Given the following:
for fn_name <- [:foo, :bar, :baz] do
defmacro unquote(fn_name)(do: inner) do
fn_name = unquote(fn_name) # <--- Why?
quote do
IO.puts "#{unquote(fn_name)} called"
unquote(inner)
end
end
end
What's the reason for fn_name = unquote(fn_name)? If I omit this line, it's a compile error. What's the reason for this "double" unquoting?
Let's simplify the example a little bit:
for fn_name <- [:foo, :bar, :baz] do
defmacro unquote(fn_name)(do: inner) do
fn_name = unquote(fn_name) # <--- Why?
quote do
{unquote(fn_name), unquote(inner)}
end
end
end
In the example above, because quote is returning a tuple with two unquoted elements, it is equivalent to:
for fn_name <- [:foo, :bar, :baz] do
defmacro unquote(fn_name)(do: inner) do
fn_name = unquote(fn_name) # <--- Why?
{fn_name, inner}
end
end
Now it is easier to understand what happens if you don't unquote(fn_name) before: the variable fn_name simply wouldn't exist inside the macro definition. Remember that all defs (def, defp, defmacro, etc) start a new variable scope, so if you want to use fn_name inside, you need to define it somehow.
The other property we are seeing in this code is that Elixir will stop unquoting when it sees a quote. So in the quote above, unquote won't be unquoted when the macro is defined but rather when the macro is executed, which also explains why the variable is required to be defined inside the macro.
It's because of Hygiene.
Elixir has the concept of macro hygiene. Hygiene means that variables, imports, and aliases that you define in a macro do not leak into the caller’s own definitions.
for fn_name <- [:foo, :bar, :baz] do
defmacro unquote(fn_name)(do: inner) do
fn_name = unquote(fn_name) # <-- This is macro's context
quote do
IO.puts "#{unquote(fn_name)} called" # <-- This is caller's context
unquote(inner)
end
end
end
You should read Hygiene Protects the Caller’s Context from Chris McCord's Metaprogramming Elixir book
Related
Let's say I want to define a series of structs that will be used as parametric types for some other struct later on. For instance, I would like to have something like
abstract type Letter end
struct A <: Letter end
struct B <: Letter end
...etc...
The idea I've had is to define a macro which takes a string of the name of the struct I want to create and defines it as well as some basic methods. I would then execute the macro in a loop for all names and get all my structs defined at compile time. Something like this:
const LETTERS = ["A","B","C","D"]
abstract type Letter end
macro _define_type(ex)
lines = Vector{Expr}()
push!(lines, Meta.parse("export $ex"))
push!(lines, Meta.parse("struct $ex <: Letter end"))
push!(lines, Meta.parse("string(::Type{$ex}) = \"$ex\""))
return Expr(:block,lines...)
end
#_define_type "A"
for ex in LETTERS
#_define_type ex
end
The first way of executing the macro (with a single string) works and does what I want. However, when I execute the macro in a loop it does not. It tells me some variables are declared both as local and global variables.
Can someone explain what is happening? I believe it may be solved by a proper use of esc, but I can't figure out how.
Thanks!
Edit: Thank you all for the amazing responses! Got my code running!
I think this is what you are trying to do:
module MyModule
abstract type Letter end
const LETTERS = ["A", "B", "C", "D"]
for letter in LETTERS
sym = Symbol(letter)
#eval begin
export $sym
struct $sym <: Letter end
Base.string(::Type{$sym}) = $letter
end
end
end
using .MyModule
string(A) # "A"
See the Code Generation section for more details:
https://docs.julialang.org/en/v1/manual/metaprogramming/#Code-Generation
Okay, the problem here is that in Julia for loops introduce a separate, local scope, while struct definitions need to be in the global scope. So your macro fails because it creates struct definitions in the local scope of the for loop.
A way to get around this is to use #eval, to ensure your struct definitions are evaluated in the global scope. In that case, you don't need to create a macro for that, just have a simple loop like this:
abstract type Letter end
const LETTERS = [:A, :B, :C, :D, :E]
for ex in LETTERS
#eval struct $ex <: Letters end
end
You can even put that loop in a function and it will still work. The defined structs can have fields, as #eval covers the entire code block that follows it.
Note that LETTERS must contain symbols rather than strings for this to work correctly. It's easy enough to convert a vector of strings into a vector of symbols using Symbol.(vector_of_strings).
While there are other ways of achieving what you want to do I believe the core issue is understanding the nature of macros. From the docs (emphasis mine):
Macros are necessary because they execute when code is parsed, therefore, macros allow the programmer to generate and include fragments of customized code before the full program is run.
