what is the parameter to ns ? the documentation says something, but it's not clear (to me, at least)
my-new-namespace=> (doc ns)
-------------------------
clojure.core/ns
([name docstring? attr-map? references*])
Macro
Sets *ns* to the namespace named by name (unevaluated), creating it
if needed.
...the rest of the documentation was not copied
the confusion comes from all the other functions that play with namespaces:
user=> (find-ns 'my-new-namespace)
nil
user=> (remove-ns 'my-new-namespace)
nil
user=> (create-ns 'my-new-namespace)
#<Namespace my-new-namespace>
user=> (ns 'my-new-namespace)
java.lang.ClassCastException: clojure.lang.PersistentList cannot be cast to clojure.lang.Symbol (NO_SOURCE_FILE:26)
user=> (ns my-new-namespace)
nil
my-new-namespace=>
find-ns, create-ns, remove-ns take my-new-namespace quoted, while ns takes my-new-namespace unquoted
so, what is the deal ? why some get a quoted form and others get and unquoted form of my-new-namespace ?
what is, in each case, my-new-namespace ?
ps: think i figured out the answer to this as i was writing the question here, but it seemed and interesting exercise, so the question still got posted :D
ns is a macro and under the hood if quotes the name to save you the trouble:
user> (macroexpand '(ns foo))
(do (clojure.core/in-ns (quote foo)) (clojure.core/with-loading-context
(clojure.core/refer (quote clojure.core))))
the other functions cant get away with unquoted arguments because they are functions, and functions have there arguments evaluated first.
the ns macro, as a macro, gets to look at and twiddle its arguments before they are evaluated so it can get at the my-new-namespace before the reader attempts to look it up as a variable name.
in short this saves wear and tare on your keyboard and wrists ;) Just kidding, the in both cases what the function receives is a symbol they just differ in what you have to do to pass them that symbol.
for comparason; if you where to go ahead and pass (ns 'my-new-namespace) it would get the symbol 'my-new-namespace quote and all! you likely don't want the quote to be part of the name space.
if you want to make it consistent you could write wrapper macros that take unquoted symbols for the rest of the namespace functions that simply quote the first argument and return a call to the real function (remember that macros return the code to be called)
(defmacro my-find-ns [name]
`(find-ns (quote ~name)))
though this would break "the first rule of macro club"
the error about the argument being a list is caused by it attempting to use (quote my-new-namespace) as a name
user> (macroexpand '(ns 'foo))
(do (clojure.core/in-ns (quote (quote foo))) (clojure.core/with-loading-context
(clojure.core/refer (quote clojure.core))))
Related
I'm studying Clojure, and I've read that in Clojure a function definition is just data, i.e. parameters vector is just an ordinary vector. If that's the case, why can I do this
(def add (fn [a b]
(+ a b)))
but not this
(def vector-of-symbols [a b])
?
I know I normally would have to escape symbols like this:
(def vector-of-symbols [`a `b])
but why don't I have to do it in fn/defn? I assume this is due to fn/defn being macros. I tried examining their source, but they are too advanced for me so far. My attempts to recreate defn also fail, and I'm not sure why (I took example from a tutorial):
(defmacro defn2 [name param & body]
`(def ~name (fn ~param ~#body)))
(defn2 add [a b] (+ a b)) ;;I get "Use of undeclared Var app.core/defn2"
Can someone please explain, how exactly does Clojure turn data structures, especially symbols, into code? And what am I missing about the macro example?
Update Apparently, macro does not work because my project is actually in Clojurescript (in Clojure it does work). I did not think it matters, but as I progress - I discover more and more things that somehow don't work for me in with Clojurescript.
Update 2 This helps: https://www.clojurescript.org/about/differences
A function is a first-class citizen as other data in Clojure.
To define a vector you use (vector ...) or reader has syntaxic sugar [...], for a list it's (list ...) or '(...) the quote not to evaluate the list as a function call, for a set (set ...) or #{...}.
So the factory function for a function is fn (in fact fn*, that comes from Java core of Clojure, fn is a series of macros to manage to destructure and all).
