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While macro bind_quoted breaks on first iteration

How to create a dynamic function name using Elixir macro?Elixir - Calling / Invoking Macros - UndefinedFunctionErrorWriting a custom assertion macroDynamic function definition with macroShould tuples be passed to macro?Elixir macros, quoting and unquotingUnquoting argument in macro definition hangs function callElixir: application quits immediately, or doesn't receive input when running distillery package. Why does it work that way, and how to work around it?Elixir: required inside a macroIO.inspect inside elixir Macros

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0

Maybe I am misunderstanding the use of bind_quoted but take this simple while loop macro:

defmodule Loop do
 defmacro while(expression, do: block) do
 quote bind_quoted: [expression: expression, block: block] do
 for _ <- Stream.cycle([:ok]) do
 if(expression) do
 block
 end
 end
 end
 end
end

Usage:

Interactive Elixir (1.8.0) - press Ctrl+C to exit (type h() ENTER for help) 
iex(1)> c "loop.ex" 
[Loop] 
iex(2)> import Loop
iex(3)> while 1 == 1 do 
...(3)> IO.puts "hi"
...(3)> end
hi

I would expect an infinite loop of "hi" but instead only get one iteration. If I remove the bind_quoted and simple unquote() each argument, it works as expected. Any ideas?

Thanks

elixir

share|improve this question

edited Mar 25 at 19:15

Brett Beatty

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asked Mar 25 at 18:09

BotonomousBotonomous

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add a comment | 

0

Maybe I am misunderstanding the use of bind_quoted but take this simple while loop macro:

defmodule Loop do
 defmacro while(expression, do: block) do
 quote bind_quoted: [expression: expression, block: block] do
 for _ <- Stream.cycle([:ok]) do
 if(expression) do
 block
 end
 end
 end
 end
end

Usage:

Interactive Elixir (1.8.0) - press Ctrl+C to exit (type h() ENTER for help) 
iex(1)> c "loop.ex" 
[Loop] 
iex(2)> import Loop
iex(3)> while 1 == 1 do 
...(3)> IO.puts "hi"
...(3)> end
hi

I would expect an infinite loop of "hi" but instead only get one iteration. If I remove the bind_quoted and simple unquote() each argument, it works as expected. Any ideas?

Thanks

elixir

share|improve this question

edited Mar 25 at 19:15

Brett Beatty

1,81111 gold badge1414 silver badges2424 bronze badges

asked Mar 25 at 18:09

BotonomousBotonomous

1,22911 gold badge1111 silver badges3333 bronze badges

add a comment | 

Maybe I am misunderstanding the use of bind_quoted but take this simple while loop macro:

defmodule Loop do
 defmacro while(expression, do: block) do
 quote bind_quoted: [expression: expression, block: block] do
 for _ <- Stream.cycle([:ok]) do
 if(expression) do
 block
 end
 end
 end
 end
end

Usage:

Interactive Elixir (1.8.0) - press Ctrl+C to exit (type h() ENTER for help) 
iex(1)> c "loop.ex" 
[Loop] 
iex(2)> import Loop
iex(3)> while 1 == 1 do 
...(3)> IO.puts "hi"
...(3)> end
hi

I would expect an infinite loop of "hi" but instead only get one iteration. If I remove the bind_quoted and simple unquote() each argument, it works as expected. Any ideas?

Thanks

elixir

share|improve this question

edited Mar 25 at 19:15

Brett Beatty

1,81111 gold badge1414 silver badges2424 bronze badges

asked Mar 25 at 18:09

BotonomousBotonomous

1,22911 gold badge1111 silver badges3333 bronze badges

defmodule Loop do
 defmacro while(expression, do: block) do
 quote bind_quoted: [expression: expression, block: block] do
 for _ <- Stream.cycle([:ok]) do
 if(expression) do
 block
 end
 end
 end
 end
end

Interactive Elixir (1.8.0) - press Ctrl+C to exit (type h() ENTER for help) 
iex(1)> c "loop.ex" 
[Loop] 
iex(2)> import Loop
iex(3)> while 1 == 1 do 
...(3)> IO.puts "hi"
...(3)> end
hi

share|improve this question

edited Mar 25 at 19:15

Brett Beatty

1,81111 gold badge1414 silver badges2424 bronze badges

asked Mar 25 at 18:09

BotonomousBotonomous

1,22911 gold badge1111 silver badges3333 bronze badges

share|improve this question

edited Mar 25 at 19:15

Brett Beatty

1,81111 gold badge1414 silver badges2424 bronze badges

asked Mar 25 at 18:09

BotonomousBotonomous

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edited Mar 25 at 19:15

Brett Beatty

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edited Mar 25 at 19:15

Brett Beatty

1,81111 gold badge1414 silver badges2424 bronze badges

asked Mar 25 at 18:09

BotonomousBotonomous

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asked Mar 25 at 18:09

BotonomousBotonomous

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1 Answer
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0

When you use bind_quoted, it evaluates the pieces you pass in. So the IO.puts "hi" is evaluated in [expression: expression, block: block], not inside your if block. You're just injecting an :ok there.

