I'm having some problem writing a LISP function. The function is defined as
(defun foo (arg1 &optional cont))
(cond ((null arg1) nil)
((= 0 cont) arg1)
((do_something))
((recursive call))))
When i call the function with cont everything works fine, but when I call it just with arg1 the error returned is:
Error: in ZEROP of (NIL) arguments should be of type NUMBER
I guess something is wrong in the condition ((= 0 cont) arg1), can you help me solve this problem?
Thanks
The = function, along with some other ones, expect exclusively numbers.
You need to use EQL or a more general equality comparison (equal, equalp) when you expect NIL to be a valid input; here, NIL is expected because it is the default value of the optional argument.
You could also provide a numerical default value to cont:
... &optional (cont 0) ...
... which might be the correct approach if cont has no reason to be anything else than a number.
Related
I am trying to check whether a boolean variable (boolVariable) is True (T) using the following code:
(defvar boolVariable T)
(if (= boolVariable T)
(print 'TRUE)
)
However, I get the following error:
=: T is not a number
This seems strange, considering that I thought that you can check whether variables equal booleans in Lisp?
Common Lisp's = compares only numbers and neither t nor boolVariable is number.
There are some other equality predicates like eq, eql, equal or equalp, but in this case, you can just use value of bool-variable (renamed in kebab-case):
(defvar bool-variable t)
(if bool-variable (print 'TRUE))
If with only then-form can be also replaced with when:
(defvar bool-variable t)
(when bool-variable (print 'TRUE))
I am new to lisp and I have a problem, I'm trying to find the number in the list but it is not working. I haven't made the return statement yet
(defun num (x 'y)
(if (member x '(y)) 't nil))
(write (num 10 '(5 10 15 20)))
My output just outputs the nil instead of doing the function and I'm confused of what I am doing wrong.
Solution
(defun member-p (element list)
"Return T if the object is present in the list"
(not (null (member element list))))
The not/null pattern is equivalent to (if (member element list) t nil) but is more common.
In fact, you do not really need this separate function,
member is good enough.
The -p suffix stands for predicate, cf. integerp and upper-case-p.
Your code
You cannot quote lambda list elements, so you need to replace defun num (x 'y) with defun num (x y)
You need not quote t
Quoting '(y) makes no sense, replace it with y.
You do not need to write the function call, the REPL will do it for you.
See also
When to use ' (or quote) in Lisp?
Can you program without REPL on Lisp?
You are almost certainly expected to not just use member, but to write a function which does what you need (obviously in real life you would just use member because that's what it's for).
So. To know if an object is in a list:
if the list is empty it's not;
if the head of the list is equal to the object it is;
otherwise it is in the list if it's in the tail of the list.
And you turn this into a function very straightforwardly:
(defun num-in-list-p (n l)
;; is N in L: N is assumed to be a number, L a list of numbers
(cond ((null l)
nil)
((= n (first l))
t)
(t
(num-in-list-p n (rest l)))))
You could use the built in position function which will return the index of the number if it is in the list:
(position 1 '(5 4 3 2 1))
If you want to define your own function:
CL-USER> (defun our-member(obj lst)
(if(zerop (length lst))
nil
(if(equal(car lst)obj)
T
(our-member obj (cdr lst)))))
OUR-MEMBER
CL-USER> (our-member 1 '(5 4 3 2 1))
T
CL-USER> (our-member 99 '(1 2 3 4 5))
NIL
We can create a function called "our-member" that will take an object (in your case a number) and a list (in your case a list of numbers) as an argument. In this situation our "base-case" will be whether or not the length of the list is equal to zero. If it is and we still haven't found a match, we will return nil. Otherwise, we will check to see if the car of the list (the first element in the list) is equal to the obj that we passed. If so, we will return T (true). However, if it is not, we will call the function again passing the object and the cdr of the list (everything after the car of the list) to the function again, until there are no items left within the list. As you can see, The first example of a call to this function returns T, and the second example call returns NIL.
What makes this utility function a good example is that it essentially shows you the under workings of the member function as well and what is going on inside.
How can I break a function execution in LISP if I get a certain value?
For example, I have a main function like this:
(defun recognize-a (arg input)
(if (equal (recognize-b arg input) '())
T
NIL
))
I want to break the function recognize-b in case the input is an empty list, without passing any values to the main function:
(defun recognize-b (fa input)
(if (equal input '())
<<<WANTED BREAK>>>
(<Else branch>)))
You can use ERROR to signal an error from RECOGNIZE-B when INPUT is empty.
(defun recognize-b (arg input)
(when (emptyp input)
(error "INPUT is empty!"))
;; Do whatever the function normally does...
:return-value-from-b)
I'll just return :RETURN-VALUE-FROM-B since I don't know what the function is supposed to do. You could define an error type to signal, but by default ERROR will signal a SIMPLE-ERROR.
To handle the error in RECOGNIZE-A, you can use HANDLER-CASE.
(defun recognize-a (arg input)
(handler-case (recognize-b arg input)
(simple-error () t)))
This simply returns the value from RECOGNIZE-B if there was no error, or T if there was.
(recognize-a 10 '(1 2)) ;=> :RETURN-VALUE-FROM-B
(recognize-a 10 '()) ;=> T
There is a good introduction to the condition system in the book Practical Common Lisp, Chapter 19. Beyond Exception Handling: Conditions and Restarts.
