If I want to print function definition for a symbol, what should I do?
If I understand correctly, you want a function print-function such that after
(define (foo x) (cons x x))
it will behave as
> (print-function foo)
(lambda (x) (cons x x))
Standard Scheme doesn't have a facility for that. The reason is Scheme implementations may, and generally do, compile functions into a different representation (bytecode, machine code).
Some Schemes may keep the function definition around; check your implementation's manual.
Related
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.
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.
(define make (lambda (x) (lambda (y) (cons x (list y)))))
(let ((x 7)
(p (make 4)))
(cons x (p 0)))
I'm new to Scheme and functional program, so I am a bit clunky with walking through programs, but I get that if I used deep binding this program will return (7 4 0). Makes sense. What would this program do using shallow binding? I get this may sound dumb but is the p in the line with cons a redefinition? So in that case, we would return (7 0)?
Basically, I understand the concept of deep v. shallow binding, but I feel like I'm jumbling it up when looking at Scheme because I'm not crazy familiar with it.
Deep or shallow binding is an implementational technique and can not be observed from inside the program. The difference for the programmer is between lexical and dynamic scoping rules, but both can be implemented with any of the two techniques (i.e. one notion has got nothing to do with the other).
Deep or shallow refers to the choice of stack frame to hold a given outer scoped variable's binding. In deep binding there is a chain of frames to be accessed until the correct frame is entered holding the record for the variable; in shallow binding all bindings are present in one, shallow environment. See also "rerooting" (which only makes sense in the context of shallow binding implementation of lexical scoping).
To your specific question, under lexical scoping rules your code would return (7 4 0) and under dynamic - (7 7 0), because the call ((lambda(y) (list x y)) 0) is done inside the dynamic scope of x=7 binding (as a side note, (cons x (list y)) is the same as (list x y)):
x = 7
p = (lambda (y) (list x y)) ; x=4 is unused, in p=(make 4)
(cons 7 (p 0)) == (list 7 7 0) ; 'x' in this line and in lambda body for p
; both refer to same binding that is
; in effect, i.e. x=7
NB same terms (deep/shallow binding) are used in other language(s) now with completely different meaning (they do have something to do with the scoping rules there), which I don't care to fully understand. This answer is given in the context of Scheme.
Reference: Shallow Binding in LISP 1.5 by Baker, Henry G. Jr., 1977.
See this wikipedia article for a discussion on scoping (it mentions lexical/dynamic scoping and deep/shallow binding) bearing in mind that Scheme is lexically scoped. Will Ness' answer provides additional information.
For now, let's see step-by-step what's happening in this snippet of code:
; a variable called x is defined and assigned the value 7
(let ((x 7)
; make is called and returns a procedure p, inside its x variable has value 4
(p (make 4)))
; 7 is appended at the head of the result of calling p with y = 0
(cons x (p 0)))
=> '(7 4 0)
Notice that in the second line a closure is created in the lambda returned by make, and the variable x inside will be assigned the value 4. This x has nothing to do with the outer x, because Scheme is lexically scoped.
The last line is not a redefinition, as mentioned in the previous paragraph the x inside make is different from the x defined in the let expression.
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