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!

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.

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

Wow thanks. I already tried to use unquote but I got errors because I didn’t wrap it first in “quote do”. The correct code is this

defmacro my_if(expr, do: if_block) do
  quote do
    if unquote(expr), do: unquote(if_block), else: nil
  end
end

Thank you very much for your help.

I faced the same issue will working through Metaprogramming Elixir. my_if returns true irrespective of the condition when invoked without the else block. I think it’s an errata, I supposed the line should be:

defmacro my_if(expr, do: block) do
    quote do
      if unquote(expr), do: unquote(block), else: nil
    end
  end

Below is the full snippet from the book on page 22, maybe @chrismccord can clarify, I could be wrong with my assumption.

defmodule ControlFlow do
   defmacro my_if(expr, do: if_block), do: if(expr, do: if_block, else: nil)
   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