I'm new to lisp and I am simply trying to have two functions called at once if a conditional returns true.
(cond
((equals (first expression) "+")
(function1 parameter)
(function2 parameter)))
In the above code, I just want function1 and function2 to be called. Any thoughts?
Common Lisp
EQUALS does not exist, EQUAL does
COND already does what you want
COND allows several calls after the test:
(cond ((equal (first expression) "+")
(do-something ...)
(do-something-more ...)))
It is called Lisp, not LISP
read a Beginner Introduction
use a Reference
Yes, progn like this:
(cond
((equals (first expression) "+")
(progn
(function1 paramter)
(function2 parameter))))
cond takes one expression to evaluate if true. In this use progn (with its arguments) is that one expression. progn, subsequently take n expressions and evaluates them.
Related
In Common Lisp, a macro definition must have been seen before the first use. This allows a macro to refer to itself, but does not allow two macros to refer to each other. The restriction is slightly awkward, but understandable; it makes the macro system quite a bit easier to implement, and to understand how the implementation works.
Is there any Lisp family language in which two macros can refer to each other?
What is a macro?
A macro is just a function which is called on code rather than data.
E.g., when you write
(defmacro report (x)
(let ((var (gensym "REPORT-")))
`(let ((,var ,x))
(format t "~&~S=<~S>~%" ',x ,var)
,var)))
you are actually defining a function which looks something like
(defun macro-report (system::<macro-form> system::<env-arg>)
(declare (cons system::<macro-form>))
(declare (ignore system::<env-arg>))
(if (not (system::list-length-in-bounds-p system::<macro-form> 2 2 nil))
(system::macro-call-error system::<macro-form>)
(let* ((x (cadr system::<macro-form>)))
(block report
(let ((var (gensym "REPORT-")))
`(let ((,var ,x)) (format t "~&~s=<~s>~%" ',x ,var) ,var))))))
I.e., when you write, say,
(report (! 12))
lisp actually passes the form (! 12) as the 1st argument to macro-report which transforms it into:
(LET ((#:REPORT-2836 (! 12)))
(FORMAT T "~&~S=<~S>~%" '(! 12) #:REPORT-2836)
#:REPORT-2836)
and only then evaluates it to print (! 12)=<479001600> and return 479001600.
Recursion in macros
There is a difference whether a macro calls itself in implementation or in expansion.
E.g., a possible implementation of the macro and is:
(defmacro my-and (&rest args)
(cond ((null args) T)
((null (cdr args)) (car args))
(t
`(if ,(car args)
(my-and ,#(cdr args))
nil))))
Note that it may expand into itself:
(macroexpand '(my-and x y z))
==> (IF X (MY-AND Y Z) NIL) ; T
As you can see, the macroexpansion contains the macro being defined.
This is not a problem, e.g., (my-and 1 2 3) correctly evaluates to 3.
However, if we try to implement a macro using itself, e.g.,
(defmacro bad-macro (code)
(1+ (bad-macro code)))
you will get an error (a stack overflow or undefined function or ...) when you try to use it, depending on the implementation.
Here's why mutually recursive macros can't work in any useful way.
Consider what a system which wants to evaluate (or compile) Lisp code for a slightly simpler Lisp than CL (so I'm avoiding some of the subtleties that happen in CL), such as the definition of a function, needs to do. It has a very small number of things it knows how to do:
it knows how to call functions;
it knows how to evaluate a few sorts of literal objects;
it has some special rules for a few sorts of forms – what CL calls 'special forms', which (again in CL-speak) are forms whose car is a special operator;
finally it knows how to look to see whether forms correspond to functions which it can call to transform the code it is trying to evaluate or compile – some of these functions are predefined but additional ones can be defined.
So the way the evaluator works is by walking over the thing it needs to evaluate looking for these source-code-transforming things, aka macros (the last case), calling their functions and then recursing on the results until it ends up with code which has none left. What's left should consist only of instances of the first three cases, which it then knows how to deal with.
So now think about what the evaluator has to do if it is evaluating the definition of the function corresponding to a macro, called a. In Cl-speak it is evaluating or compiling a's macro function (which you can get at via (macro-function 'a) in CL). Let's assume that at some point there is a form (b ...) in this code, and that b is known also to correspond to a macro.
So at some point it comes to (b ...), and it knows that in order to do this it needs to call b's macro function. It binds suitable arguments and now it needs to evaluate the definition of the body of that function ...
... and when it does this it comes across an expression like (a ...). What should it do? It needs to call a's macro function, but it can't, because it doesn't yet know what it is, because it's in the middle of working that out: it could start trying to work it out again, but this is just a loop: it's not going to get anywhere where it hasn't already been.
