dokuzbir
Need help about quote and unquote
This code is from metaprograming elixir book. I dont understand when should use unquote. In first function there is no unqote to execute expression but in second function there is unquote why ?
defmodule ControlFlow do
# Why there is no unquote here ?
defmacro my_if(expr, do: if_block), do: if(expr, do: if_block, else: nil)
# Why there is unquote here ?
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
endMarked As Solved
aethereus
Your no-quote version of my_if does not work. I tested it with the following code:
ControlFlow.my_if 1 > 2, do: IO.puts("This should not be printed.")
And I saw that the sentence “This should not be printed.” is actually printed. This is because when you pass 1 > 2 to expr, it gets automatically quoted (becomes AST), and the AST is {:>, [context: Elixir, import: Kernel], [1, 2]} and is not false or nil, so the test of if(expr, ...) passes, and the if_block (which is also an AST) gets returned by the macro my_if.
If you test it with
ControlFlow.my_if false, do: IO.puts("This should not be printed.")
You get the correct behavior by accident because false is an atom, and atoms are “Elixir literals” that have exactly the same form in code and in AST.
Also Liked
gregvaughn
It’s calling Kernel.if/2 which is a macro that handles the quoting itself. But if you’re going to rely on Kernel.if then why write my_if? That’s why I think it’s a typo.
OvermindDL1
Should be:
quote(do: my_while(expression, do: block))
Also, your Stream.cycle bit is not ever going to end, it will loop forever?
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