is it possible to give procedure definition as an argument and then somehow run it in the program? For example if I call (program 'write-hello '((procedure write-hello ('Hello.)))) . How can I read the procedure from argument? I spent few hours on it and can't really find any solution to it as I'm new to Scheme.
Thanks
Functions are first class citizens in Scheme: they are values that can be passed just like the more traditional data types, and the parameter name can be used as if there were a function with that name defined in the traditional way.
If you have defined a function in the normal way (using define), then you can just pass the name. But you can also make a function object using lambda, which can be used like any other value (bound to names, passed as an argument) in addition to its function-like abilities (i.e. applied to arguments).
The following are equivalent:
(define (plus a b) (+ a b))
(define plus_a (lambda (a b) (+ a b)))
And if you have the following (notice how op is being used):
(define (do_op op a b) (op a b))
Then these would also be equivalent to each other:
(do_op + 5 6)
(do_op plus 5 6)
(do_op plus_a 5 6)
(do_op (lambda (a b) (+ a b)) 5 6)
For Scheme you just pass the lambda definition like (lambda (n) (* n n)) eg.
(define (display-result procedure value)
(display (procedure value))
(display-result (lambda (n) (* n n)) 5) ; displays 25
If a procedure is bound to a name, like + or defined with define you just use that name in place of the lambda expression.
If you are after making an interpreter of some language then you need to implement this feature yourself by managing your own environment. It's not simple and if you're a newbie it will take more than hours to complete. I recomment watching the SICP videos.
Related
I was trying to understand the concept of dynamic/static scope with deep and shallow binding. Below is the code-
(define x 0)
(define y 0)
(define (f z) (display ( + z y))
(define (g f) (let ((y 10)) (f x)))
(define (h) (let ((x 100)) (g f)))
(h)
I understand at dynamic scoping value of the caller function is used by the called function. So using dynamic binding I should get the answer- 110. Using static scoping I would get the answer 0. But I got these results without considering shallow or deep binding. What is shallow and deep binding and how will it change the result?
There's an example in these lecture notes 6. Names, Scopes, and Bindings: that explains the concepts, though I don't like their pseudo-code:
thres:integer
function older(p:person):boolean
return p.age>thres
procedure show(p:person, c:function)
thres:integer
thres:=20
if c(p)
write(p)
procedure main(p)
thres:=35
show(p, older)
As best I can tell, this would be the following in Scheme (with some, I hope, more descriptive names:
(define cutoff 0) ; a
(define (above-cutoff? person)
(> (age person) cutoff))
(define (display-if person predicate)
(let ((cutoff 20)) ; b
(if (predicate person)
(display person))))
(define (main person)
(let ((cutoff 35)) ; c
(display-if person above-cutoff?)))
With lexical scoping the cutoff in above-cutoff? always refers to binding a.
With dynamic scoping as it's implemented in Common Lisp (and most actual languages with dynamic scoping, I think), the value of cutoff in above-cutoff?, when used as the predicate in display-if, will refer to binding b, since that's the most recent on on the stack in that case. This is shallow binding.
So the remaining option is deep binding, and it has the effect of having the value of cutoff within above-cutoff? refer to binding c.
Now let's take a look at your example:
(define x 0)
(define y 0)
(define (f z) (display (+ z y))
(define (g f) (let ((y 10)) (f x)))
(define (h) (let ((x 100)) (g f)))
(h)
I'm going to add some newlines so that commenting is easier, and use a comment to mark each binding of each of the variables that gets bound more than once.
(define x 0) ; x0
(define y 0) ; y0
(define (f z) ; f0
(display (+ z y)))
(define (g f) ; f1
(let ((y 10)) ; y1
(f x)))
(define (h)
(let ((x 100)) ; x1
(g f)))
Note the f0 and f1 there. These are important, because in the deep binding, the current environment of a function passed as an argument is bound to that environment. That's important, because f is passed as a parameter to g within f. So, let's cover all the cases:
With lexical scoping, the result is 0. I think this is the simplest case.
With dynamic scoping and shallow binding the answer is 110. (The value of z is 100, and the value of y is 10.) That's the answer that you already know how to get.
Finally, dynamic scoping and deep binding, you get 100. Within h, you pass f as a parameter, and the current scope is captured to give us a function (lambda (z) (display (+ z 0))), which we'll call ff for sake of convenience. Once, you're in g, the call to the local variable f is actually a call to ff, which is called with the current value of x (from x1, 100), so you're printing (+ 100 0), which is 100.