Well, there's a horrible trick you could do to avoid this. The infinite regress above happens because the evaluator is trying to expand all of the macros ahead of time, and so there's no base to the recursion. But let's assume that the definition of a's macro function has code which looks like this:
(if <something>
(b ...)
<something not involving b>)
Rather than doing the expand-all-the-macros-first trick, what you could do is to expand only the macros you need, just before you need their results. And if <something> turned out always to be false, then you never need to expand (b ...), so you never get into this vicious loop: the recursion bottoms out.
But this means you must always expand macros on demand: you can never do it ahead of time, and because macros expand to source code you can never compile. In other words a strategy like this is not compatible with compilation. It also means that if <something> ever turns out to be true then you'll end up in the infinite regress again.
Note that this is completely different to macros which expand to code which involves the same macro, or another macro which expands into code which uses it. Here's a definition of a macro called et which does that (it doesn't need to do this of course, this is just to see it happen):
(defmacro et (&rest forms)
(if (null forms)
't
`(et1 ,(first forms) ,(rest forms))))
(defmacro et1 (form more)
(let ((rn (make-symbol "R")))
`(let ((,rn ,form))
(if ,rn
,rn
(et ,#more)))))
Now (et a b c) expands to (et1 a (b c)) which expands to (let ((#:r a)) (if #:r #:r (et b c))) (where all the uninterned things are the same thing) and so on until you get
(let ((#:r a))
(if #:r
#:r
(let ((#:r b))
(if #:r
#:r
(let ((#:r c))
(if #:r
#:r
t))))))
Where now not all the uninterned symbols are the same
And with a plausible macro for let (let is in fact a special operator in CL) this can get turned even further into
((lambda (#:r)
(if #:r
#:r
((lambda (#:r)
(if #:r
#:r
((lambda (#:r)
(if #:r
#:r
t))
c)))
b)))
a)
And this is an example of 'things the system knows how to deal with': all that's left here is variables, lambda, a primitive conditional and function calls.
One of the nice things about CL is that, although there is a lot of useful sugar, you can still poke around in the guts of things if you like. And in particular, you still see that macros are just functions that transform source code. The following does exactly what the defmacro versions do (not quite: defmacro does the necessary cleverness to make sure the macros are available early enough: I'd need to use eval-when to do that with the below):
(setf (macro-function 'et)
(lambda (expression environment)
(declare (ignore environment))
(let ((forms (rest expression)))
(if (null forms)
't
`(et1 ,(first forms) ,(rest forms))))))
(setf (macro-function 'et1)
(lambda (expression environment)
(declare (ignore environment))
(destructuring-bind (_ form more) expression
(declare (ignore _))
(let ((rn (make-symbol "R")))
`(let ((,rn ,form))
(if ,rn
,rn
(et ,#more)))))))
There have been historic Lisp systems that allow this, at least in interpreted code.
We can allow a macro to use itself for its own definition, or two or more macros to mutually use each other, if we follow an extremely late expansion strategy.
That is to say, our macro system expands a macro call just before it is evaluated (and does that each time that same expression is evaluated).
(Such a macro expansion strategy is good for interactive development with macros. If you fix a buggy macro, then all code depending on it automatically benefits from the change, without having to be re-processed in any way.)
Under such a macro system, suppose we have a conditional like this:
(if (condition)
(macro1 ...)
(macro2 ...))
When (condition) is evaluated, then if it yields true, (macro1 ...) is evaluated, otherwise (macro2 ...). But evaluation also means expansion. Thus only one of these two macros is expanded.
This is the key to why mutual references among macros can work: we are able rely on the conditional logic to give us not only conditional evaluation, but conditional expansion also, which then allows the recursion to have ways of terminating.
For example, suppose macro A's body of code is defined with the help of macro B, and vice versa. And when a particular invocation of A is executed, it happens to hit the particular case that requires B, and so that B call is expanded by invocation of macro B. B also hits the code case that depends on A, and so it recurses into A to obtain the needed expansion. But, this time, A is called in a way that avoids requiring, again, an expansion of B; it avoids evaluating any sub-expression containing the B macro. Thus, it calculates the expansion, and returns it to B, which then calculates its expansion returns to the outermost A. A finally expands and the recursion terminates; all is well.
What blocks macros from using each other is the unconditional expansion strategy: the strategy of fully expanding entire top-level forms after they are read, so that the definitions of functions and macros contain only expanded code. In that situation there is no possibility of conditional expansion that would allow for the recursion to terminate.