So the macro in your loop does not "see" the values "A", "B", "C" and "D", it sees the expression: ex. To demonstrate this try using #macroexpand:
julia> #macroexpand #_define_type ex
quote
export ex
struct ex <: Main.Letter
#= none:1 =#
end
var"#11#string"(::Main.Type{Main.ex}) = begin
#= none:1 =#
"ex"
end
end
As you can see the actual value of the variable ex does not matter. With this in mind let's look at the actual error you get. You can reproduce it like this:
julia> for ex in ["A"]
struct ex <: Letter
end
end
ERROR: syntax: variable "ex" declared both local and global
Stacktrace:
[1] top-level scope
# REPL[52]:1
You can probably see that this is not what you want, but why this specific error? The reason is that structs are implicitly global while the loop variable is local.
Here is a possible solution that uses a macro that takes a variable number of arguments instead (I also switched to providing the expression directly instead of as a string):
abstract type Letter end
macro _define_types(exprs...)
blocks = map(exprs) do ex
name = string(ex)
quote
export $ex
struct $ex <: Letter end
Base.string(::Type{$ex}) = $name
end
end
Expr(:block, blocks...)
end
#_define_types A
#_define_types B C D
Given the below code, the line use Composite, user_opts: user_opts ends up as [{:user_opts, [line: 3, counter: {MockUserNode1, 2}], Automaton.Node}] inside the using(opts) as opts. Is there any way to inject that code inside the outer macro?
defmacro __using__(user_opts) do
a =
if Enum.member?(Composite.types(), user_opts[:node_type]) do
IO.inspect(user_opts)
quote bind_quoted: [user_opts: user_opts] do
use DynamicSupervisor
use Composite, user_opts: user_opts
end
else
quote do: use(Action)
end
end
Answering the question stated: it’s perfectly possible to call macros from inside other macros, the just inject the AST recursively in the end.
defmodule DeeplyUsed do
defmacro __using__(opts) do
quote bind_quoted: [opts: opts] do
opts
end
end
end
defmodule Used do
defmacro __using__(opts) do
quote bind_quoted: [opts: opts] do
use DeeplyUsed, opts: opts
end
end
end
defmodule Using do
use Used, line: 3, counter: {MockUserNode1, 2}
end
That said, your issue is induced.
Sidenote: [{:user_opts, _, Automaton.Node}] looks indeed very suspicious there, that’s not how keyword lists are being quoted. Start with explicit unquoting and logging what comes to user_opts there.
defmacro __using__(user_opts) do
IO.inspect(user_opts, label: "Outside")
quote do
IO.inspect(unquote(user_opts), label: "Inside")
use Composite, user_opts: unquote(user_opts)
end
end
Consider these two macro definitions:
macro createTest1()
quote
function test(a = false)
a
end
end |> esc
end
macro createTest2()
args = :(a = false)
quote
function test($args)
a
end
end |> esc
end
According to the builtin Julia facilities they should both evaluate to the same thing when expanded:
println(#macroexpand #createTest1)
begin
function test(a=false)
a
end
end
println(#macroexpand #createTest2)
begin
function test(a = false)
a
end
end
Still I get a parse error when trying to evaluate the second macro:
#createTest2
ERROR: LoadError: syntax: "a = false" is not a valid function argument name
It is a space in the second argument list. However, that should be correct Julia syntax. My guess is that it interprets the second argument list as another Julia construct compared to the first. If that is the case how do I get around it?
The reason that the second macro is failing as stated in my question above. It looks correct when printed however args is not defined correctly and Julia interprets it as an expression which is not allowed. The solution is to instead define args according to the rules for function parameters. The following code executes as expected:
macro createTest2()
args = Expr(:kw, :x, false)
quote
function test($(args))
a
end
end |> esc
end
I have a macro which gets a module name as parameter and I want to call a function on that module to get some data in order to generate the quote block.
Example:
defmacro my_macro(module) do
data = apply(module, :config, [])
# do something with data to generate the quote do end
end
Obviously, this doesn't work because the parameter value is quoted. I could fetch the data inside the quote block and act accordingly but that would put the whole logic inside the module that uses my macro which is quite dirty. I want to inject as little code as possible.
You can extract the module out by pattern matching with its quoted form: {:__aliases__, _, list} where list is a list of atoms which when concatenated with a dot (use Module.concat/1) produces the full module name.
defmodule A do
defmacro my_macro({:__aliases__, _, list}) do
module = Module.concat(list)
module.foo()
end
end
defmodule B do
def foo do
quote do
42
end
end
end
defmodule C do
import A
IO.inspect my_macro B
end
Output:
42
Can you please have a look at my Macro?
I am getting undefined function number/0 error, and I can't figure it out why.
defmodule DbUtil do
defmacro __using__(opts) do
quote do
import unquote(__MODULE__)
#before_compile unquote(__MODULE__)
end
end
defmacro __before_compile__(%{module: definition} = _env) do
quote do
import Ecto.Query
def last do
from x in unquote(definition), order_by: [desc: x.id], limit: 1
end
# This dumps error
def limits(number) do
from a in unquote(definition), limit: ^unquote(number)
end
end
end
end
You don't need to unquote number. unquote is used when you want to inject a variable present outside the quote block. Since number is defined inside the quote, you don't need to unquote. The following should work for you:
def limits(number) do
from a in unquote(definition), limit: ^number
end