I am trying an example on Chapter 4 of SICP (part of writing the LISP interpreter)
(define (definition-value exp)
(if (symbol? (cadr exp))
(caddr exp)
(make-lambda
(cdadr exp) ; formal parameters
(cddr exp) ; body
)
)
)
(define (make-lambda parameters body)
(cons 'lambda (cons parameters body))
)
I Tested it, definition-value on '(define (double x) (+ x x))) should return a lambda function
( (definition-value '(define (double x) (+ x x))) 10)
Racket outputs
procedure application: expected procedure, given: (lambda (x) (+ x x)); arguments were: 10
Isn't "(lambda (x) (+ x x))" a procedure? Or it is a reference? If it is a reference, any way to "dereference" it?
definition-value returns the value in the definition expression given to it as an argument:
(definition-value '(define x 42))
=> 42
(definition-value '(define (qq x) (+ x y 42)))
=> (make-lambda '(x) '((+ x y 42)))
=> '(lambda (x) (+ x y 42))
You can't call the quoted list as a function, as you do: ( '(lambda (x) (+ x y 42)) 10) is invalid. It is not a function, it is just an s-expression.
definition-value is part of an interpreter. This interpreter is the way to "dereference", i.e. interpret function definitions. Different interpreters can have different ways to interpret same function definitions, giving different semantics to the resulting languages.
Evaluation of expressions must be done in context - they appear inside certain lexical scope (area in code where a variable is visible), which gives rise to environments (also, this). In the example above, y is defined in some enclosing scope in the program being interpreted. Trying to interpret that expression in REPL by calling Racket's eval, what value would y have?
I figured the answer, if execute a Racket script in file, racket interpreter doesn't know the namespace, however, the REPL knows it. The solution is to add this line at the beginning of the file
(define ns (make-base-namespace))
Then pass ns to eval when using it
(eval <what ever code reference here> ns)
That will make my above mentioned examples work.
Related
I'm having trouble figuring out how to go about creating a function that can take a series of the same function as arguments with the last argument as an operand. For example:
(func sqrt sqrt sqrt 390625)
The call above should return 5 as (sqrt 390625) > (sqrt 625) > (sqrt 25) > 5
I'm having trouble figuring out the exact way I should write this as any way I have tried has given me errors or achieved an infinite loop.
This the code is have so far:
(define func
(lambda L
(cond ( (equal? (length L) 2) ((car L) (cadr L)) ) ;; If the list consists of only 2 elements, carry out the function (element 1) onto the operand (element 2)
( #t (apply (car L) (func (cdr L))) ) ;; otherwise, apply the function (1st element) onto the rest of the list
)
)
)
The first condition works, for example returning 5 if i call (func sqrt 25), however the recursive call is throwing errors.
I would appreciate any help with this.
The OP doesn't provide a definition for chain, so that part is unclear, but I think that a fundamental problem here is that there is no recursive call to func; further, apply isn't used in the right position.
Instead of using (equal (length L) 2) as a base case, it might be nicer to make recursive calls as long as the first element in the input is a procedure, or otherwise just return the element:
#lang racket
(define multi-call
(lambda args
(let ((arg (car args)))
(if (procedure? arg)
(arg (apply multi-call (cdr args)))
arg))))
Here, when arg is a procedure, then it is applied to the result of calling multi-call recursively on the remaining arguments. Note that multi-call takes an arbitrary number of arguments, wrapping them in the list args. The reduction step provides (cdr args), which is a list of the remaining arguments. This means that apply should be used to call multi-call on those remaining arguments because multi-call expects an arbitrary number of arguments, not a list of arguments.
multi-call.rkt> (multi-call sqrt sqrt sqrt 390625)
5
I have a code which takes a list and returns all possible permutations by the parameter result.
But when I compile I have an error which says *** - =: (1+ INDEX) is not a number.
Is this message true or I messed up the code generally?
I am new to lisp I can looking for a fix and also open to suggestions from fucntional programmers.
