Here's a function that asks a number and returns the value if its type is indeed a number and else executes the function again:
(defun ask-number ()
(format t "Please enter a number.~%")
(let ((val (read)))
(if (numberp val)
val
(ask-number))))
I understand that after the value is read, it is labelled as val and the whole ((val (read))) is an argument of let. What I don't understand is, why the if-statement is nested within let. I would have assumed that the program should be something like this:
(defun ask-number ()
(format t "Please enter a number.~%")
(let ((val (read))))
(if (numberp val)
val
(ask-number)))
which leads to an error. I am not sure why this happens.
The reason the if is inside the let is that the val you've created with the let is only valid within the let; once you exit the let, val doesn't exist any more.
let is syntactic sugar for creating and instantly calling a lambda expression, so your let expression is basically the same as:
((lambda (val)
(if (numberp val)
val
(ask-number)))
(read))
Related
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.
I'm doing a tutorial on emacs lisp, and it's talking about the let function.
;; You can bind a value to a local variable with `let':
(let ((local-name "you"))
(switch-to-buffer-other-window "*test*")
(erase-buffer)
(hello local-name)
(other-window 1))
I don't understand the role of the double parentheses after let in the first line. What are they doing that a single set wouldn't do? Running that section without them, I get an error: Wrong type argument: listp, "you".
You can introduce multiple variables there. The outer parentheses delimit the list of bindings, the inner the individual binding form.
(let ((foo "one")
(bar "two"))
(frobnicate foo bar))
There are not "double parens".
Presumably, you are thinking of (let ((foo...)...)), and you mean the (( that come after let? If so, consider this:
(let (a b c) (setq a 42)...)
IOW, let declares local variables. It may also bind them. In the previous sexp, it declares a, b, and c, but it doesn't bind any of them, leaving it to the let body to give them values.
An example that declares two variables but binds only one of them (a):
(let ((a 42) b) ... (setq b ...) ...)
According to gnu.org, it looks like you can construct and initialize multiple variables with one let statement, so the double parenthesis is there to allow the separation between the variables.
If the varlist is composed of two-element lists, as is often the case, the template for the let expression looks like this:
(let ((variable value)
(variable value)
…)
body…)
The let special form takes a list of bindings: (let (<binding-form> ...) <body>).
The binding form is one of <symbol> (denoting a variable bound to the value nil) or a list (<symbol> <value>) (where value is computed when the let is entered).
The difference between let and let* is how the "value" bits are executed. For plain let, they're executed before any of the values are bound:
(let ((a 17)
(b 42))
(let ((a b) ; Inner LET
(b a))
(list a b)))
Whereas let* executes the binding forms one after another. Both have their places, but you can get by with only using let since (let* (<form1> <form2>...) is equivalent to (let (<form1>) (let (<form2>) ...))
I'm trying to write a macro that will let me streamline the definition of multiple top-level variables in one single expression.
The idea was to make it work similar to how let works:
(defparameters ((*foo* 42)
(*bar* 31)
(*baz* 99)))
I tried using the following, but it doesn't seem to do anything.
(defmacro defparameters (exprs)
(dolist (expr exprs)
(let ((name (car expr))
(exp (cadr expr)))
`(defparameter ,name ,exp))))
I've tried using macroexpand but it doesn't seem to expand at all.
What am I doing wrong? and how can I fix it?
The return value of a dolist is given by its optional third argument, so your macro returns the default of nil.
Macros only return one form, so when you have multiple things, such as your series of defparameters, you need to wrap them all in some form and return that. progn will be suitable here. For Example:
(defmacro defparameters (exprs)
`(progn ,#(loop for (name exp) in exprs
collect `(defparameter ,name ,exp))))
I have various functions and I want to call each function with the same value. For instance,
I have these functions:
(defun OP1 (arg) ( + 1 arg) )
(defun OP2 (arg) ( + 2 arg) )
(defun OP3 (arg) ( + 3 arg) )
And a list containing the name of each function:
(defconstant *OPERATORS* '(OP1 OP2 OP3))
So far, I'm trying:
(defun TEST (argument) (dolist (n *OPERATORS*) (n argument) ) )
I've tried using eval, mapcar, and apply, but these haven't worked.
This is just a simplified example; the program that I'm writing has eight functions that are needed to expand nodes in a search tree, but for the moment, this example should suffice.
Other answers have provided some idiomatic solutions with mapcar. One pointed out that you might want a list of functions (which *operators* isn't) instead of a list of symbols (which *operators* is), but it's OK in Common Lisp to funcall a symbol. It's probably more common to use some kind of mapping construction (e.g., mapcar) for this, but since you've provided code using dolist, I think it's worth looking at how you can do this iteratively, too. Let's cover the (probably more idiomatic) solution with mapping first, though.
Mapping
You have a fixed argument, argument, and you want to be able to take a function function and call it with that `argument. We can abstract this as a function:
(lambda (function)
(funcall function argument))
Now, we want to call this function with each of the operations that you've defined. This is simple to do with mapcar:
(defun test (argument)
(mapcar (lambda (function)
(funcall function argument))
*operators*))
Instead of operators, you could also write '(op1 op2 op3) or (list 'op1 'op2 'op3), which are lists of symbols, or (list #'op1 #'op2 #'op3) which is a list of functions. All of these work because funcall takes a function designator as its first argument, and a function designator is
an object that denotes a function and that is one of: a symbol (denoting the function named by that symbol in the global environment), or a function (denoting itself).
