I have the following macro:
(defmacro anon-mac [value]
#(+ value 1))
I would expect for this to behave as such:
((anon-mac 1) 1) ;=> 2
However I get this error:
IllegalArgumentException No matching ctor found for class user$anon_mac$fn__10767 clojure.lang.Reflector.invokeConstructor (Reflector.java:163)
What should I do to be able to have this macro return an anonymous function that works as I would expect?
The answer must be a macro. Seeing as my question concerns the ability of macros to return anonymous functions
Why must the answer be a macro? In my case it is because I do not want this conversion to be called more than once at compile time wherever it is found. If I were to have this conversion in a for loop that called it 200 times, with a function the conversion would be run 200 times. However seeing as macros edit the code itself it would only be run once for that for loop.
I simply needed to escape the function while evaluating the inner variable like so:
(defmacro anon-mac [value] `#(+ % ~value))
Not sure what you are using it for, but if you wanted to go with a function, you might find partial helpful, as it provides the behavior that you are after.
(defn anon-partial [val] (partial + val))
((anon-partial 1) 1) ;;=> 2
There are also some helpful examples of partial at clojuredocs.org.
Related
Recently I came across a use for eval within a macro, which I understand is a bit of a faux pas but let's ignore that for now. What I found surprising, was that eval was able to resolve global vars at macroexpansion time. Below is a contrived example, just to illustrate the situation I'm referring to:
(def list-of-things (range 10))
(defmacro force-eval [args]
(apply + (eval args)))
(macroexpand-1 '(force-eval list-of-things))
; => 45
I would have expected args to resolve to the symbol list-of-things inside force-eval, and then list-of-things to be evaluated resulting in an error due to it being unbound:
"unable to resolve symbol list-of-things in this context"
However, instead list-of-things is resolved to (range 10) and no error is thrown - the macroexpansion succeeds.
Contrast this with attempting to perform the same macroexpansion, but within a local binding context:
(defmacro force-eval [args]
(apply + (eval args)))
(let [list-of-things (range 10)]
(macroexpand-1 '(force-eval list-of-things)))
; => Unable to resolve symbol: list-of-thingss in this context
Note in the above examples I'm assuming list-of-things is not previously bound, e.g. a fresh REPL. One final example illustrates why this is important:
(defmacro force-eval [args]
(apply + (eval args)))
(def list-of-things (range 10 20))
(let [list-of-thing (range 10)]
(macroexpand-1 '(force-eval list-of-things)))
; => 145
The above example shows that the locals are ignored, which is expected behavior for eval, but is a bit confusing when you are expecting the global to not be available at macroexpansion time either.
I seem to have a misunderstanding about what exactly is available at macroexpansion time. I had previously thought that essentially any binding, be it global or local, would not be available until runtime. Apparently this is an incorrect assumption. Is the answer to my confusion simply that global vars are available at macroexpansion time? Or am I missing some further nuance here?
Note: this related post closely describes a similar problem, but the focus there is more on how to avoid inappropriate use of eval. I'm mainly interested in understanding why eval works in the first example and by extension what's available to eval at macroexpansion time.
Of course, vars must be visible at compile time. That's where functions like first and + are stored. Without them, you couldn't do anything.
But keep in mind that you have to make sure to refer to them correctly. In the repl, *ns* will be bound, and so a reference to a symbol will look in the current namespace. If you are running a program through -main instead of the repl, *ns* will not be bound, and only properly qualified vars will be found. You can ensure that you qualify them correctly by using
`(force-eval list-of-things)
instead of
'(force-eval list-of-things)
Note I do not distinguish between global vars and non-global vars. All vars in Clojure are global. Local bindings are not called vars. They're called locals, or bindings, or variables, or some combination of those words.
Clojure is designed with an incremental compilation model. This is poorly documented.
In C and other traditional languages, source code must be compiled, then linked with pre-compiled libraries before the final result can be executed. Once execution begins, no changes to the code can occur until the program is terminated, when new source code can be compiled, linked, then executed. Java is normally used in this manner just like C.
With the Clojure REPL, you can start with zero source code in a live executing environment. You can call existing functions like (+ 2 3), or you can define new functions and variables on the fly (both global & local), and redefine existing functions. This is only possible because core Clojure is already available (i.e. clojure.core/+ etc is already "installed"), so you can combine these functions to define your own new functions.
The Clojure "compiler" works just like a giant REPL session. It reads and evaluates forms from your source code files one at a time, incrementally adding them the the global environment. Indeed, it is a design goal/requirement that the result of compiling and executing source code is identical to what would occur if you just pasted each entire source code file into the REPL (in proper dependency order).
Indeed, the simplest mental model for code execution in Clojure is to pretend it is an interpreter instead of a traditional compiler.
