Is there a way to only get one of the results returned from values? I tried doing (define x (first (values 1 2))) but that didn't work. Is the only way to write something like (define-values (x dont-care) (values 1 2)) and define an extra variable?
Is the only way to write something like (define-values (x dont-care) (values 1 2)) and define an extra variable?
The only ways to get one of the values out of a multiple-valued thing all involve defining extra variables like dont-care here. However, you can use scoping, either local-scope or macro-scope, to mitigate this and hide them away.
You can use let-values to limit the scope like this:
(define x (let-values ([(x dont-care) (values 1 2)]) x))
Or you can use a macro to "hide" the other variables:
(define-syntax-rule (define-first-of-two-values x expr/2vs)
(define-values (x dont-care) expr/2vs))
(define-first-of-two-values x (values 1 2))
After this definition x will be available, but dont-care will not because its limited to the scope of the macro definition.
You can also use an existing macro such as match-define-values, with the _ wildcard for the values you don't care about.
(match-define-values (x _) (values 1 2))
The match-define-values macro actually expands into something very similar to the first let-values example above.
I've been using this macro:
(define-syntax-rule (first-value expr)
(call-with-values (λ () expr)
(λ (result . ignored) result)))
> (first-value (values 10 20 30 40 50))
10
This works with any number of values. See call-with-values for documentation.
Related
CL-USER> (a-sum 0 3)
->> 6
I wrote this program :
(defun a-sum (x y)
(if (and (> x -1) (> y -1))
(do ((i 0 (1+ i))
(sum 0)
(num x))
((equal i (+ (- y x) 1)))
(setq sum (+ sum num))
(setq num (+ num 1))
sum)
(print " NOPE")))
put if I run it in the terminal it returns nil and not the answer stated above;
can someone help with the problem so it returns the value then Boolean.
DO,DO* Syntax
The entry for DO,DO* says that the syntax is as follows:
do ({var | (var [init-form [step-form]])}*)
(end-test-form result-form*)
declaration*
{tag | statement}*
The body is used as a list of statements and no intermediate value in this body is used as the result form of the do form. Instead, the do form evaluates as the last expression in result-form*, which defaults to nil.
(do ((i 0 (1+ i))
(sum 0)
(num x))
((equal i (+ (- y x) 1))
;;; RESULT FORMS HERE
)
(setq sum (+ sum num)) ;; (*)
(setq num (+ num 1)) ;; (*)
sum ;; (*)
)
All the expressions marked commented (*) above are used for side-effects only: the result of their evaluation is unused and discarded.
Problem statement
It is not clear to me what Σpi=ni means, and your code does not seem to compute something that could be expressed as that mathematical expression.
One red flag for example is that if (+ (- y x) 1) is negative (i.e. if y < x-1, for example y=1,x=3), then your loop never terminates because i, which is positive or null, will never be equal to the other term which is negative.
I would try to rewrite the problem statement more clearly, and maybe try first a recursive version of your algorithm (whichever is easier to express).
Remarks
Please indent/format your code.
Instead of adding setq statements in the body, try to see if you can define them in the iteration clauses of the loop (since I'm not sure what you are trying to achieve, the following example is only a rewrite of your code):
(do ((i 0 (1+ i))
(sum 0 (+ sum num)
(num x (1+ num))
(... sum))
Consider what value(s) a function returns. It's the value of the last form evaluated. In your case, that appears to be a do or maybe a setq or print (It's difficult to read as it's formatted now, and I don't have question edit privileges).
In short, the form that's returning the value for the function looks to be one evaluated for side-effects instead of returning a value.
I'm new to lisp programming and I'm trying to create a program that accept six numbers and check whether each number is either odd or even number.
(princ"Input six number: ")
(setq a(read))
(setq b(read))
(setq c(read))
(setq d(read))
(setq e(read))
(setq f(read))
(format t "~% ~d" a)
(format t "~% ~d" b)
(format t "~% ~d" c)
(format t "~% ~d" d)
(format t "~% ~d" e)
(format t "~% ~d" f)
(if(= 0(mod a 2))
(print"even")
(print"odd"))
(if(= 0(mod b 2))
(print"even")
(print"odd"))
(if(= 0(mod c 2))
(print"even")
(print"odd"))
(if(= 0(mod d 2))
(print"even")
(print"odd"))
(if(= 0(mod e 2))
(print"even")
(print"odd"))
(if(= 0(mod f 2))
(print"even")
(print"odd"))
(terpri)
You have a lot of code that looks like this:
(if(= 0(mod ... 2))
(print"even")
(print"odd"))
(this might be a copy/paste problem but in your question they are indented more and more to the right. They are however not nested, they are all at the same depth (they are toplevel expressions) so by convention they should not be indented).
A first step would be to factor them using a function like this one:
(defun check-even-odd (number)
(if (= 0 (mod number 2))
(print "even")
(print "odd")))
The above defines a function named check-even-odd of one parameter number, and it applies the same logic you originally had for any arbitrary number.
