I am playing around with Elixir macros - specifically macros that call themselves which is something that I do often in Scheme. I have created a little test macro below however it just hangs iex - nothing is printed to console. Does anyone have any insight into why and what could be done to correct it?
defmodule RecMac do
defmacro test_rec(x) do
quote do
IO.puts("Started?")
if(unquote(x) < 1) do
IO.puts("Done?")
"done"
else
IO.puts("Where are we")
IO.puts(unquote(x))
RecMac.test_rec(unquote(x) - 1)
end
end
end
end
EDIT!!
OK, so it turns out you can define recursive macros where there is a structural difference to match on (eg lists). The following is working for me. And to confirm #Aleksei Matiushkin below, the above will not work and does indeed not work in scheme!
defmacro test_rec([h | t]) do
quote do
IO.inspect([unquote(h) | unquote(t)])
RecMac.test_rec(unquote(t))
end
end
defmacro test_rec([]) do
quote do
IO.puts "Done"
end
end
end
I am very happy to have dug into this as I learned something about two languages!
TL;DR: this is impossible.
Macros in elixir are not what you expect them to be. When a compiler sees a macro, it calls it during a compilation time and injects the AST it returned in the place of where it was called. That said, recursive macros would always lead to the infinite loop at the compilation stage.
Put IO.puts("something") before quote do instruction and you’ll see it to be printed infinitely, once per subsequent call to expand the macro.
You can use #compile {:inline, test_rec: 1} to achieve the behaviour you are after. For better understanding macros, you probably should read Macros section in the Elixir Guide in general and this excerpt in particular:
Macros are harder to write than ordinary Elixir functions and it’s considered to be bad style to use them when they’re not necessary. So write macros responsibly.
Actually you can do kind of recursion, but point is to think what you are doing to avoid calling macro inside quote do [....] end. Your example with IO.puts would be something like this
defmodule RecMac do
defmacro test_rec(list) do
{:__block__, [], quote_value(list)}
end
defp quote_value([]), do: []
defp quote_value([h | t]) do
if h < 1 do
[]
else
ast = quote do
IO.puts("#{unquote(h)}")
end
[ast | quote_value(t)]
end
end
end
you will notice two things, I do use recursion in macro but outside quote using private function quote_value/1 and that function has logic that "stops" after it finds value lower than 1. All "quotes" are put into list and trick is to put this list into tuple {:__block__, [], put_quote_list_here}
Now note that this macro wont compile if list parameter of test_rec is not know upfront (during compile time), so you need to call macro test_rec(["a", "b", 0, 100, 200]) so compiler knows size and elements of that list.
BTW, I used body optimized recursion, but you can easily add accumulator and convert this into tail optimized recursion.
Related
I'm trying to do what a Lisp hacker would call a "symbol macro". To wit, here's what I'm using now:
global ghypers = Dict()
macro hyp(variable, value); ghypers[:($variable)] = :($value); end
macro hyp(variable); ghypers[:($variable)]; end
#hyp foo 5
println(ghypers)
println(#hyp foo)
println(#hyp(foo)+1)
So far so good, but the last thing is ugly, I want to do this:
#foo+1
Sort of like this:
macro foo(); ghypers[:foo]; end
println(#foo)
println(#foo()+1)
Close, and works, but not quite what I wanted, which, again, is:
println(#foo+1)
MethodError: no method matching #foo(::LineNumberNode, ::Module, ::Expr)
Except that that doesn't work.
Lisp has the concept of a symbol macro, where you can bind a macro expansion to a symbol, like foo, so that symbol would expand the to value.
Now, an obvious but equally bad (in the unhygienic sense) way to do this would simply be to bind the global var foo to the value, but I don't want global (or, rather, I want them localized in ghypers).
Is there any way to do the thing I'm seeking in Julia?
(<flame> Having been a happy lisp camper for 40+ years, Julia is the first programming language that I can say I actually like even a little. But not having grok'ed the importance of homoiconicity, their macro system is ... well, to my eyes, quite a mess, where Lisp's is clear as rain water. </flame> :-)
There are two direct ways. If the thing is hygienic and referentially transparent, just use a const value. That won't work in your example.
