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)
1 Like

I suspect there’s a typo there. The first my_if clause should be:

defmacro my_if(expr, do: if_block), do: my_if(expr, do: if_block, else: nil)

so it delegates to the 2nd clause

But it works when it is if.

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.

1 Like

So in a macro we can use another macro without quote. Thanks i got it

I tried a smilar approach for while as first if function, but my_second_while function freezes iex. my_while function is working as expected

defmodule Loop do
	defmacro my_while(expression, do: block) do
		quote do
			for _ <- Stream.cycle([:ok]) do
				if unquote(expression) do
				# break out of loop

    #Same as first if function 
	defmacro my_second_while(expression, do: block) do
		my_while(expression, do: block)


require Loop
Loop.my_second_while true do
IO.puts "looping!"

Should be:

quote(do: my_while(expression, do: block))

Also, your Stream.cycle bit is not ever going to end, it will loop forever?

1 Like

Yes i guess so but why first my_if function works without unquote? what is difference? That example from book yes never ends

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.