(fn args body)
is a function call that returns a function, where args is a vector of argument(s) event. empty and body is a series of Clojure expressions to be evaluated with args bind to the environment. If nothing is to be evaluated it returns nil. There is also a syntactic sugar #(...) with %x as argument x and % as argument 1.
(fn ...) return a value that is a function. So
(def my-super-function (fn [a b c d] (println "coucou") (+ a b c d)))
binds the symbol my-super-function with the anonymous function returned by (fn [a b c d] (println "coucou") (+ a b c d)).
(def my_vector [1 2 3])
binds the symbol my_vector with the vector [1 2 3]
List of learning resources: https://github.com/io-tupelo/clj-template#documentation
As #jas said, your defn2 macro looks fine.
The main point is that macros are an advanced feature that one almost never needs. A macro is equivalent to a compiler extension, and that is almost never the best solution to a problem. Also keep in mind that functions can do some things macros can't.
Another point: the syntax-quote (aka backquote) ` is very different from a single quote '. In your example you want the single quote for ['a 'b]. Even better would be to quote the entire vector form '[a b].
As to your primary question, it is poorly explained how source-file text is converted into code. This is a 2-step process. The Clojure Reader consumes text string data (from a file or a literal string) and produces data structures like lists, vectors, strings, numbers, symbols. The Clojure compiler takes these data structures as input and produces java byte code that can be executed.
It is confusing because, when printed, one can't tell the difference between the text representation of a vector [1 2 3] and the text string that is input to the reader [1 2 3]. Ideally it would be color-coded or something. This problem doesn't exist in Java, etc since they don't have macros and hence there is no confusion between the source code (text) and the data structures used by a macro (not text).
For a more detailed answer on creating macros in Clojure, please see this answer.
I would like to create a nil function that takes any number of symbols and sets them all to nil.
(defun clean (as many args as given by user)
(setq each-arg nil)
)
(clean x y z)
How to do this 'cleanly'?
Since you're not quoting the arguments, it has to be a macro:
(defmacro clean (&rest symbols)
`(progn
,#(mapcar (lambda (sym) (list 'setq sym 'nil))
symbols)))
Similar idea as Dmitry, but generates slightly less code:
(defmacro clean (&rest variables)
`(setq ,#(loop for var in variables nconc (list var nil))))
(macroexpand '(clean a b c d))
;; (setq a nil b nil c nil d nil)
Regarding your other questions:
simple, but time consuming way to know is to move the point to the function that you don't know and C-h f or M-xdescribe-function this will put the function name in the prompt (if it is indeed an Emacs Lisp function) and show the description and, if availably the location in the source code.
I'll try to explain, but I'm no language reference :)
defmacro - is similar to function, but it doesn't evaluate arguments. Macros are executed when your code is read and compiled into the bytecode. Their primary goal is to generate other code.
lambda - is a macro that creates an anonymous function and returns it.
mapcar - is a high-order function that applies a function to all elements of the list in succession and collects the result into a list in the order it applied the function.
&rest - is a special keyword in the function's lambda-list (i.e. the definition of parameters) which means literally that the identifier following this symbol is a list of all arguments on the right of it).
,# is a special operator used in macros, inside back-quote macros. It instructs the reader that the expression following it must be evaluated, treated as list, and all of its conses must be appended to the form that is being currently parsed.
There's something I don't understand about anonymous functions using the short notation #(..)
The following works:
REPL> ((fn [s] s) "Eh")
"Eh"
But this doesn't:
REPL> (#(%) "Eh")
This works:
REPL> (#(str %) "Eh")
"Eh"
What I don't understand is why (#(%) "Eh") doesn't work and at the same time I don't need to use str in ((fn [s] s) "Eh")
They're both anonymous functions and they both take, here, one parameter. Why does the shorthand notation need a function while the other notation doesn't?
#(...)
is shorthand for
(fn [arg1 arg2 ...] (...))
(where the number of argN depends on how many %N you have in the body). So when you write:
#(%)
it's translated to:
(fn [arg1] (arg1))
Notice that this is different from your first anonymous function, which is like:
(fn [arg1] arg1)
Your version returns arg1 as a value, the version that comes from expanding the shorthand tries to call it as a function. You get an error because a string is not a valid function.
Since the shorthand supplies a set of parentheses around the body, it can only be used to execute a single function call or special form.
