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
end
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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.
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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
unquote(block)
else
# break out of loop
end
end
end
end
#Same as first if function
defmacro my_second_while(expression, do: block) do
my_while(expression, do: block)
end
end
iex
require Loop
Loop.my_second_while true do
IO.puts "looping!"
end
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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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.
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