 = 4
Why is this so? I find it quite confusing to remember how many parentheses to put around various items.
Firstly, note that let syntax contains two parts, both of which can have zero or more elements. It binds zero or more variables, and evaluates zero or more forms.
All such Lisp forms create a problem: if the elements are represented as a flat list, there is an ambiguity: we don't know where one list ends and the other begins!
(let <var0> <var1> ... <form0> <form1> ...)
For instance, suppose we had this:
(let (a 1) (b 2) (print a) (list b))
What is (print a): is that the variable print being bound to a? Or is it form0 to be evaluated?
Therefore, Lisp constructs like this are almost always designed in such a way that one of the two lists is a single object, or possibly both. In other words: one of these possibilities:
(let <var0> <var1> ... (<form0> <form1> ...))
(let (<var0> <var1> ...) (<form0> <form1> ...))
(let (<var0> <var1> ...) <form0> <form1> ...)
Traditional Lisp has followed the third idea above in the design of let. That idea has the benefit that the pieces of the form are easily and efficiently accessed in an interpreter, compiler or any code that processes code. Given an object L representing let syntax, the variables are easily retrieved as (cadr L) and the body forms as (cddr L).
Now, within this design choice, there is still a bit of design freedom. The variables could follow a structure similar to a property list:
(let (a 1 b 2 c 3) ...)
or they could be enclosed:
(let ((a 1) (b 2) (c 3)) ...)
The second form is traditional. In the Arc dialect of Lisp designed Paul Graham, the former syntax appears.
The traditional form has more parentheses. However, it allows the initialization forms to be omitted: So that is to say if the initial value of a variable is desired to be nil, instead of writing (a nil), you can just write a:
;; These two are equivalent:
(let ((a nil) (b nil) (c)) ...)
(let (a b c) ...)
This is a useful shorthand in the context of a traditional Lisp which uses the symbol nil for the Boolean false and for the empty list. We have compactly defined three variables that are either empty lists or false Booleans by default.
Basically, we can regard the traditional let as being primarily designed to bind a simple list of variables as in (let (a b c) ...) which default to nil. Then, this syntax is extended to support initial values, by optionally replacing a variable var with a (var init) pair, where init is an expression evaluated to specify its initial value.
In any case, thanks to macros, you can have any binding syntax you want. In more than one program I have seen a let1 macro which binds just one variable, and has no parentheses. It is used like this:
(let1 x 2 (+ x 2)) -> 4
In Common Lisp, we can define let1 very easily like this:
(defmacro let1 (var init &rest body)
`(let ((,var ,init)) ,#body))
If we restrict let1 to have a one-form body, we can then write the expression with obsessively few parentheses;
(let1 x 2 + x 2) -> 4
That one is:
(defmacro let1 (var init &rest form)
`(let ((,var ,init)) (,#form)))
Remember that let allows you to bind multiple variables. Each variable binding is of the form (variable value), and you collect all the bindings into a list. So the general form looks like
(let ((var1 value1)
(var2 value2)
(var3 value3)
...)
body)
That's why there are two parentheses around x 2 -- the inner parentheses are for that specific binding, the outer parentheses are for the list of all bindings. It's only confusing because you're only binding one variable, it becomes clearer with multiple variables.
try to figure out how to use "append" in Scheme
the concept of append that I can find like this:
----- part 1: understanding the concept of append in Scheme-----
1) append takes two or more lists and constructs a new list with all of their elements.
2) append requires that its arguments are lists, and makes a list whose elements are the elements of those lists. it concatenates the lists it is given. (It effectively conses the elements of the other lists onto the last list to create the result list.)
3) It only concatenates the top-level structure ==> [Q1] what does it mean "only concatenates the top-level"?
4) however--it doesn't "flatten" nested structures.
==> [Q2] what is "flatten" ? (I saw many places this "flatten" but I didn't figure out yet)
==> [Q3] why append does not "flatten" nested structures.
---------- Part 2: how to using append in Scheme --------------------------------
then I looked around to try to use "append" and I saw other discussion
based on the other discussion, I try this implementation
[code 1]
(define (tst-foldr-append lst)
(foldr
(lambda (element acc) (append acc (list element)))
lst
'())
)
it works, but I am struggling to understand that this part ...(append acc (list element)...
what exactly "append" is doing in code 1, to me, it just flipping.
then why it can't be used other logics e.g.
i) simply just flip or
iii).... cons (acc element).....
[Q4] why it have to be "append" in code 1??? Is that because of something to do with foldr ??
again, sorry for the long question, but I think it is all related.
Q1/2/3: What is this "flattening" thing?
Scheme/Lisp/Racket make it very very easy to use lists. Lists are easy to construct and easy to operate on. As a result, they are often nested. So, for instance
`(a b 34)
denotes a list of three elements: two symbols and a number. However,
`(a (b c) 34)
denotes a list of three elements: a symbol, a list, and a number.
The word "flatten" is used to refer to the operation that turns
`(3 ((b) c) (d (e f)))
into
`(3 b c d e f)
That is, the lists-within-lists are "flattened".
The 'append' function does not flatten lists; it just combines them. So, for instance,
(append `(3 (b c) d) `(a (9)))
would produce
`(3 (b c) d a (9))
Another way of saying it is this: if you apply 'append' to a list of length 3 and a list of length 2, the result will be of length 5.
Q4/5: Foldl really has nothing to do with append. I think I would ask a separate question about foldl if I were you.