As the other answers have already very nicely pointed out, the #(%) you posted actually expands to something like (fn [arg1] (arg1)), which is not at all the same as (fn [arg1] arg1).
#John Flatness pointed out that you can just use identity, but if you're looking for a way to write identity using the #(...) dispatch macro, you can do it like this:
#(-> %)
By combining the #(...) dispatch macro with the -> threading macro it gets expanded to something like (fn [arg1] (-> arg1)), which expands again to (fn [arg1] arg1), which is just want you wanted. I also find the -> and #(...) macro combo helpful for writing simple functions that return vectors, e.g.:
#(-> [%2 %1])
When you use #(...), you can imagine you're instead writing (fn [args] (...)), including the parentheses you started right after the pound.
So, your non-working example converts to:
((fn [s] (s)) "Eh")
which obviously doesn't work because the you're trying to call the string "Eh". Your example with str works because now your function is (str s) instead of (s). (identity s) would be the closer analogue to your first example, since it won't coerce to str.
It makes sense if you think about it, since other than this totally minimal example, every anonymous function is going to call something, so it'd be a little foolish to require another nested set of parens to actually make a call.
If you're in doubt what your anonymous function gets converted to, you can use the macroexpand procedure to get the representation. Remember to quote your expression before passing it to macroexpand. In this case we could do:
(macroexpand '#(%))
# => (fn* [p1__281#] (p1__281#))
This might print different names for p1__281# which are representations of the variables %.
You can also macroexpand the full invocation.
(macroexpand '(#(%) "Eh"))
# => ((fn* [p1__331#] (p1__331#)) "Eh")
Converted to more human readable by replacing the cryptic variable names by short names. We get what the accepted answers have reported.
# => ((fn* [s] (s)) "Eh")
Resources:
https://clojure.org/guides/weird_characters#_n_anonymous_function_arguments
https://clojuredocs.org/clojure.core/macroexpand
I am trying to determine whether a given argument within a macro is a function, something like
(defmacro call-special? [a b]
(if (ifn? a)
`(~a ~b)
`(-> ~b ~a)))
So that the following two calls would both generate "Hello World"
(call-special #(println % " World") "Hello")
(call-special (println " World") "Hello")
However, I can't figure out how to convert "a" into something that ifn? can understand. Any help is appreciated.
You might want to ask yourself why you want to define call-special? in this way. It doesn't seem particularly useful and doesn't even save you any typing - do you really need a macro to do this?
Having said that, if you are determined to make it work then one option would be to look inside a and see if it is a function definition:
(defmacro call-special? [a b]
(if (#{'fn 'fn*} (first a))
`(~a ~b)
`(-> ~b ~a)))
This works because #() function literals are expanded into a form as follows:
(macroexpand `#(println % " World"))
=> (fn* [p1__2609__2610__auto__]
(clojure.core/println p1__2609__2610__auto__ " World"))
I still think this solution is rather ugly and prone to failure once you start doing more complicated things (e.g. using nested macros to generate your functions)
First, a couple of points:
Macros are simply functions that receive as input [literals, symbols, or collections of literals and symbols], and output [literals, symbols, or collections of literals and symbols]. Arguments are never functions, so you could never directly check the function the symbol maps to.
(call-special #(println % " World") "Hello") contains reader macro code. Since reader macros are executed before regular macros, you should expand this before doing any more analysis. Do this by applying (read-string "(call-special #(println % \" World\") \"Hello\")") which becomes (call-special (fn* [p1__417#] (println p1__417# "world")) "Hello").
While generally speaking, it's not obvious when you would want to use something when you should probably use alternative methods, here's how I would approach it.
You'll need to call macroexpand-all on a. If the code eventually becomes a (fn*) form, then it is guaranteed to be a function. Then you can safely emit (~a ~b). If it macroexpands to eventually be a symbol, you can also emit (~a ~b). If the symbol wasn't a function, then an error would throw at runtime. Lastly, if it macroexpands into a list (a function call or special form call), like (println ...), then you can emit code that uses the thread macro ->.
