voger
Can someone please explain this 'my_if' code in the Metaprogramming Elixir book?
I just finished “Programming Elixir” and now I am reading “Metaprogramming Elixir”. I can’t say I comprehend everything. I try to understand but I mostly pretend I understand and keep reading. While the previous examples were somehow manageable to follow I am stuck in recreating the if macro. Here is the code.
defmodule ControlFlow do
defmacro my_if(expr, do: if_block) do
if(expr, do: if_block, else: nil)
end
defmacro my_if(expr, do: if_block, else: else_block) do
quote do
case unquote(expr) do
result when result in [false, nil] -> unquote(else_block)
_ -> unquote(if_block)
end
end
end
end
Can someone please be patient and explain with simple words what is happening in this code?
I understand what quote and unquote do. I don’t understand why the two defmacro definitions.
Now that I write the question I realize the first defmacro is just for the case we have a simple if without else. The second defmacro handles the if/else case. But still I can’t understand how they work.
Marked As Solved
chrismccord
Bah, yes it is indeed erratta. You’re the first to catch this in like 8 years! ![]()
If you give in a non AST literal, it’s always true because we are passing the ast to if in the first clause, not the expanded expression. So for example, 1 > 2 will be the ast tuple, instead of the boolean, which is always truthy. The code should have quoted the first clause. As @LostKobrakai said tho, a better example wouldn’t proxy to Elixir’s if in the first place:
iex(2)>
nil
iex(3)> defmodule ControlFlow do
...(3)> defmacro my_if(expr, [{:do, if_block} | _] = opts) do
...(3)> quote do
...(3)> case unquote(expr) do
...(3)> result when result in [false, nil] -> unquote(opts[:else])
...(3)> _ -> unquote(if_block) end
...(3)> end
...(3)> end
...(3)> end
{:module, ControlFlow,
<<70, 79, 82, 49, 0, 0, 6, 208, 66, 69, 65, 77, 65, 116, 85, 56, 0, 0, 0, 183,
0, 0, 0, 20, 18, 69, 108, 105, 120, 105, 114, 46, 67, 111, 110, 116, 114,
111, 108, 70, 108, 111, 119, 8, 95, 95, 105, ...>>, {:my_if, 2}}
iex(4)> require ControlFlow
ControlFlow
iex(5)> ControlFlow.my_if 1 > 2, do: :here
nil
iex(6)> ControlFlow.my_if 3 > 2, do: :here
:here
iex(7)>
Also Liked
brainbag
If you understand that code like this:
if something do
x
else
y
end
is simply syntactic sugar for this:
if something, do: x, else: y
and that is simply syntactic sugar for this:
if(something, [do: x, else: y])
and THAT is simply syntactic sugar for this:
if(something, [{:do, x}, {:else, y}])
then it might be a bit clearer. Remember that Elixir allows you to drop the List [] if you’re passing in a keyword list.
Some more detail: your first my_if function takes two arguments, one of which is an explicit keyword list of do:. Since else isn’t specified, it just passes nil in for the else. That will run when you call code like this:
my_if x == 2 do
x
end
# or
my_if(x == 2, do: x)
# which, without the sugar, is actually this:
my_if(x == 2, [{:do, x}])
Your second my_if function does a case statement on the expression (something in my example). Inside of the case statement, it checks to see if the result evaluation is one of false or nil, and then returns the else_block. Otherwise, it returns the if_block.
my_if x == 2 do
x
else
y
end
# or
my_if(x == 2, do: x, else: y)
# which, without the sugar, is actually this:
my_if(x == 2, [{:do, x}, {:else, y}])
# and this is, conceptually, what your my_if function does
case x == 2 do
false -> y
nil -> y
_ -> x
end
You might check the “Keywords and maps” section of the documentation for further reference: http://elixir-lang.org/getting-started/keywords-and-maps.html
Hope that helps!
brainbag
No, because you forgot something very important in your first definition. ![]()
If you do an IO.inspect in that first my_if, what is the value of expr? It’s not what you think!
When you figure that out, think about how you “transform” values so they’re usable by macros, and then look at your first my_if again and see what you’re missing.
kodepett
Hi, kindly find below - elixir v1.13.3
iex(92)> ControlFlow.my_if false, do: :here
nil
iex(93)> ControlFlow.my_if true, do: :here
:here
iex(94)> ControlFlow.my_if 1 > 2, do: :here
:here
iex(95)> ControlFlow.my_if 1 == 2, do: :here
:here
Popular in Questions
Other popular topics
Categories:
Sub Categories:
Forums
Popular Tags
- #ecto
- #liveview
- #troubleshooting
- #learning-elixir
- #deployment
- #library
- #erlang
- #testing
- #genserver
- #mix
- #absinthe
- #remote-other
- #otp
- #plug
- #how-to-question
- #macros
- #postgres
- #channels
- #elixirconf
- #exunit
- #discussion
- #code-sync
- #javascript
- #podcasts
- #onsite
- #dialyzer
- #docker
- #authentication
- #umbrella
- #full-time-contract
- #podcasts-by-brainlid
- #ecto-query
- #elixir-ls
- #phoenix_html
- #iex
- #blog-post
- #graphql
- #genstage
- #ai
- #websockets
- #supervisor
- #advent-of-code
- #elixirconf-us
- #distillery
- #processes
- #forms
- #api
- #metaprogramming
- #security
- #performance