The rest of your code can be simplified as:
(check-even-odd a)
(check-even-odd b)
(check-even-odd c)
(check-even-odd d)
(check-even-odd e)
(check-even-odd f)
Now, you can define two additional global variables:
(defparameter total-even 0)
(defparameter total-odd 0)
Each of them holds a sum, and is initialized to 0.
You can rewrite the check-even-odd function as follows to update the counters. First of all, let's just rewrite the current code by using cond, since we are going to need to perform multiple actions in each case, and if only accepts one expression for each branch (the combination if and progn is a bit ugly):
(defun check-even-odd (number)
(cond
((= 0 (mod number 2))
(print "even"))
(t
(print "odd"))))
This above behaves as the original code.
In order to increment a variable by a certain amount, you can use INCF:
(defun check-even-odd (number)
(cond
((= 0 (mod number 2))
(print "even")
(incf total-even number))
(t
(print "odd")
(incf total-odd number))))
When you execute your whole script, the total will be initialized to zero, then each call to check-even-odd will add the number to the appropriate counter.
Notes:
You may find other places where you can use functions to abstract duplicated code
You should use defparameter instead of setq when initializing a, b, etc, because otherwise the variables are not declared and calling setq on undeclared variables is not standard
In fact, it is possible to rewrite the whole program without having any global state, this could be a good next exercise
You can generalize for less or more numbers instead of 6, you would need to write a loop or a recursive function to repeat the same code an arbitrary amount of time
Input/output may need to be flushed (see finish-output, clear-input) otherwise you could experience strange behavior when the underlying stream is buffered.
How do I get this piece of macro to function as intended? -- I'd like to capture p from the lexical environment without having to send it to the macro as an argument.
(define-syntax-rule (fi a b)
(if p a b)) ;--->capture `p` from lexical env
(let ((p #t))
(fi 1 2))
Bonus thanks -- How would I do the same in CL?
In Common Lisp a macro is simply a function that takes as input the list structure of the code and returns a list structure representing the new code.
(defmacro fi (a b)
`(if p ,a ,b))
So if you were to use fi like this:
(let ((p t)) ; Common Lisp uses 't' for truth.
(fi 1 2))
It is as if you had typed:
(let ((p t))
(if p 1 2))
To see how you would get this expansion, imagine fi was a function and you gave it the arguments of 1 and 2.
(fi 1 2) => (if p 1 2)
Then took the list structure it returned and substituted it with the call to fi.
The example you give is simple because the arguments evaluate to themselves. If you had something more complicated like the expressions (* 1 1) and (+ 1 1), the actual list structure is passed in (the value of a is the list (* 1 1), and the value of b is the list (+ 1 1))
(fi (* 1 1) (+ 1 1)) => (if p (* 1 1) (+ 1 1))
You can't capture local bindings with syntax-rules. You can use syntax-case for that, though:
(define-syntax fi
(lambda (stx)
(syntax-case stx ()
((_ a b)
(with-syntax ((p (datum->syntax stx #'p)))
#'(if p a b))))))
However, using datum->syntax to capture identifiers of a fixed name like this is not ideal. If you're using Racket, it's better to use syntax parameters for this.
For Scheme implementations that don't have syntax-case but have explicit renaming, you might write the macro this way:
(define-syntax fi
(er-macro-transformer
(lambda (exp rename compare)
`(,(rename 'if) p ,(cadr exp) ,(caddr exp)))))
Some people find it simpler, but the onus is on you to rename everything that you're not intentionally capturing. In this case, we're explicitly renaming if; for most other macros that use lambda, let, etc., those all must be renamed.
Problem: How to handle a catch-all parameter after & in a macro, when the arguments to be passed are sequences, and the catch-all variable needs to be dealt with as a sequence of sequences? What gets listed in the catch-all variable are literal expressions.
This is a macro that's intended to behave roughly Common Lisp's mapc, i.e. to do what Clojure's map does, but only for side-effects, and without laziness:
(defmacro domap [f & colls]
`(dotimes [i# (apply min (map count '(~#colls)))]
(apply ~f (map #(nth % i#) '(~#colls)))))
I've come to realize that this is not a good way to write domap--I got good advice about that in this question. However, I'm still wondering about the tricky macro problem that I encountered along the way.