Comments
As I said, I think the deep binding is sort of unusual, and I don't know whether many languages actually implement that. You could think of it as taking the function, checking whether it has any free variables, and then filling them in with values from the current dynamic environment. I don't think this actually gets used much in practice, and that's probably why you've received some comments asking about these terms. I do see that it could be useful in some circumstances, though. For instance, in Common Lisp, which has both lexical and dynamic (called 'special') variables, many of the system configuration parameters are dynamic. This means that you can do things like this to print in base 16 (since *print-radix* is a dynamic variable):
(let ((*print-radix* 16))
(print value))
But if you wanted to return a function that would print things in base 16, you can't do:
(let ((*print-radix* 16))
(lambda (value)
(print value)))
because someone could take that function, let's call it print16, and do:
(let ((*print-radix* 10))
(print16 value))
and the value would be printed in base 10. Deep binding would avoid that issue. That said, you can also avoid it with shallow binding; you just return
(lambda (value)
(let ((*print-radix* 16))
(print value)))
instead.
All that said, I think that this discussion gets kind of strange when it's talking about "passing functions as arguments". It's strange because in most languages, an expression is evaluated to produce a value. A variable is one type of expression, and the result of evaluating a variable is the expression of that variable. I emphasize "the" there, because that's how it is: a variable has a single value at any given time. This presentation of deep and shallow binding makes it gives a variable a different value depending on where it is evaluated. That seems pretty strange. What I think would make much more sense is if the discussions were about what you get back when you evaluate a lambda expression. Then you could ask "what will the values of the free variables in the lambda expression be"? The answer, in shallow binding, will be "whatever the dynamic values of those variables are when the function is called later. The answer, in deep binding, is "whatever the dynamic values of those variables are when the lambda expression is evaluated."
Then we wouldn't have to consider "functions being passed as arguments." The whole "functions being passed as arguments" is bizarre, because what happens when you pass a function as a parameter (capturing its dynamic environment) and whatever you're passing it to then passes it somewhere else? Is the dynamic environment supposed to get re-bound?
Related Questions and Answers
Dynamic Scoping - Deep Binding vs Shallow Binding
Shallow & Deep Binding - What would this program print?
Dynamic/Static scope with Deep/Shallow binding (exercises) (The answer to this question mentions that "Dynamic scope with deep binding is much trickier, since few widely-deployed languages support it.")
I am wondering how the substitution model can be used to show certain things about infinite streams. For example, say you have a stream that puts n in the nth spot and so on inductively. I define it below:
(define all-ints
(lambda ((n <integer>))
(stream-cons n (all-ints (+ 1 n)))))
(define integers (all-ints 1))
It is pretty clear that this does what it is supposed to, but how would someone go about proving it? I decided to use induction. Specifically, induction on k where
(last (stream-to-list integers k))
provides the last value of the first k values of the stream provided, in this case integers. I define stream-to-list below:
(define stream-to-list
(lambda ((s <stream>) (n <integer>))
(cond ((or (zero? n) (stream-empty? s)) '())
(else (cons (stream-first s)
(stream-to-list (stream-rest s) (- n 1)))))))
What I'd like to prove, specifically, is the property that k = (last (stream-to-list integers k)) for all k > 1.
Getting the base case is fairly easy and I can do that, but how would I go about showing the "inductive case" as thoroughly as possible? Since computing the item in the k+1th spot requires that the previous k items also be computed, I don't know how this could be shown. Could someone give me some hints?
In particular, if someone could explain how, exactly, streams are interpreted using the substitution model, I'd really appreciate it. I know they have to be different from the other constructs a regular student would have learned before streams, because they delay computation and I feel like that means they can't be evaluated completely. In turn this would man, I think, the substitution model's apply eval apply etc pattern would not be followed.
stream-cons is a special form. It equalent to wrapping both arguments in lambdas, making them thunks. like this:
(stream-cons n (all-ints (+ 1 n))) ; ==>
(cons (lambda () n) (lambda () (all-ints (+ n 1))))
These procedures are made with the lexical scopes so here n is the initial value while when forcing the tail would call all-ints again in a new lexical scope giving a new n that is then captured in the the next stream-cons. The procedures steam-first and stream-rest are something like this:
(define (stream-first s)
(if (null? (car s))
'()
((car s))))
(define (stream-rest s)
(if (null? (cdr s))
'()
((cdr s))))
Now all of this are half truths. The fact is they are not functional since they mutates (memoize) the value so the same value is not computed twice, but this is not a problem for the substitution model since side effects are off limits anyway. To get a feel for how it's really done see the SICP wizards in action. Notice that the original streams only delayed the tail while modern stream libraries delay both head and tail.
Just started with Scheme. I'm having problem with printing on console.
A simple list printing example:
(define factorial
(lambda (n)
(cond
((= 0 n) 1)
(#t (* n (factorial (- n 1)))))))
I want to print n, every time the function is called. I figured that I can't do that within the same function? Do I need to call another function just so I can print?