And eval in a macro makes no sense.
Because:
a macro already implicitely contains an eval
at the very final step.
If you use macroexpand-1, you make visible how the code was manipulated in the macro before the evocation of the implicite eval inside the macro.
An eval in a macro is an anti-pattern which might indicate that you should use a function instead of a macro - and in your examle this is exactly the case.
So your aim is to dynamically (in run-time) evoke sth in a macro. This you can only do through an eval applied over a macro call OR you should rather use a function.
(defmacro force-eval [args]
(apply + (eval args)))
;; What you actually mean is:
(defn force-eval [args]
(apply + args))
;; because a function in lisp evaluates its arguments
;; - before applying the function body.
;; That means: args in the function body is exactly
;; `(eval args)`!
(def list-of-things (range 10))
(let [lit-of-things (range 10 13)]
(force-eval list-of-things))
;; => 45
;; so this is exactly the behavior you wanted!
The point is, your construct is a "bad" example for a macro.
Because apply is a special function which allows you to
dynamically rearrange function call structures - so it has
some magic of macros inside it - but in run-time.
With apply you can do quite some meta programming in some cases when you just quote some of your input arguments.
(Try (force-eval '(1 2 3)) it returns 6. Because the (1 2 3) is put together with + at its front by apply and then evaluated.)
The second point - I am thinking of this answer I once gave and this to a dynamic macro call problem in Common Lisp.
In short: When you have to control two levels of evaluations inside a macro (often when you want a macro inject some code in runtime into some code), you need too use eval when calling the macro and evaluate those parts in the macro call which then should be processed in the macro.
I've been into Clojure lately and have avoided macros up until now, so this is my first exposure to them. I've been reading "Mastering Clojure Macros", and on Chapter 3, page 28, I encountered the following example:
user=> (defmacro square [x] `(* ~x ~x))
;=> #'user/square
user=> (map (fn [n] (square n)) (range 10))
;=> (0 1 4 9 16 25 36 49 64 81)
The context is the author is explaining that while simply passing the square macro to map results in an error (can't take value of a macro), wrapping it in a function works because:
when the anonymous function (fn [n] (square n)) gets compiled, the
square expression gets macroexpanded, to (fn [n] (clojure.core/* n
n)). And this is a perfectly reasonable function, so we don’t have any
problems with the compiler
This makes sense to me if we assume the body of the function is evaluated before runtime (at compile, or "definition" time) thus expanding the macro ahead of runtime. However, I always thought that function bodys were not evaluated until runtime, and at compile time you would basically just have a function object with some knowledge of it's lexical scope (but no knowledge of its body).
I'm clearly mixed up on the compile/runtime semantics here, but when I look at this sample I keep thinking that square won't be expanded until map forces it's call, since it's in the body of the anonymous function, which I thought would be unevaluated until runtime. I know my thinking is wrong, because if that was the case, then n would be bound to each number in (range 10), and there wouldn't be an issue.
I know it's a pretty basic question, but macros are proving to be pretty tricky for me to fully wrap my head around at first exposure!
Generally speaking function bodies aren't evaluated at compile time, but macros are always evaluated at compile time because they're always expanded at compile time whether inside a function or not.
You can write a macro that expands to a function, but you still can't refer/pass the macro around as if it were a function:
(defmacro inc-macro [] `(fn [x#] (inc x#)))
=> #'user/inc-macro
(map (inc-macro) [1 2 3])
=> (2 3 4)
defmacro is expanded at compile time, so you can think of it as a function executed during compilation. This will replace every occurrence of the macro "call" with the code it "returns".
You may consider a macros as a syntax rule. For example, in Scheme, a macros is declared with define-syntax form. There is a special step in compiler that substitutes all the macros calls into their content before compiling the code. As a result, there won't be any square calls in your final code. Say, if you wrote something like
(def value (square 3))
the final version after expansion would be
(def value (clojure.core/* 3 3))
There is a special way to check what will be the body of your macros after being expanded:
user=> (defmacro square [x] `(* ~x ~x))
#'user/square
user=> (macroexpand '(square 3))
(clojure.core/* 3 3)
That's why a macros is an ephemeral thing that lives only in source code but not in the compiled version of it. That's why it cannot be passed as a value or referenced somehow.
The best rule regarding macroses is: try to avoid them until you really need them in your work.
Here are two macros I had written
(defmacro hello [x] '(+ 1 2))
&
(defmacro hello [x] (eval '(+ 1 2)))
On macroexpanding the first one, I get (+ 1 2), and while macroexpanding the second, I get the value 3. Does this mean the addition happened at compile time? How is that even possible? What if instead of '(+ 1 2) I had written a function that queries a db. Would it query the db at compile time?