Otherwise you have to use the macro in call syntax: #foo(). There's no around that, it's how Julia syntax work. Although I won't recommend doing it either.
But a nicer alternative, IMHO, would be an "anaphoric context macro", something like:
#withhyper (stuff) begin
println(foo+1)
end
where foo is a name escaped inside some local scope. Expand the block to a variation of
let foo = setup_hyper(stuff, ...)
println(foo+1)
end
I'm learning control structures from https://elixirschool.com/en/lessons/basics/control-structures/
and I noticed that it mentions
Chances are you’ve encountered if/2 before, and if you’ve used Ruby you’re familiar with unless/2. In Elixir they work much the same way but they are defined as macros, not language constructs. You can find their implementation in the Kernel module.
so what's the difference between a language construct and a macro in Elixir and when is it necessary to write a macro?
A macro is a way to program a programming language. Simply put, a macro is a way to generate program code, instead of writing it yourself all the time.
A language construct on the other hand (sometimes called "special form"), is something that is at the core of elixir itself. An oversimplification could be that the implementation of if is not done in Elixir, but in the language in which Elixir is implemented.
Suppose you want to use the mentioned unless.
Edit: unless is available in Elixir. But let's assume for the remainder that it is not.
In Elixir, there is no unless available in the language. José Valim did not implement it. But you can always write something that has the same semantics: a negated if.
We would like to have this, but we don't:
unless sun_shines() do
open_umbrella()
end
But we only have an if and a not, so we can write:
if not sun_shines() do
open_umbrella()
end
Secondly, a macro is a special kind of function, but its parameters are code, and the result of executing a macro is code as well. Assuming we have the unless macro, it takes in a condition (i.e., sun_shines()), and a body (i.e., open_umbrella()), and returns if not sun_shines(), do: open_umbrella(). So a macro is a function that works at the level of your "dead code" and generates "dead code".
You might think that this is just too stupid to write a macro for. That's true. But these types of problems happen more often than you think, and then a macro is a great solution to that problem. It's just a way to program your programming language.
An example implementation of the unless macro has been provided by Aleksei Matiushkin:
defmodule MyMacros do
defmacro unless(ast, do: block) do
quote do
if not unquote(ast) do
unquote(block)
end
end
end
end
Here you can clearly see that you give it an AST (Abstract Syntax Tree), and it will transform it to another AST (quote), and inject that in the place where you called the macro. Note that this all happens at compile time. Your program is not being executed at this point!
For example, suppose you have the above module available, and this is your program:
defmodule MyProgram do
def my_function(x) do
unless sun_shining() do
open_umbrella()
end
end
end
After compilation, and before execution, your program will look like this:
defmodule MyProgram do
def my_function(x) do
if not sun_shining() do
open_umbrella()
end
end
end
This phase is what we call macro expansion phase.
As an extra, here you can find two actual macros used in Elixir and ExUnit respectively.
https://github.com/elixir-lang/elixir/blob/d48b16cf549eca0629449a47cc5574a7170706c3/lib/ex_unit/lib/ex_unit/assertions.ex#L104
https://github.com/elixir-lang/elixir/blob/13ced80fcda1bea69037aacd4b052a0c44b4be61/lib/elixir/lib/stream/reducers.ex#L58
Note: I keep adding more and more information to this answer. The answer actually deserves a whole book and Metaprogramming Elixir by Chris McCord is the best one.
I'm trying to learn Common Lisp, and found something unexpected (to me) when trying something out in the repl. Based on order of execution in most programming languages, and the great first class function support I'd always heard about from lisp, I'd think the following should work:
((if t 'format) t "test")
In Ruby I can do:
if true
Object.method(:puts)
end.call("test")
My thinking in how the above lisp code should work is that it should evaluate the inner lisp form, return format, then begin evaluating the outer lisp form, with format then being the first atom. I'd read that the first form needs to be a symbol, so I also tried ((if t format) t "test") even though my initial thought was that this would try to evaluate format before returning from the inner form.
I've noticed that sometimes lisp forms need to be preceded by #' in order for their results to be callable, but using (#'(if t 'format) t "test") doesn't work either. I'm sure I'm just misunderstanding something basic as I'm pretty new to lisp, but what's going on here?
Common Lisp doesn't evaluate the first element of an expression normally. It has to be either a literal symbol naming a function, or a lambda expression.