;; Creates permutatiions of a given list and returns it via parameter
(defun create-permuations (source)
(setf result (list))
(create-permuations-helper source 0 '() result)
result)
(defmacro create-permuations-helper (source index cur result)
(if (= (list-length cur) index)
(cons cur result)
(loop for i from 0 to (list-length cur) do
(create-permuations-helper source (1+ index)
(append cur (list (nth i source))) result))))
99% of times when a compiler reports an error you can trust it to be true. Here Index is the list (1+ index), literally the 1+ symbol followed by the index symbol. This is so because you are using a macro, and macros operate on code.
In your macro, you do not return a form to be evaluated, you execute code during macro-expansion that depends on itself. That alone is an undefined behaviour. For example:
(defmacro a (x)
(if (plusp x)
(a (- x 1))
nil))
In the body of a, you want to expand code using a recursive call to itself. But the macro is not yet fully known and cannot be until the whole macro is defined.
Maybe the particular lisp implementation binds a to the macro function in body of the macro, which is a strange thing to do, or you evaluated the definition twice. The first time the compiler assumes a is an unknown function, then binds a to a macro, and the second time it tries to expand the macro.
Anyway macro are not supposed to be recursive.
In the example, since the macro does not evaluate its argument, the nested call to the macro is given the literal expression (- x 1), and not its actual value, which cannot be known anyway since x is unknown. You are crossing a level of abstraction here by trying to evaluate things at macroexpansion time.
But, macros can expand into code that refers to themselves.
(defmacro a (x)
(if (plusp x)
`(b (a ,(- x 1)))
nil))
Now, (a 2) expands into (b (a 1)), which itself macroexpands into (b (b (a 0))), and finally reaches a fixpoint which is (b (b nil)).
The difference is that the macro produces a piece of code and returns, which the compiler macroexpands again, whereas in the first example, the macro must already be expanded in the body of its own definition.
Possible implementation
One way to solve your problem is to define a local function that has access to a variable defined in your main function. Then, the local function can set it, and you do not need to pass a variable by reference (which is not possible to do):
(defun permut (list)
(let (result)
(labels ((recurse (stack list)
(if list
(dolist (x list)
(recurse (cons x stack)
(remove x list :count 1)))
(push stack result))))
(recurse nil list))
result))
Alternatively, you can split the process in two; first, define permut-helper, which is a higher-order function that takes a callback function; it generates permutations and calls the callback for each one:
(defun permut-helper (stack list callback)
(if list
(dolist (x list)
(permut-helper (cons x stack)
(remove x list :count 1)
callback))
(funcall callback stack)))
You call it with a function that pushes results into a list of permutations:
(defun permut (list)
(let (result)
(flet ((add-result (permutation)
(push permutation result)))
(permut-helper nil list #'add-result))
result))
During the execution of my code I get the following errors in the different Scheme implementations:
Racket:
application: not a procedure;
expected a procedure that can be applied to arguments
given: '(1 2 3)
arguments...:
Ikarus:
Unhandled exception
Condition components:
1. &assertion
2. &who: apply
3. &message: "not a procedure"
4. &irritants: ((1 2 3))
Chicken:
Error: call of non-procedure: (1 2 3)
Gambit:
*** ERROR IN (console)#2.1 -- Operator is not a PROCEDURE
((1 2 3) 4)
MIT Scheme:
;The object (1 2 3) is not applicable.
;To continue, call RESTART with an option number:
; (RESTART 2) => Specify a procedure to use in its place.
; (RESTART 1) => Return to read-eval-print level 1.
Chez Scheme:
Exception: attempt to apply non-procedure (1 2 3)
Type (debug) to enter the debugger.
Guile:
ERROR: In procedure (1 2 3):
ERROR: Wrong type to apply: (1 2 3)
Chibi:
ERROR in final-resumer: non procedure application: (1 2 3)
Why is it happening
Scheme procedure/function calls look like this:
(operator operand ...)
Both operator and operands can be variables like test, and + that evaluates to different values. For a procedure call to work it has to be a procedure. From the error message it seems likely that test is not a procedure but the list (1 2 3).
All parts of a form can also be expressions so something like ((proc1 4) 5) is valid syntax and it is expected that the call (proc1 4) returns a procedure that is then called with 5 as it's sole argument.
Common mistakes that produces these errors.