A macro injects arbitrary code into the compiler. Usually, the purpose is to "pre-process" custom code like (1 + 2) into something Clojure understands like (+ 1 2). However, you could include anything (including DB access) into the compilation phase if you really want to. After all the compiler is just a piece of software running on a general-purpose computer. Since it is open-source, you could modify the compiler code directly to do anything.
Using a macro is just a more convenient way of modifying/extending the base compiler code, that is optimized for extending the core Clojure language. However, macros are not limited to that use-case (if you really want to get crazy).
There is a similar ability using the C++ expression template mechanism, which is a Turing Complete compiler pre-processor. A famous example was to use the compiler to compute the first several prime numbers as "error" messages. See http://aszt.inf.elte.hu/~gsd/halado_cpp/ch06s04.html#Static-metaprogramming
Macro body is executed during compile time and its return value is used to replace its usage in the code and this new code will be compiled.
Thus your macro code:
(defmacro hello [x] (eval '(+ 1 2)))
will actually execute eval with the form '(+ 1 2) during compilation and the result value of that expression (3) will be returned as the result of the macro replacing its usage (e.g. (hello 0)).
This question similar to say, In clojure, how to apply 'and' to a list?, but solution doesn't apply to my question. every? function returns just a boolean value.
or is a macro, so this code is invalid:
(apply or [nil 10 20]) ; I want to get 10 here, as if 'or was a function.
Using not-every? I will get just boolean value, but I want to preserve semantics of or - the "true" element should be a value of the expression. What is idiomatic way to do this?
My idea so far:
(first (drop-while not [nil 10 20])) ; = 10
you might be able to use some for this:
(some identity [nil 10 20]) ; = 10
Note that this differs from or if it fails
(some identity [false]) ; = nil
(or false) ; = false
A simple macro:
(defmacro apply-or
[coll]
`(or ~#coll))
Or even more abstract
(defmacro apply-macro
[macro coll]
`(~macro ~#coll))
EDIT: Since you complained about that not working in runtime here comes a version of apply-macro that works in runtime. Compare it with answers posted here: In clojure, how to apply a macro to a list?
(defmacro apply-macro-in-runtime
[m c]
`(eval (concat '(~m) ~c)))
Notice that this version utilizes that parameters are passed unevaluated (m is not evaluated, if this was a function, it would throw because a macro doesn't have a value) it uses concat to build a list containing of a list with the quoted macro-name and whatever the evaluation of form c (for coll) returns so that c as (range 5) would be fully evaluated to the list that range returns. Finally it uses eval to evaluate the expression.
Clarification: That obviously uses eval which causes overhead. But notice that eval was also used in the answer linked above.
Also this does not work with large sequences due to the recursive definition of or. It is just good to know that it is possible.
Also for runtime sequences it makes obviously more sense to use some and every?.
This is not my 'production code' but a simplication of the problem for illustration purposes. Also, the title of this question is misleading because it brings to mind the ~# expansion, which I understand, and which may not necessarily be the problem. Please suggest a better question title if you can.
Given a macro with the following form:
(defmacro my-add [x & ys] `(+ ~x ~#ys))
Now let's say we have a list:
(def my-lst '(2 3))
Now I want a function that uses my-add that I can pass my-lst to as an arg, i.e.
(call-my-add 1 my-lst)
I define the function in what would seem to be the obvious way:
(defn call-my-add [x ys]
(apply my-add (cons x ys)))
But:
java.lang.Exception: Can't take value of a macro: #'user/call-my-add (repl-1:60)
I've tried all sorts of wild tricks to get the call-my-add function to work using evals, applies, and even defining call-my-add as a macro, but they all give similar ClassCastExceptions.
Is there any way out of this?
No. Macros do not, cannot, will never, have access to the actual run-time values contained in their arguments, so cannot splice them into the expansion. All they get is the symbol(s) you pass to them, in this case my-list. The "way around this" is to define my-add as a function, and then (optionally) have a macro that calls that function in order to generate its code.
I wrote a blog post about this semi-recently that you might find enlightening.
You could do it with evals if you wanted to, but that is a terrible idea in almost every case:
(let [my-list '(1 2)]
(eval `(my-add 5 ~#my-list)))
What a great example showing that macros are not first class citizens in Clojure (or any Lisp that I know of). They cannot be applied to functions, stored in containers, or passed to functions etc. In exchange for this they get to control when and if their arguments are evaluated.
What happens at macro expansion time must stay in macro expansion time. So if my-add is being evaluated at macro expansion time and you want to use apply then you need... another macro; to do the applying.
(defmacro call-my-add [x ys]
`(my-add ~#(cons x ys)))
Macros are somewhat contagious in this way.
PS: I'm not at my repl so please edit if you see a bug in this example (or I'll fix it when I get back)