markdev
Are macros lazily evaluated?
I am watching this talk by Jesse Anderson at ElixirConf about how macros work. He creates a macro (if statement), then contrasts it with a functional example.
Macro Version:
defmodule MyMacros do
defmacro my_if(condition, do: do_clause, else: else_clause) do
quote do
case unquote(condition) do
true -> unquote(do_clause)
false -> unquote(else_clause)
end
end
end
end
Function version:
defmodule MyFunctions do
def my_if(condition, do: do_clause, else: else_clause) do
case condition do
true -> do_clause
false -> else_clause
end
end
end
If you were to call my_if 1 + 2 == 3, do: "this", else: "that", you get the same result from either module. But if evaluation order matters in the do/else blocks, the behavior is different.
iex(17)> MyMacros.my_if 1 + 2 == 3, do: IO.puts("this"), else: IO.puts("that")
this
:ok
iex(18)> MyFunctions.my_if 1 + 2 == 3, do: IO.puts("this"), else: IO.puts("that")
this
that
:ok
Is this considered an example of lazy evaluation, or is this deferred evaluation typically described some other way?
Marked As Solved
tmbb
Your macro isn’t perdorming any kind of evaluation, lazy or strict. The macro works because it expands into a case special form, which does allow you to pick branches. In a sense I guess the case macros is doing something that can be considered lazy evaluation.
But your if macro is not performing lazy evaluation. It’s just expanding into another syntax form.
Also Liked
blatyo
A macro is replaced with the AST that it returns. So, this:
MyMacros.my_if 1 + 2 == 3, do: IO.puts("this"), else: IO.puts("that")
Turns into this at compile time:
case 1 + 2 == 3 do
true -> IO.puts("this")
false -> IO.puts("that")
end
Like @tmbb said, case is lazy. Not the macro.
The function appears non-lazy, because the arguments have to be evaluated before they are passed to MyFunctions.my_if/2.
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