Note, by the way, that a macro system which expands late doesn't recursively expand macros in a macro expansion. Suppose (mac1 x y) expands into (if x (mac2 y) (mac3 y)). Well, that's all the expansion that is done for now: the if that pops out is not a macro, so expansion stops, and evaluation proceeds. If x yields true, then mac2 is expanded, and mac3 is not.
I am relatively new to Lisp and I was trying to do a linear search on LISP. But I haven't been able to do so. I am always getting an error that says that "IF has too few parameters".
(setq a '(8 6 2 3 9 5 1))
(LET (key))
(setq key (read))
(loop
(if(= (first a) (key)))
(return t)
(return NIL)
(setq a (rest a))
)
Many problems in your code:
Globally setq an undefined variable
(let (key)) alone does nothing. If you want to define a global variable, use defparameter or defvar.
You if has only a test, and no branches. The special operator if takes a condition, a then expression and an optional else expression: (if test then [else])
If you intended to have your return inside the if, your linear search would stop at the first comparison, because of (return NIL). Indeed, what you would have written would be equivalent to (return (= (first a) key)) and the loop would not even be needed in that case. Maybe you intended to use return to return a value from the if, but if is an expression an already evaluates as a value. return exits the loop (there is an implicit (block NIL ...) around the loop).
(setq a (rest a)) is like (pop a) and would indeed be the right thing to do if you did not already returned from loop at this point.
Just to be sure, be aware that = is for comparing numbers.
The beginning of your code can be written as:
(let ((a '(8 6 2 3 9 5 1))
(key (read)))
(linear-search key a)
Then, how you perform linear-search depends on what you want to learn. There are built-in for this (find, member). You can also use some with a predicate. Loop has a thereis clause. You can even try with reduce or map with a return-from. If you want to learn do or tagbody, you will have an occasion to use (pop a).
I am trying to emulate the single namespace of scheme within common lisp, with a macro (based on Doug Hoyte's) that expands to a lambda, where every use of an f! symbol (similar to Doug Hoyte's o! and g! symbols) in the function position expands to the same expression, but with funcall added in the function position of each invocation. For example:
(fplambda (f!z x) (f!z x x))
would expand to:
(LAMBDA (F!Z X) (FUNCALL F!Z X X))
The macro currently looks like this:
(defmacro fplambda (parms &body body)
(let ((syms (remove-duplicates
(remove-if-not #'f!-symbol-p
(flatten body)))))
`(lambda ,parms
(macrolet ,(mapcar
(lambda (f)
`(,f (&rest parmlist) `(funcall ,',f ',#parmlist)))
syms))
,#body)))
but given the above input, it expands (as far as I can see) to this:
(LAMBDA (F!F X)
(MACROLET ((F!F (&REST PARMLIST) `(FUNCALL ,'F!F ',#PARMLIST))))
(F!F X X))
In the macrolet definition, F!F should not be quoted or unquoted, and parmlist should just be unquoted. What is going on?
Thanks in advance!
Your definition is mostly right. You just made two pretty simple mistakes. The first one being a mismatched paren. The macrolet does not include the body (in the output the macrolet and the body are at the same level of indentation).
As for the nested backquote, the only mistake is the quote before parmlist. Other than that everything else is correct. The comma and quote before F!F is actually correct. From the hyperspec:
"An implementation is free to interpret a backquoted form F1 as any form F2 that, when evaluated, will produce a result that is the same under equal as the result implied by the above definition". Since the inner backquote has not been expanded yet, it does not have to be free of quotes and unquotes. The expression `(,'x) is actually the same as `(x).
Nested backquotes are notoriously complicated. What is probably the easiest way to understand them is to read Steele's explanation of them.
Edit:
The answer to your question about whether it is possible to use a fplambda expression in the function position is no. From the part of the hyperspec that deals with the evaluation of code: "If the car of the compound form is not a symbol, then that car must be a lambda expression, in which case the compound form is a lambda form.". Since the car of the form, (fplambda ...), is not a lambda expression, your code is no longer valid Common Lisp code.
There is a workaround to this that I figured out, but it's kind of ugly. You can define a reader macro that will allow you to write something like ([fplambda ...] ...) and have it read as
((LAMBDA (&REST #:G1030) (APPLY (FPLAMBDA ...) #:G1030)) ...)
which would do what you want. Here is code that will allow you to do that:
(set-macro-character #\[ 'bracket-reader)
(set-macro-character #\] (get-macro-character #\)))
(defun bracket-reader (stream char)
"Read in a bracket."
(declare (ignore char))
(let ((gargs (gensym)))
`(lambda (&rest ,gargs)
(apply ,(read-delimited-list #\] stream t)
,gargs))))
The only other solution I can think of would be to use some sort of code walker (I can't help you there).