If you want to call a function determined dynamically, you need to use the FUNCALL function:
(funcall (if t 'format) t "test")
This is analogous to the need to use the .call() method in Ruby.
What you tried would work in some other Lisp dialects, such as Scheme.
I am experimenting with Macros in elixir. Therefore, the code I'm about to show should certainly be done with simple functions but.. I'm experimenting !
I want to define 2 macros (A and B) and make A use B to experiment with macro expansion.
When I use A, I get a compile error saying that the function B is undefined.
Here is the code:
defmodule MyMacros do
defmacro print_expr(expr) do
quote do
IO.puts(unquote(expr))
end
end
defmacro print_hashes_around(expr) do
quote do
IO.puts "###"
print_expr(unquote(expr))
IO.puts "###"
end
end
end
defmodule MyModule do
require MyMacros
def my_print(expr) do
MyMacros.print_hashes_around(expr)
end
end
MyModule.my_print("hello world")
And here is the compile error:
macro_test.exs:17: warning: redefining module MyModule
** (CompileError) macro_test.exs:21: function print_expr/1 undefined
(stdlib) lists.erl:1336: :lists.foreach/2
macro_test.exs:17: (file)
(elixir) lib/code.ex:307: Code.require_file/2
The way I (mis)understand things:
By requiring MyMacros, the module MyModule should know the existence of both macros. Therefore I should be able to use any macros.
When print_hashes_around is expanded in MyModule, the compiler should find that print_expr is also a macro. Therefore, another expansion should happen.
What seems to happen is that second expansion does not happen. Therefore the compiler looks for a function definition that does not exist.
Am I right ?
As suggested in slack, prefixing print_expr with MyMacros. fixes it. I still don't understand why. MyModule requires MyMacros so both Macros should be known and expandable... When I look at the definition of unless, it uses if, not Kernel.if.
By requiring MyMacros, the module MyModule should know the existence of both macros. Therefore I should be able to use any macros.
The misunderstanding is here. :) require only makes the module available to the compiler, it does not import the module functions. If you used import MyModule then it would work.
However, it would be best to fix the issue by prefixing the module name, because then you allow developers using your code to use your macros explicitly (with require) or by importing them.
Another option is to avoid multiple macro invocations like this:
defmodule MyMacros do
defmacro print_expr(expr) do
quoted_print_expr(expr)
end
defmacro print_hashes_around(expr) do
quote do
IO.puts "###"
unquote(quoted_print_expr(expr))
IO.puts "###"
end
end
defp quoted_print_expr(expr) do
quote do
IO.puts(unquote(expr))
end
end
end
Suppose I have this macro definition in a module:
module Example
export #example_macro
macro example_macro(a)
quote
local r = RemoteRef()
put!(r, $(esc(a)))
remotecall_fetch(2, (r) -> fetch(r), r)
end
end
end
And here is its expansion:
julia> include("Example.jl")
julia> using Example
julia> macroexpand(quote #example_macro a end)
quote # none, line 1:
begin # /.../Example.jl, line 7:
local #121#r = Example.RemoteRef() # line 8:
Example.put!(#121#r,a) # line 9:
Example.remotecall_fetch(2,(r) -> Example.fetch(r),#121#r)
end
end
Every single one of globally available functions (like put! or fetch) are prefixed with the name of the module. I understand that this is needed for the macro to be hygienic - if, say, fetch was redefined in the module in which #example_macro is called, and fetch was inserted into the expansion as is, it wouldn't work correctly.
However, this also requires Example module to be available not only in the main process, but also on the second worker (since remotecall_fetch needs to execute Example.fetch on it). I don't want it - after all, fetch is a basic function available on all workers by default.
So, is there a way to disable prefixing all identifiers with the name of the current module? I think this would mean turning the macro non-hygienic as it is impossible to decide where some identifier (like fetch) is defined on macro expansion phase, and that's fine for me.
Since this is a pretty profound question, I think that you should give the Julia devs themselves a chance to answer it by asking on julia-users.
Currently, you can completely circumvent macro hygiene by wrapping the whole quote block in your macro in an esc(...) (don't forget to take away the esc around a), but I would in general advise against it - then you are on your own.