Printing in Scheme works by calling display (and possibly, newline).
Since you want to call it sequentially before/after something else (which, in a functional (or in the case of Scheme, functional-ish) language only makes sense for the called functions side-effects), you would normally need to use begin, which evaluates its arguments in turn and then returns the value of the last subexpression. However, lambda implicitly contains such a begin-expression.
So in your case, it would go like this:
(lambda (n)
(display n) (newline)
(cond [...]))
Two remarks:
You can use (define (factorial n) [...]) as a shorthand for (define factorial (lambda (n) [...])).
The way you implement factorial forbids tail call-optimization, therefore the program will use quite a bit of stack space for larger values of n. Rewriting it into a optimizable form using an accumulator is possible, though.
If you only want to print n once, when the user calls the function, you will indeed need to write a wrapper, like this:
(define (factorial n)
(display n) (newline)
(inner-factorial n))
And then rename your function to inner-factorial.
Can anybody explain an example in Paul Graham's ANSI Common Lisp page 110?
The example try to explain the use &rest and lambda to create functional programming facilities. One of them is a function to compose functional arguments. I cannot find anything explaining how it worked. The code is as follows:
(defun compose (&rest fns)
(destructuring-bind (fn1 . rest) (reverse fns)
#'(lambda (&rest args)
(reduce #'(lambda (v f) (funcall f v))
rest
:initial-value (apply fn1 args)))))
The usage is:
(mapcar (compose #'list #'round #'sqrt)
'(4 9 16 25))
The output is:
((2) (3) (4) (5))
Line 2 and 6 look especially like magic to me.
The compose function returns a closure that calls each of the functions from last to first, passing on the result of each function call to the next.
The closure resulting from calling (compose #'list #'round #'sqrt) first calculates the square root of its argument, rounds the result to the nearest integer, then creates a list of the result. Calling the closure with say 3 as argument is equivalent to evaluating (list (round (sqrt 3))).
The destructuring-bind evaluates the (reverse fns) expression to get the arguments of compose in reverse order, and binds its first item of the resulting list to the fn1 local variable and the rest of the resulting list to the rest local variable. Hence fn1 holds the last item of fns, #'sqrt.
The reduce calls each the fns functions with the accumulated result. The :initial-value (apply fn1 args) provides the initial value to the reduce function and supports calling the closure with multiple arguments. Without the requirement of multiple arguments, compose can be simplified to:
(defun compose (&rest fns)
#'(lambda (arg)
(reduce #'(lambda (v f) (funcall f v))
(reverse fns)
:initial-value arg)))
destructuring-bind combines destructors with binding. A destructor is a function that lets you access a part of a data structure. car and cdr are simple destructors to extract the head and tail of a list. getf is a general destructor framework. Binding is most commonly performed by let. In this example, fns is (#'list #'round #'sqrt) (the arguments to compose), so (reverse fns) is (#'sqrt #'round #'list). Then
(destructuring-bind (fn1 . rest) '(#'sqrt #'round #'list)
...)
is equivalent to
(let ((tmp '(#'sqrt #'round #'list)))
(let ((fn1 (car tmp))
(rest (cdr tmp)))
...))
except that it doesn't bind tmp, of course. The idea of destructuring-bind is that it's a pattern matching construct: its first argument is a pattern that the data must match, and symbols in the pattern are bound to the corresponding pieces of the data.
So now fn1 is #'sqrt and rest is (#'round #'list). The compose function returns a function: (lambda (&rest args) ...). Now consider what happens when you apply that function to some argument such as 4. The lambda can be applied, yielding
(reduce #'(lambda (v f) (funcall f v))
'(#'round #'list)
:initial-value (apply #'sqrt 4)))
The apply function applies fn1 to the argument; since this argument is not a list, this is just (#'sqrt 4) which is 2. In other words, we have
(reduce #'(lambda (v f) (funcall f v))
'(#'round #'list)
:initial-value 2)
Now the reduce function does its job, which is to apply #'(lambda (v f) (funcall f v)) successively to the #'round and to #'list, starting with 2. This is equivalent to
(funcall #'list (funcall #'round 2))
→ (#'list (#'round 2))
→ '(2)
Okay, here goes:
It takes the functions given, reverses it (in your example, it becomes (#'sqrt #'round #'list)), then sticks the first item into fn1, and the rest into rest. We have: fn1 = #'sqrt, and rest = (#'round #'list).
Then it performs a fold, using (apply sqrt args) (where args are the values given to the resulting lambda) as the initial value, and with each iteration grabbing the next function from rest to call.
For the first iteration you end up with (round (apply sqrt args)), and the second iteration you end up with (list (round (apply sqrt args))).