Final advice: go check out htdp.org .
Q1: It means that sublists are not recursively appended, only the top-most elements are concatenated, for example:
(append '((1) (2)) '((3) (4)))
=> '((1) (2) (3) (4))
Q2: Related to the previous question, flattening a list gets rid of the sublists:
(flatten '((1) (2) (3) (4)))
=> '(1 2 3 4)
Q3: By design, because append only concatenates two lists, for flattening nested structures use flatten.
Q4: Please read the documentation before asking this kind of questions. append is simply a different procedure, not necessarily related to foldr, but they can be used together; it concatenates a list with an element (if the "element" is a list the result will be a proper list). cons just sticks together two things, no matter their type whereas append always returns a list (proper or improper) as output. For example, for appending one element at the end you can do this:
(append '(1 2) '(3))
=> '(1 2 3)
But these expressions will give different results (tested in Racket):
(append '(1 2) 3)
=> '(1 2 . 3)
(cons '(1 2) '(3))
=> '((1 2) 3)
(cons '(1 2) 3)
=> '((1 2) . 3)
Q5: No, cons will work fine here. You wouldn't be asking any of this if you simply tested each procedure to see how they work. Please understand what you're using by reading the documentation and writing little examples, it's the only way you'll ever learn how to program.
'and' in Scheme will ignore the error 'division by 0', like (and (negative? (random 100)) (/ 1 0)) returns #f.
How does it do that?
i (define (∧ b₀ b₁) (if (not b₀) #f b₁)), (∧ (negative? (random 100)) (/ 1 0)) still goes into a 'division by 0' error.
You can't define and directly as a function because Scheme is strict--this means that function arguments are always evaluated before being passed to the function.
However, you can define a proper short-circuiting and using a macro. Here's the simplest version in Racket:
(define-syntax-rule (my-and a b) (if a b #f))
or the equivalent form using syntax-rules which is in standard Scheme:
(define-syntax my-and
(syntax-rules ()
[(_ a b) (if a b #f)]))
A macro is not a normal function. Instead, it's a syntactic transformation that runs at compile-time. When you use your new and in your code, it gets "expanded" to the corresponding if expression. So:
(my-and #f (/ 1 0))
gets transformed into
(if #f (/ 1 0) #f)
before your program is run. Since if is built into the language, this has the correct behavior.
Since macros are not functions, it also means you can't pass around as arguments. So you can't write a fold using and directly--you'd have to wrap it into a lambda.
To be more faithful to the original and, you could define my-and to take an arbitrary number of arguments by making the macro recursive. To define a "rest parameter" in a macro, you use the special ... keyword:
(define-syntax my-and
(syntax-rules ()
[(_) #t]
[(_ a) a]
[(_ a b ...) (if a (my-and b ...) #f)]))
If you were using a lazy language like Racket's #lang lazy or Haskell instead, you would not need to use macros here. You could just define and directly:
#lang lazy
(define (and a b) (if a b #f))
or in Haskell:
and a b = if a then b else False
and it would have the correct behavior, as a normal function. You would be able to pass this and to a fold, and it would even stop evaluating the list as soon as it encountered a False! Take a look:
Prelude> foldl and True [True, False, error "fail"]
False
(error in Haskell errors out just like 1/0. Since Haskell is statically typed, the arguments to and have to be booleans so you can't use 1/0 directly.)
Like most languages, Scheme's logical AND uses short circuit evaluation, which means its right operand will only be evaluated if the left operand is true. If the left operand is false, then the result of the expression must be false regardless of the value of the right operand, so if the left operand evaluates to false, it returns false immediately, without evaluating the right operand at all.
To be precise, here's the language from the spec (I'm section 4.2.1 of R5RS, but this isn't an area that's likely to change must between revisions of the spec):
(and <test1> ... )
The <test> expressions are evaluated from left to right, and the value of the first expression that evaluates to a false value (see section 6.3.1) is returned. Any remaining expressions are not evaluated.
Boolean shortcutting. The first argument to "and" evaluates to false, therefore the result must be false, and there's no reason for it to evaluate the second argument (and therefore incur the error).
In Structure and Interpretation of Computer Programs (SICP) Section 2.2.3 several functions are defined using:
(accumulate cons nil
(filter pred
(map op sequence)))
Two examples that make use of this operate on a list of the fibonacci numbers, even-fibs and list-fib-squares.
The accumulate, filter and map functions are defined in section 2.2 as well. The part that's confusing me is why the authors included the accumulate here. accumulate takes 3 parameters:
A binary function to be applied
An initial value, used as the rightmost parameter to the function
A list to which the function will be applied
An example of applying accumulate to a list using the definition in the book:
(accumulate cons nil (list 1 2 3))
=> (cons 1 (cons 2 (cons 3 nil)))
=> (1 2 3)
Since the third parameter is a list, (accumulate cons nil some-list) will just return some-list, and in this case the result of (filter pred (map op sequence)) is a list.
Is there a reason for this use of accumulate other than consistency with other similarly structured functions in the section?
I'm certain that those two uses of accumulate are merely illustrative of the fact that "consing elements to construct a list" can be treated as an accumulative process in the same way that "multiplying numbers to obtain a product" or "summing numbers to obtain a total" can. You're correct that the accumulation is effectively a no-op.
(Aside: Note that this could obviously be a more useful operation if the output of filter and input of accumulate was not a list; for example, if it represented a lazily generated sequence.)