Trying to group expressions or create a block
(if (< a b)
((proc1)
(proc2))
#f)
When the predicate/test is true Scheme assumes will try to evaluate both (proc1) and (proc2) then it will call the result of (proc1) because of the parentheses. To create a block in Scheme you use begin:
(if (< a b)
(begin
(proc1)
(proc2))
#f)
In this (proc1) is called just for effect and the result of teh form will be the result of the last expression (proc2).
Shadowing procedures
(define (test list)
(list (cdr list) (car list)))
Here the parameter is called list which makes the procedure list unavailable for the duration of the call. One variable can only be either a procedure or a different value in Scheme and the closest binding is the one that you get in both operator and operand position. This would be a typical mistake made by common-lispers since in CL they can use list as an argument without messing with the function list.
wrapping variables in cond
(define test #t) ; this might be result of a procedure
(cond
((< 5 4) result1)
((test) result2)
(else result3))
While besides the predicate expression (< 5 4) (test) looks correct since it is a value that is checked for thurthness it has more in common with the else term and whould be written like this:
(cond
((< 5 4) result1)
(test result2)
(else result3))
A procedure that should return a procedure doesn't always
Since Scheme doesn't enforce return type your procedure can return a procedure in one situation and a non procedure value in another.
(define (test v)
(if (> v 4)
(lambda (g) (* v g))
'(1 2 3)))
((test 5) 10) ; ==> 50
((test 4) 10) ; ERROR! application: not a procedure
Undefined values like #<void>, #!void, #<undef>, and #<unspecified>
These are usually values returned by mutating forms like set!, set-car!, set-cdr!, define.
(define (test x)
((set! f x) 5))
(test (lambda (x) (* x x)))
The result of this code is undetermined since set! can return any value and I know some scheme implementations like MIT Scheme actually return the bound value or the original value and the result would be 25 or 10, but in many implementations you get a constant value like #<void> and since it is not a procedure you get the same error. Relying on one implementations method of using under specification makes gives you non portable code.
Passing arguments in wrong order
Imagine you have a fucntion like this:
(define (double v f)
(f (f v)))
(double 10 (lambda (v) (* v v))) ; ==> 10000
If you by error swapped the arguments:
(double (lambda (v) (* v v)) 10) ; ERROR: 10 is not a procedure
In higher order functions such as fold and map not passing the arguments in the correct order will produce a similar error.
Trying to apply as in Algol derived languages
In algol languages, like JavaScript and C++, when trying to apply fun with argument arg it looks like:
fun(arg)
This gets interpreted as two separate expressions in Scheme:
fun ; ==> valuates to a procedure object
(arg) ; ==> call arg with no arguments
The correct way to apply fun with arg as argument is:
(fun arg)
Superfluous parentheses
This is the general "catch all" other errors. Code like ((+ 4 5)) will not work in Scheme since each set of parentheses in this expression is a procedure call. You simply cannot add as many as you like and thus you need to keep it (+ 4 5).
Why allow these errors to happen?
Expressions in operator position and allow to call variables as library functions gives expressive powers to the language. These are features you will love having when you have become used to it.
Here is an example of abs:
(define (abs x)
((if (< x 0) - values) x))
This switched between doing (- x) and (values x) (identity that returns its argument) and as you can see it calls the result of an expression. Here is an example of copy-list using cps:
(define (copy-list lst)
(define (helper lst k)
(if (null? lst)
(k '())
(helper (cdr lst)
(lambda (res) (k (cons (car lst) res))))))
(helper lst values))
Notice that k is a variable that we pass a function and that it is called as a function. If we passed anything else than a fucntion there you would get the same error.
Is this unique to Scheme?
Not at all. All languages with one namespace that can pass functions as arguments will have similar challenges. Below is some JavaScript code with similar issues:
function double (f, v) {
return f(f(v));
}
double(v => v * v, 10); // ==> 10000
double(10, v => v * v);
; TypeError: f is not a function
; at double (repl:2:10)
// similar to having extra parentheses
function test (v) {
return v;
}
test(5)(6); // == TypeError: test(...) is not a function
// But it works if it's designed to return a function:
function test2 (v) {
return v2 => v2 + v;
}
test2(5)(6); // ==> 11
Could someone explain to me what's going on in this very simple code snippet?