You can also cover the cases such as when the form macroexpands into a data structure, but you haven't specified the desired behavior.
a in your macro is just a clojure list data structure (it is not a function yet). So basically you need to check whether the data structure a will result is a function or not when it is evaluated, which can be done like show below:
(defmacro call-special? [a b]
(if (or (= (first a) 'fn) (= (first a) 'fn*))
`(~a ~b)
`(-> ~b ~a)))
By checking whether the first element of the a is symbol fn* or fn
which is used to create functions.
This macro will only work for 2 cases: either you pass it a anonymous function or an expression.
can i somehow find all functions/macros that take a specific type of parameter ?
for example, what function accepts a Namespace object as parameter ?
(this is because i can create a namespace and store it in a var, but i don't know where i could use that var; what function might i pass that var to ?)
here is some code:
user=> (def working-namespace (create-ns 'my-namespace))
#'user/working-namespace
;i created a namspace and want to use it later
user=> (class working-namespace)
clojure.lang.Namespace
; out of curiosity i found out that "working-namespace" is a Namespace object
user=> (ns working-namespace)
nil
working-namespace=>
; but how do i switch to it ? this didn't do what i wanted...
user=> (refer working-namespace)
java.lang.ClassCastException: clojure.lang.Namespace cannot be cast to clojure.lang.Symbol (NO_SOURCE_FILE:0)
; this did not work either as my expectations
user=> (the-ns working-namespace)
#<Namespace my-namespace>
user=> (class (the-ns working-namespace))
clojure.lang.Namespace
; great, this gave me the same thing, a Namespace
hence the question: how do i use it dynamically (that's why i had put my namespace into a var) ? how do i get something useful for me from a var that points to a namespace ?
i can try look around for functions that make use of a Namespace object or that convert it to something else. i did and only found "intern". searching by hand not seems not that promising
what if i have this problem a million time ? is there an automated way to get me what i'm looking for without having to ask around each time ?
In Clojure 1.2 and previous function arguments dont have types. every function argument is an object. So the question really becomes "how do i find functions that will cast the object I pass them into this type. so searching for type hints will find some of them, though it wont get you everything. I wish it where more possible to answer this in general.
starting with 1.3 (current dev branch 9/2010) function paramerters and return types can have a defined type and will be passed/returned as that type instead of being cast to object and then cast on the other side. This drops one of the zeros from the exacution time of numerical functions with the important limitation that it only works for :static functions and only with direct calls (ie: not through map/reduce/filter/etc.) There is not a lot published on this change yet though it has the important breaking change that integers are no longer boxed by default and integer (actually Long) overflow throws an exception. you can read more here
(defn ^:static fib ^long [^long n]
(if (<= n 1)
1
(+ (fib (dec n)) (fib (- n 2)))))
so after 1.3 is released and widely adopted you will see code with more commonly defined types because they will offer a big speed benefit and then you will be able to find more functions by argument type though still not all of them.
At the same lecture where I learned about function argument types, Rich mentioned plans in the distant Clojure future (after 'Clojure in Clojure') about better support for exposing the compiler internals to tools such as IDEs. So there is hope that someday you will get a real answer to this question.
Dynamic languages make this slightly more difficult in practice and a lot harder in theory.
You already got a good answer from Arthur, so I'll only answer the "how do i get something useful for me from a var that points to a namespace ?". From (doc ns), note that it says unevaluated:
user=> (doc ns)
-------------------------
clojure.core/ns
([name docstring? attr-map? references*])
Macro
Sets *ns* to the namespace named by name (unevaluated), creating it
Now there's something you could do with in-ns if you want (the whole namespace object -> string -> symbol conversion is probably stupid, but enough to illustrate my point):
user=> (in-ns (symbol (str working-namespace)))
#<Namespace my-namespace>
my-namespace=>
I don't think you can do that without a lot of hackery in a dynamic language. If you want to now what function that take namespaces look at the documentation of the namespace stuff.
For example cleaning namespaces or reload them.
You wrote:
user=> (ns working-namespace)
nil
working-namespace=>
; but how do i switch to it ? this didn't do what i wanted...
But you did switch to the working-namespace namespace (that's why the prompt changed), so I'm not clear as to what "you wanted".
As I noted earlier, you need to present the ultimate problem are you trying to solve. It's entirely likely that messing with namespace objects won't be the solution.