I try to do this directly to the interpret:
(setf example (coerce "blablabla" 'list))
and works fine. Infact (car example) returns #\b
but if I try this:
(defun myfun (string) ( (setf example (coerce string 'list))))
(myfun "blablabla")
I don't get the same!
How can I fix?
Remove the extra parentheses around the setf in the defun:
(defun myfun (string)
(setf example (coerce string 'list)))
Now you'll get the same. Note that the outer parenthesis have meaning. In Lisp, either it is quoted, or must be a function call. If the first element is, as in this case, a list, it cannot be a function to call, hence the error.
In elisp I can evaluate or as a function just like +.
(or nil 0 nil) ==> 0
(+ 1 0 1) ==> 2
I can use apply to apply + to a list
(apply '+ '(1 0 1)) ==> 2
So, I would think or would work the same way, but it doesn't.
(apply 'or '(nil 0 nil)) ==> error: (invalid-function or)
I imagine this comes from some internal magic used to implement the short-circuit evaluation. How can I use apply to execute the or operation over a list?
P.S. my desired application is to find out whether any elements on the command line match a particular pattern, so the important part of what I am writing is:
(apply 'or (mapcar (lambda (x) (string-match-p "pattern" x)) command-line-args))
But it doesn't work
The problem is that or is a macro (which is the "internal magic" in question), and you're right that that's done so it can do short-circuiting. If or was a function, then calling it would need to follow the usual rules for evaluating a function call: all the arguments would need to get evaluated before the call is made.
See also this question -- it's about Scheme but it's the exact same issue.
As for a solution, you should probably use some, as in:
(some (lambda (x) (string-match-p "pattern" x)) command-line-args)
Note: this uses common lisp that is not included in emacs by default. Just use (require 'cl)
If it makes you feel any better, you're in good company! This is the third question in the "Common Pitfalls" section of the Lisp FAQ:
Here's the simple, but not necessarily satisfying, answer: AND and OR are
macros, not functions; APPLY and FUNCALL can only be used to invoke
functions, not macros and special operators.
...and Eli is of course right on the money with his suggestion to use SOME:
The Common Lisp functions EVERY and SOME can be used to get the
functionality you intend when trying to apply #'AND and #'OR.
(The FAQ and this answer are mostly about Common Lisp but in this case if you omit the # character the answer is the same.)
If you don't care performance, use (eval (cons 'or '(nil 0 nil)))
When I was trying to 'apply' a macro to an argument list, I got an error that the function is unbound, which means that, 'apply' only receives a function, instead of a macro, as its first argument.
In order to fix this, I wrote a new function 'apply-macro' as follows:
(defun apply-macro (macro arg-list)
(eval
`(,macro ,#(loop for arg in arg-list
collect `(quote ,arg)))))
For instance, I wrote a macro to concatenate multiple lists together:
(defmacro conc-lists (&rest lists)
`(funcall #'concatenate 'list ,#lists))
e.g.
(conc-lists '(a b) '(c d) '(e f)) ;;=> (A B C D E F)
Now try 'apply-macro':
(apply-macro 'conc-lists '((a b) (c d) (e f)))
It works and returns the same output.
In fact, it will be expanded into:
(eval
(conc-lists (quote (a b)) (quote (c d)) (quote (e f))))
You can also pass a form to a macro:
(apply-macro 'conc-lists (maplist #'list '(a b c)))
;;=> ((A B C) (B C) (C))
Get back to your question, it's solved:
(apply-macro 'or '(nil 0 nil)) ;;=> 0
I'm only guessing here, but I think or might be one of these 20-odd 'functions' in lisp that are not really functions, since they don't always evaluate all parameters.
This makes sense to make or one of these, since if you have found one TRUE, you can stop searching. In other words, or is not a function as a form of optimization. Still only guessing though.
Eli Barzilay's answer is correct and idiomatic. I want to provide an alternative answer based on dash.el, the library I use to write terse functional-style code, when I have to work with lists. or returns the first non-nil element, nil otherwise, due to short-circuiting. Therefore simply use -first:
(-first 'identity '(nil 0 1 nil)) ; 0
(-first 'identity '(nil nil)) ; nil
identity function simply returns its argument. Which is clever, because -first applies a predicate until it returns non-nil. identity returns non-nil if the argument is itself non-nil. If you simply want to test whether there is non-nil elements in a list, use -any? instead:
(-any? 'identity '(nil 0 1 nil)) ; t
(-any? 'identity '(nil nil)) ; nil