(defun test-a ()
(let ((x '(nil)))
(setcar x (cons 1 (car x)))
x))
Upon a calling (test-a) for the first time, I get the expected result: ((1)).
But to my surprise, calling it once more, I get ((1 1)), ((1 1 1)) and so on.
Why is this happening? Am I wrong to expect (test-a) to always return ((1))?
Also note that after re-evaluating the definition of test-a, the return result resets.
Also consider that this function works as I expect:
(defun test-b ()
(let ((x '(nil)))
(setq x (cons (cons 1 (car x))
(cdr x)))))
(test-b) always returns ((1)).
Why aren't test-a and test-b equivalent?
The Bad
test-a is self-modifying code. This is extremely dangerous. While the variable x disappears at the end of the let form, its initial value persists in the function object, and that is the value you are modifying. Remember that in Lisp a function is a first class object, which can be passed around (just like a number or a list), and, sometimes, modified. This is exactly what you are doing here: the initial value for x is a part of the function object and you are modifying it.
Let us actually see what is happening:
(symbol-function 'test-a)
=> (lambda nil (let ((x (quote (nil)))) (setcar x (cons 1 (car x))) x))
(test-a)
=> ((1))
(symbol-function 'test-a)
=> (lambda nil (let ((x (quote ((1))))) (setcar x (cons 1 (car x))) x))
(test-a)
=> ((1 1))
(symbol-function 'test-a)
=> (lambda nil (let ((x (quote ((1 1))))) (setcar x (cons 1 (car x))) x))
(test-a)
=> ((1 1 1))
(symbol-function 'test-a)
=> (lambda nil (let ((x (quote ((1 1 1))))) (setcar x (cons 1 (car x))) x))
The Good
test-b returns a fresh cons cell and thus is safe. The initial value of x is never modified. The difference between (setcar x ...) and (setq x ...) is that the former modifies the object already stored in the variable x while the latter stores a new object in x. The difference is similar to x.setField(42) vs. x = new MyObject(42) in C++.
The Bottom Line
In general, it is best to treat quoted data like '(1) as constants - do not modify them:
quote returns the argument, without evaluating it. (quote x) yields x.
Warning: quote does not construct its return value, but just returns
the value that was pre-constructed by the Lisp reader (see info node
Printed Representation). This means that (a . b) is not
identical to (cons 'a 'b): the former does not cons. Quoting should
be reserved for constants that will never be modified by side-effects,
unless you like self-modifying code. See the common pitfall in info
node Rearrangement for an example of unexpected results when
a quoted object is modified.
If you need to modify a list, create it with list or cons or copy-list instead of quote.
See more examples.
PS1. This has been duplicated on Emacs.
PS2. See also Why does this function return a different value every time? for an identical Common Lisp issue.
PS3. See also Issue CONSTANT-MODIFICATION.
I found the culprit is indeed 'quote. Here's its doc-string:
Return the argument, without evaluating it.
...
Warning: `quote' does not construct its return value, but just returns
the value that was pre-constructed by the Lisp reader
...
Quoting should be reserved for constants that will
never be modified by side-effects, unless you like self-modifying code.
I also rewrote for convenience
(setq test-a
(lambda () ((lambda (x) (setcar x (cons 1 (car x))) x) (quote (nil)))))
and then used
(funcall test-a)
to see how 'test-a was changing.
It looks like the '(nil) in your (let) is only evaluated once. When you (setcar), each call is modifying the same list in-place. You can make (test-a) work if you replace the '(nil) with (list (list)), although I presume there's a more elegant way to do it.
(test-b) constructs a totally new list from cons cells each time, which is why it works differently.
I'm trying to write functions that wrap another function but I'm not sure how to pass parameters correctly while maintaining a sensible lambda-list.
E.g. if I have a function
(defun f (x &key y z) ...)
I want to write something like
(defun g (x &key y z)
(h (f x :y y :z z)))
This isn't satisfactory because I want to call f from g with the exact arguments g was called with, which doesn't happen (e.g. I don't want to supply keyword arguments to f that weren't supplied to g by the caller).