Interestingly, only the initial function (sqrt in your case) is allowed to take multiple arguments. The rest of the functions are called with single arguments only, even if any particular function in the chain does a multiple-value return.
This example stumped me for a day. I could finally understand it by renaming some of the arguments and commenting each line before it made sense. Below is what helped me explain it to myself.
In the book example using the call:
(mapcar (compose #'list #'round #'sqrt) '(4 9 16 25))
The parameter functions becomes (#'LIST #'ROUND #'SQRT)
(defun compose (&rest functions)
(destructuring-bind (fx . fxs) (reverse functions)
;; fx becomes #'SQRT
;; fxs becomes '(#'ROUND #'LIST)
#'(lambda (&rest args) ; This is the function returned as result.
;; The args parameter will be (4) on the mapcar's first
;; iteration on the (4 9 16 25) list passed in the call:
;; (mapcar #'(compose #'List #'round #'sqrt) '(4 9 16 25)) => ((2) (3) (4) (5))
;; or e.g. the (4) in (funcall (compose #'list #'sqrt '(4)) => (2.0)
;; Note that args is not ((#'ROUND #'LIST)).
(reduce #'(lambda (x y) (funcall y x))
;; fxs is (#'ROUND #'LIST) - captuted as closure since it is now
;; locally unbound.
fxs
;; Initial value is: (apply #'SQRT '(4) => 2.0.
;; In Paul Graham's example, the mapcar passes
;; each square number individually.
;; The reverse order of parameters in the second lambda
;; first invokes: (ROUND 2.0) => 2
;; and then invokes: (LIST 2) => (2)
:initial-value (apply fx args)))))
I don't know if you would call it the canonical formulation, but to bind a local function I am advised by the GNU manual to use 'flet':
(defun adder-with-flet (x)
(flet ( (f (x) (+ x 3)) )
(f x))
)
However, by accident I tried (after having played in Scheme for a bit) the following expression, where I bind a lambda expression to a variable using 'let', and it also works if I pass the function to mapcar*:
(defun adder-with-let (x)
(let ( (f (lambda (x) (+ x 3))) )
(car (mapcar* f (list x)) ))
)
And both functions work:
(adder-with-flet 3) ==> 6
(adder-with-let 3) ==> 6
Why does the second one work? I cannot find any documentation where 'let' can be used to bind functions to symbols.
Unlike Scheme, Emacs Lisp is a 2-lisp, which means that each symbol has two separate bindings: the value binding and the function binding. In a function call (a b c d), the first symbol (a) is looked up using a function binding, the rest (b c d) are looked up using the value binding. Special form let creates a new (local) value binding, flet creates a new function binding.
Note that whether value or function binding is used for lookup depends on the position in the (a b c d) function call, not on the type of the looked-up value. In particular, a value binding can resolve to function.
In your first example, you function-bind f (via flet), and then do a function lookup:
(f ...)
In your second example, you value-bind f to a function (via let), and then use a value lookup:
(... f ...)
Both work because you use the same kind of binding and lookup in each case.
http://en.wikipedia.org/wiki/Common_Lisp#Comparison_with_other_Lisps
I did a quick search of the Emacs lisp manual and couldn't find any reference to 'flet, which isn't terribly surprising since that is a part of cl - the common-lisp package.
let will do a local binding as well, but it won't bind to the "function cell" for that symbol.
i.e. This works:
(let ((myf (lambda (x) (list x x))))
(eval (list myf 3)))
but
(let ((myf (lambda (x) (list x x))))
(myf 3))
fails with the error: "Lisp error: (void-function myf)"
flet on the other hand, does do the binding to the function cell, so this works:
(flet ((myf (x) (list x x)))
(myf 3))
Notice the difference being that flet allows you to use the symbol myf directly, whereas the let does not - you have to use some indirection to get the function out of the "value cell" and apply that appropriately.
In your example, the 'mapcar' did the equivalent to my use of 'eval.
#d11wq there is `funcall' for this purpose. The following works:
(defun adder-with-let (x)
(let ((f #'(lambda (x) (+ x 3))))
(funcall f 3)))
(adder-with-let 3) ;=> 6
You don't have to use flet if you do not want to. You place a function in the function cell of a local symbol defined using let as in the following example:
(let ((ALocalSymbol))
(fset 'ALocalSymbol (lambda (x) (* 2 x)))
(ALocalSymbol 4)
)
Evaluating this will return 8. Do notice the quote in front of ALocalSymbol in (let ((ALocalSymbol))...). While setq quotes symbols, fset does not.
flet is a syntactic sugar of sorts. Using a plain-old let to define nil-valued symbols, allows you to choose which "cell" of a symbol to set. You could use setq to set the symbol's value cell or fset to set the function cell.
Hope this helps,
Pablo