In Common Lisp, the special operator quote makes whatever followed by un-evaluated, like
(quote a) -> a
(quote {}) -> {}
But why the form (quote ()) gives me nil? I'm using SBCL 1.2.6 and this is what I got in REPL:
CL-USER> (quote ())
NIL
More about this problem: This is some code from PCL Chapter 24
(defun as-keyword (sym)
(intern (string sym) :keyword))
(defun slot->defclass-slot (spec)
(let ((name (first spec)))
`(,name :initarg ,(as-keyword name) :accessor ,name)))
(defmacro define-binary-class (name slots)
`(defclass ,name ()
,(mapcar #'slot->defclass-slot slots)))
When the macro expand for the following code:
(define-binary-class id3-tag
((major-version)))
is
(DEFCLASS ID3-TAG NIL
((MAJOR-VERSION :INITARG :MAJOR-VERSION :ACCESSOR MAJOR-VERSION)))
which is NIL rather than () after the class name ID3-TAG.
nil and () are two ways to express the same concept (the empty list).
Traditionally, nil is used to emphasize the boolean value "false" rather than the empty list, and () is used the other way around.
The Common LISP HyperSpec says:
() ['nil], n. an alternative notation for writing the symbol nil, used
to emphasize the use of nil as an empty list.
Your observation is due to an object to having more than one representation. In Common Lisp the reader (that reads code and reads expressions) parses text to structure and data. When it's data the writer can print it out again but it won't know exactly how the data was represented when it was initially read in. The writer will print one object exactly one way, following defaults and settings, even though there are several representations for that object.
As you noticed nil, NIL, nIL, NiL, ... ,'nil, 'NIL, (), and '() are all read as the very same object. I'm not sure the standard dictates exactly how it's default representation out should be so I guess some implementations choose one of NIL, nil or maybe even ().
With cons the representation is dependent on the cdr being a cons/nil or not:
'(a . nil) ; ==> (a)
'(a . (b . c)) ; ==> (a b . c)
'(a . (b . nil)) ; ==> (a b)
With numbers the reader can get hints about which base you are using. If no base is used in the text it will use whatever *read-base* is:
(let ((*read-base* 2)) ; read numbers as boolean
(read-from-string "(10 #x10)")) ; ==> (2 16)
#x tells the reader to interpret the rest as a hexadecimal value. Now if your print-base would have been 4 the answer to the above would have been visualized as (2 100).
To sum it up.. A single value in Common Lisp may have several good representations and all of them would yield the very same value. How the value is printed will follow both implementation, settings and even arguments to the functions that produce them. Neither what it accepts as values in or the different ways it can visualize the value tells nothing about how the value actually gets stored internally.
From the Question How do I pass a function as a parameter to in elisp? I know how to pass a function as a parameter to a function. But we need to go deeper...
Lame movie quotes aside, I want to have a function, which takes a function as a parameter and is able to call itself [again passing the function which it took as parameter]. Consider this snippet:
(defun dummy ()
(message "Dummy"))
(defun func1 (func)
(funcall func))
(defun func2 (func arg)
(message "arg = %s" arg)
(funcall func)
(func2 'func (- arg 1)))
Calling (func1 'dummy) yields the expected output:
Dummy
"Dummy"
Calling (func2 'dummy 4) results in an error message:
arg = 4
Dummy
arg = 3
funcall: Symbol's function definition is void: func
I had expected four calls to dummy, yet the second iteration of func2 seems to have lost its knowledge of the function passed to the first iteration (and passed on from there). Any help is much appreciated!
There probably is a better way to do this with lexical scoping. This is more or less a translation from Rosetta Code:
(defun closure (y)
`(lambda (&rest args) (apply (funcall ',y ',y) args)))
(defun Y (f)
((lambda (x) (funcall x x))
`(lambda (y) (funcall ',f (closure y)))))
(defun factorial (f)
`(lambda (n)
(if (zerop n) 1
(* n (funcall ,f (1- n))))))
(funcall (Y 'factorial) 5) ;; 120
Here's a link to Rosetta code: http://rosettacode.org/wiki/Y_combinator with a bunch of other languages immplementing the same thing. Y-combinator is a construct, from the family of fixed-point combinators. Roughly, the idea is to eliminate the need for implementing recursive functions (recursive functions require more sophistications when you think about how to make them compile / implement in the VM). Y-combinator solves this by allowing one to mechanically translate all functions into non-recursive form, while still allowing for recursion in general.
To be fair, the code above isn't very good, because it will create new functions on each recursive step. This is because until recently, Emacs Lisp didn't have lexical bindings (you couldn't have a function capture its lexical environment), in other words, when the Emacs Lisp function is used outside the scope it was declared, the values of the bound variables will be taken from the function's current scope. In the case above such bound variables are f in the Y function and y in the closure function. Luckily, those are just symbols designating an existing function, so it is possible to mimic that behaviour using macros.
Now, what Y-combinator does:
Captures the original function into variable f.
Returns a wrapper function of one argument, which will call f, when called in its turn, used by Y-combinator to
Return a wrapper function of unbounded number of arguments which will
call the original function passing it all the arguments it was called with.
This structure also dictates you the structure of the function to be used with Y-combinator: it has to take single argument, which must be a function (which is this same function again) and return a function (of any number of arguments) which calls the function inherited from outer scope.
Well, it is known to be a little mind-boggling :)
That's because you're trying to call the function func not the function dummy.
(Hence the error "Symbol's function definition is void: func".)
You want:
(func2 func (- arg 1)))
not:
(func2 'func (- arg 1)))
You do not need to quote func in the func2 call
You are missing a recursion termination condition in func2
Here is what works for me:
(defun func2 (func arg)
(message "arg = %s" arg)
(funcall func)
(when (plusp arg)
(func2 func (- arg 1))))