This works if the collection is passed as a literal:
user=> (domap println [0 1 2])
0
1
2
nil
But doesn't work in other situations like this one:
user=> (domap println (range 3))
range
3
nil
Or this one:
user=> (def nums [0 1 2])
#'user/nums
user=> (domap println nums)
UnsupportedOperationException count not supported on this type: Symbol clojure.lang.RT.countFro (RT.java:556)
The problem is that it's literal expressions that are inside colls. This is why the macro domap works when passed a sequence of integers, but not in other situations. Notice the instances of '(nums):
user=> (pprint (macroexpand-1 '(domap println nums)))
(clojure.core/dotimes
[i__199__auto__
(clojure.core/apply
clojure.core/min
(clojure.core/map clojure.core/count '(nums)))]
(clojure.core/apply
println
(clojure.core/map
(fn*
[p1__198__200__auto__]
(clojure.core/nth p1__198__200__auto__ i__199__auto__))
'(nums))))
I've tried various combinations of ~, ~#, ', let with var#, etc. Nothing's worked. Maybe it's a mistake to try to write this as a macro, but I'd still be curious how to write a variadic macro that takes complex arguments like these.
Here is why your macro does not work:
'(~#colls) This expression creates a quoted list of all colls. E. g. if you pass it (range 3), this expression becomes '((range 3)), so the literal argument will be one of your colls, preventing evaluation of (range 3) certainly not what you want here.
Now if you would not quote (~#colls) inside the macro, of course they would become a literal function invocation like ((range 3)), which makes the compiler throw after macroexpansion time (it will try to eval ((0 1 2))).
You can use list to avoid this problem:
(defmacro domap [f & colls]
`(dotimes [i# (apply min (map count (list ~#colls)))]
(apply ~f (map #(nth % i#) (list ~#colls)))))
=> (domap println (range 3))
0
1
2
However one thing here is terrible: Inside the macro, the entire list is created twice. Here is how we could avoid that:
(defmacro domap [f & colls]
`(let [colls# (list ~#colls)]
(dotimes [i# (apply min (map count colls#))]
(apply ~f (map #(nth % i#) colls#)))))
The colls are not the only thing that we need to prevent from being evaluated multiple times. If the user passes something like (fn [& args] ...) as f, that lambda would also be compiled in every step.
Now this is the exactly the scenario where you should ask yourself why you are writing a macro. Essentially, your macro has to make sure all arguments are eval'd without transforming them in any way before. Evaluation comes gratis with functions, so let's write it as a function instead:
(defn domap [f & colls]
(dotimes [i (apply min (map count colls))]
(apply f (map #(nth % i) colls))))
Given what you want to achieve, notice there is a function to solve that already, dorun which simply realizes a seq but does not retain the head. E. g.:
`(dorun (map println (range 3)))
would do the trick as well.
Now that you have dorun and map, you can simply compose them using comp to achieve your goal:
(def domap (comp dorun map))
=> (domap println (range 3) (range 10) (range 3))
0 0 0
1 1 1
2 2 2
I decided to learn LISP today, and have been playing around with it for a bit. I wrote a simple baby function just to test my understanding, and now understand that my understanding doesn't understand as much as I had understood it to understand. :D
Anyway, here is the function. The idea is that when it is called, e.g. (esexp base x) it should return the value base^x. So (esexp 3 4) = 3^4 = 81.
(I am using the CMU implementation of Common Lisp, if that matters.)
(defun esexp (base x)
(if (= x 0)
1
(if (< x 0)
(/ esexp (base (+ x 1)) base)
(* esexp (base (+ x 1)) base))))
This doesn't work. I get errors that look like (Warning: This variable is undefined: SLBEXP) and (Error in KERNEL::UNBOUND-SYMBOL-ERROR-HANDLER: the variable SLBEXP is unbound.) So. What am I doing wrong? AND would there be a better (or more LISP-ish way) to write this function?
ETA
Corrected code:
(defun esexp (base x)
(if (= x 0)
1
(if (< x 0)
(/ (esexp base (+ x 1)) base)
(* (esexp base (- x 1)) base))))
esexp(base (+ x 1))
should be
(esexp base (+ x 1))
esexp is a function just like +. The syntax for invoking a function is
(function-name arg1 arg2 ...)
The technical explanation of the error: the compiler was parsing the code:
(/ esexp(base (+ x 1)) base)
as this:
(/ esexp (base (+ x 1)) base)
which says:
first, add 1 to the parameter x
then, invoke a function called base with the result above.
divide the value of a variable called esexp by the result above.
then, divide that result by the parameter base. (The divide operator in Common Lisp can take more than two arguments.)
You see the trick? When a word appears as the first item in a s-expression (and that s-expression isn't quoted), it's usually treated as the name of a function you want to invoke. Otherwise, it's treated as the name of a variable whose value you want to retrieve. Further, in Common Lisp a symbol like esexp can be bound to both a function value and a variable value at the same time; context or special expressions like #'esexp (which means the function) are used to figure out which one you mean.
The compiler was telling you that, though esexp was bound to a function with your defun statement, it had not yet been bound to a variable value, and therefore could not be used as such. Hence, the error.
Just a note about the code. I believe it should be
(defun esexp (base x)
(if (= x 0)
1
(if (< x 0)
(/ (esexp(base (+ x 1)) base))
(* (esexp(base (- x 1)) base))))
Otherwise the function will never terminate. (you had (* (esexp(base (+ x 1)) base)))))