Iteratively
You can do this using dolist. The [documentation for actually shows that dolist has a few more tricks up its sleeve. The full syntax is from the documentation
dolist (var list-form [result-form]) declaration* {tag | statement}*
We don't need to worry about declarations here, and we won't be using any tags, but notice that optional result-form. You can specify a form to produce the value that dolist returns; you don't have to accept its default nil. The common idiom for collecting values into a list in an iterative loop is to push each value into a new list, and then return the reverse of that list. Since the new list doesn't share structure with anything else, we usually reverse it destructively using nreverse. Your loop would become
(defun test (argument)
(let ((results '()))
(dolist (op *operators* (nreverse results))
(push (funcall op argument) results))))
Stylistically, I don't like that let that just introduces a single value, and would probably use an &aux variable in the function (but this is a matter of taste, not correctness):
(defun test (argument &aux (results '()))
(dolist (op *operators* (nreverse results))
(push (funcall op argument) results)))
You could also conveniently use loop for this:
(defun test2 (argument)
(loop for op in *operators*
collect (funcall op argument)))
You can also do somewhat succinctly, but perhaps less readably, using do:
(defun test3a (argument)
(do ((results '() (list* (funcall (first operators) argument) results))
(operators *operators* (rest operators)))
((endp operators) (nreverse results))))
This says that on the first iteration, results and operators are initialized with '() and *operators*, respectively. The loop terminates when operators is the empty list, and whenever it terminates, the return value is (nreverse results). On successive iterations, results is a assigned new value, (list* (funcall (first operators) argument) results), which is just like pushing the next value onto results, and operators is updated to (rest operators).
FUNCALL works with symbols.
From the department of silly tricks.
(defconstant *operators* '(op1 op2 o3))
(defun test (&rest arg)
(setf (cdr arg) arg)
(mapcar #'funcall *operators* arg))
There's a library, which is almost mandatory in any anywhat complex project: Alexandria. It has many useful functions, and there's also something that would make your code prettier / less verbose and more conscious.
Say, you wanted to call a number of functions with the same value. Here's how you'd do it:
(ql:quickload "alexandria")
(use-package :alexandria)
(defun example-rcurry (value)
"Calls `listp', `string' and `numberp' with VALUE and returns
a list of results"
(let ((predicates '(listp stringp numberp)))
(mapcar (rcurry #'funcall value) predicates)))
(example-rcurry 42) ;; (NIL NIL T)
(example-rcurry "42") ;; (NIL T NIL)
(defun example-compose (value)
"Calls `complexp' with the result of calling `sqrt'
with the result of calling `parse-integer' on VALUE"
(let ((predicates '(complexp sqrt parse-integer)))
(funcall (apply #'compose predicates) value)))
(example-compose "0") ;; NIL
(example-compose "-1") ;; T
Functions rcurry and compose are from Alexandria package.
I'm tring to apply closure in emacs lisp.And I find a post here:
How do I do closures in Emacs Lisp?
with some code like:
(defun foo (x) `(lambda () ,x)) (message (string (funcall (foo
66))))
But following the emacs documentation lambda should be formated like
'(lambda () x) ==> using this format ,I got an ERROR :Symbol's value as variable is void: x
When " , " is add betwenn "()" and "x" ,everything goes right .
Why?
This happens because Emacs Lisp is dynamically scoped thus foo returns a lambda where x is free. That's what the error tells you.
To do a closure in Emacs Lisp you have to use lexical-let which simulates a lexical binding and therefore allows you to make a real closure.
(defun foo (x)
(lexical-let ((x x))
(lambda () x)))
(message (string (funcall (foo 66))))
Here are some links from the Emacs Wiki:
Dynamic Binding Vs Lexical Binding
Fake Closures
Note that you could have defined x with a let like this:
(defun foo (x)
(lambda () x))
(message (string (let ((x 66)) (funcall
(foo 'i-dont-care)))))
This answer gives a bit of detail behind the first part of #Daimrod's correct answer.
Your question is why this works:
(defun foo (x) `(lambda () ,x)) (message (string (funcall (foo 66))))
and this doesn't work:
(defun foo (x) '(lambda () x)) (message (string (funcall (foo 66))))
First, the quote (') is unnecessary in the second, because in Emacs Lisp lambda forms are self-evaluating. That is, '(lambda (...) ...) generally acts the same as (lambda (...) ...).
But what does the backquote (`), instead of quote ('), do?
Inside a backquoted expression, a comma (,) means replace the next expression by its value, that is, evaluate it. So this:
`(lambda () ,x)
means create and return a list whose first element is the symbol lambda (that's not evaluated), whose second element is () (that's not evaluated), and whose third element is the value of variable x. It's equivalent to evaluating this code, which uses function list:
(list 'lambda '() x)
That's what you want: replace x by its current value (in this case, its value inside the function foo, that is, the value of foo's argument).
But this:
'(lambda () x)
(and likewise, because a lambda form is self-evaluating, (lambda () x)) returns this list: (lambda () x). And when that's evaluated using dynamic scoping (the default scoping regime in Emacs Lisp), x is unbound - it has no value. So a void-variable error is raised.