Imagine you have this code:
foo = 1
bar = 2
foo bar do
:ok
end
Would you expect it to be equivalent to:
foo = 1
bar = 2
foo
bar do
:ok
end
It isn’t because “foo” in its own line is a valid expression. Replace “foo”
by “with” and you can see why your proposed syntax doesn’t work.
If you want to write it as you proposed, you need to use parens:
foo = 1
bar = 2
foo(
bar
) do
:ok
end
--
*José Valimwww.plataformatec.com.br
<http://www.plataformatec.com.br/>Founder and Director of R&D*
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