I initially wrote something like:
(defun g (&rest f-args)
(apply #'f f-args))
And that's the effect I want, however the lambda list for g is now very cryptic and I keep having to navigate to f to see what the arguments should be.
I did come up with a solution (and it's mostly satisfactory so I posted it as an answer), but I need to be explicit with every single key argument, and with large lambda-lists (e.g. if I want to wrap drakma:http-request), it will be a pain. I hope that maybe there's a better way.
You could write a macro that defines a function by copying the lambda list from another function. The problem is that there isn't a standard way to get the lambda list, but for SBCL you can use SB-INTROSPECT:FUNCTION-LAMBDA-LIST (although that won't work with (declaim (optimize (debug 0)))). You could try reading Swank source code to see how it gets the lambda lists for various implementations.
(defmacro define-wrapper (name lambda-source &body body)
`(defun ,name ,(sb-introspect:function-lambda-list lambda-source)
,#body))
(defun f (x &key (y 3) (z 4))
(+ x y z))
(define-wrapper g f
(* 2 (f x :y y :z z)))
(f 2) ;=> 9
(g 2) ;=> 18
That's a bit ugly since the code doesn't show the variable definitions. A bit more complex solution might be to do something like
;; Requires Alexandria.
(defmacro define-wrapper (name lambda-source &body body)
(let ((lambda-list (sb-introspect:function-lambda-list lambda-source)))
(multiple-value-bind (required optional rest keywords)
(alexandria:parse-ordinary-lambda-list lambda-list)
(declare (ignore rest))
`(defun ,name ,lambda-list
,#(sublis `((_ . (,lambda-source ,#(loop for r in required collect r)
,#(loop for (name init suppliedp)
in optional collect name)
,#(loop for ((k-name name) init suppliedp)
in keywords
append (list k-name name)))))
body)))))
(defun f (x &key (y 3) (z 4))
(+ x y z))
(define-wrapper g f
(* 2 _))
Where the _ in the wrapper is replaced with a call to the function F with the given arguments. You do still have to remember that the argument variables exist and can conflict with ones you define yourself.
That passes all arguments to the function regardless of whether they were given. That might mess up a function that behaves differently depending on whether an argument was supplied or not. You could avoid that by using APPLY, but it's a bit more complex.
(defmacro define-wrapper (name lambda-source &body body)
(let ((lambda-list (sb-introspect:function-lambda-list lambda-source)))
(alexandria:with-gensyms (deparsed-arglist-sym
key-sym val-sym suppliedp-sym)
(multiple-value-bind (required optional rest keywords)
(alexandria:parse-ordinary-lambda-list lambda-list)
(declare (ignore rest))
(multiple-value-bind (body declarations docstring)
(alexandria:parse-body body :documentation t)
`(defun ,name ,lambda-list
,#(when docstring (list docstring))
,#declarations
(let ((,deparsed-arglist-sym
(nconc (loop for ,val-sym in (list ,#required) collect ,val-sym)
(loop for (,val-sym . ,suppliedp-sym)
in (list ,#(loop for (name init suppliedp)
in optional
collect (list 'cons name
(or suppliedp t))))
when ,suppliedp-sym collect ,val-sym)
(loop for (,key-sym ,val-sym ,suppliedp-sym)
in (list ,#(loop for ((kname name) init suppliedp)
in keywords
collect (list 'list kname name
(or suppliedp t))))
when ,suppliedp-sym append (list ,key-sym ,val-sym)))))
,#(sublis `((_ . (apply #',lambda-source ,deparsed-arglist-sym)))
body))))))))
(define-wrapper bar drakma:http-request
"Return the length of a response to http-request."
;; HTTP-REQUEST has some &aux variables.
(declare (ignore drakma::unparsed-uri
drakma::args))
(length _))
(bar "http://www.google.com") ;=> 11400 (14 bits, #x2C88)
I came up with this:
(defun g (x &rest f-keys &key y z)
(declare (ignorable y z))
(apply #'f x f-keys))
It's great for small lambda-lists but I hope I could do better.
I also can't see default values unless I type them explicitly.