As an example, let's define some macros that do nothing except evaluate what's passed in:

defmodule AB do
 defmacro a(block) do
 quote bind_quoted: [block: block] do
 block
 block
 :ok
 end
 end

 defmacro b(block) do
 quote do
 unquote(block)
 unquote(block)
 :ok
 end
 end
end

Ignoring some warnings from the first one, let's call each of these with a block containing IO.inspect/1 and see what happens:

iex> require AB
iex> AB.a(IO.inspect(1 + 1))
... (some warnings)
2
:ok
iex> AB.b(IO.inspect(1 + 1))
2
2
:ok

In macro a/1, the sum + inspect happens outside of the quoting, so it only gets called the once. In macro b/1, the sum + inspect happens anywhere we unquote block.

share|improve this answer

edited Mar 25 at 23:47

answered Mar 25 at 19:29

Brett BeattyBrett Beatty

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add a comment | 

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0

When you use bind_quoted, it evaluates the pieces you pass in. So the IO.puts "hi" is evaluated in [expression: expression, block: block], not inside your if block. You're just injecting an :ok there.

As an example, let's define some macros that do nothing except evaluate what's passed in:

defmodule AB do
 defmacro a(block) do
 quote bind_quoted: [block: block] do
 block
 block
 :ok
 end
 end

 defmacro b(block) do
 quote do
 unquote(block)
 unquote(block)
 :ok
 end
 end
end

Ignoring some warnings from the first one, let's call each of these with a block containing IO.inspect/1 and see what happens:

iex> require AB
iex> AB.a(IO.inspect(1 + 1))
... (some warnings)
2
:ok
iex> AB.b(IO.inspect(1 + 1))
2
2
:ok

In macro a/1, the sum + inspect happens outside of the quoting, so it only gets called the once. In macro b/1, the sum + inspect happens anywhere we unquote block.

share|improve this answer

edited Mar 25 at 23:47

answered Mar 25 at 19:29

Brett BeattyBrett Beatty

1,81111 gold badge1414 silver badges2424 bronze badges

add a comment | 

0

When you use bind_quoted, it evaluates the pieces you pass in. So the IO.puts "hi" is evaluated in [expression: expression, block: block], not inside your if block. You're just injecting an :ok there.

As an example, let's define some macros that do nothing except evaluate what's passed in:

defmodule AB do
 defmacro a(block) do
 quote bind_quoted: [block: block] do
 block
 block
 :ok
 end
 end

 defmacro b(block) do
 quote do
 unquote(block)
 unquote(block)
 :ok
 end
 end
end

Ignoring some warnings from the first one, let's call each of these with a block containing IO.inspect/1 and see what happens:

iex> require AB
iex> AB.a(IO.inspect(1 + 1))
... (some warnings)
2
:ok
iex> AB.b(IO.inspect(1 + 1))
2
2
:ok

In macro a/1, the sum + inspect happens outside of the quoting, so it only gets called the once. In macro b/1, the sum + inspect happens anywhere we unquote block.

share|improve this answer

edited Mar 25 at 23:47

answered Mar 25 at 19:29

Brett BeattyBrett Beatty

1,81111 gold badge1414 silver badges2424 bronze badges

add a comment | 

When you use bind_quoted, it evaluates the pieces you pass in. So the IO.puts "hi" is evaluated in [expression: expression, block: block], not inside your if block. You're just injecting an :ok there.

As an example, let's define some macros that do nothing except evaluate what's passed in:

defmodule AB do
 defmacro a(block) do
 quote bind_quoted: [block: block] do
 block
 block
 :ok
 end
 end

 defmacro b(block) do
 quote do
 unquote(block)
 unquote(block)
 :ok
 end
 end
end

Ignoring some warnings from the first one, let's call each of these with a block containing IO.inspect/1 and see what happens:

iex> require AB
iex> AB.a(IO.inspect(1 + 1))
... (some warnings)
2
:ok
iex> AB.b(IO.inspect(1 + 1))
2
2
:ok

In macro a/1, the sum + inspect happens outside of the quoting, so it only gets called the once. In macro b/1, the sum + inspect happens anywhere we unquote block.

share|improve this answer

edited Mar 25 at 23:47

answered Mar 25 at 19:29

Brett BeattyBrett Beatty

1,81111 gold badge1414 silver badges2424 bronze badges

defmodule AB do
 defmacro a(block) do
 quote bind_quoted: [block: block] do
 block
 block
 :ok
 end
 end

 defmacro b(block) do
 quote do
 unquote(block)
 unquote(block)
 :ok
 end
 end
end

iex> require AB
iex> AB.a(IO.inspect(1 + 1))
... (some warnings)
2
:ok
iex> AB.b(IO.inspect(1 + 1))
2
2
:ok

share|improve this answer

edited Mar 25 at 23:47

answered Mar 25 at 19:29

Brett BeattyBrett Beatty

1,81111 gold badge1414 silver badges2424 bronze badges

answered Mar 25 at 19:29

Brett BeattyBrett Beatty

1,81111 gold badge1414 silver badges2424 bronze badges

answered Mar 25 at 19:29

Brett BeattyBrett Beatty

1,81111 gold badge1414 silver badges2424 bronze badges