Imagine if bar was a toplevel function
bar = case foo of
True -> " Foo";
False -> "Bar";
Keep in mind that indentation level starts at the function name, not at the
let keyword.
On Thu, Oct 3, 2013 at 2:31 PM, Corentin Dupont
wrote:
> Hi the list,
> why do this function doesn't compile (parse
Thanks to all for your replies!
I asked the question because I often make this kind of transformations
(please don't mind the non-sensical example):
test :: Bool -> IO ()
test foo = do
bar <- case foo of
True -> return "Foo"
False -> return "Bar"
return ()
into
test :: Bool ->
On Thu, Oct 3, 2013 at 9:44 PM, Brandon Allbery wrote:
> On Thu, Oct 3, 2013 at 2:31 PM, Corentin Dupont > wrote:
>
>> test :: Bool -> IO ()
>> test foo = do
>>let bar = case foo of
>>True -> "Foo";
>>False -> "Bar"
>>return ()
>>
>> while this one does (just adding one
On Thu, Oct 3, 2013 at 2:31 PM, Corentin Dupont
wrote:
> test :: Bool -> IO ()
> test foo = do
>let bar = case foo of
>True -> "Foo";
>False -> "Bar"
>return ()
>
> while this one does (just adding one space in front of True and False):
>
> test :: Bool -> IO ()
> test foo
The first version has bar True and False all at the same indentation level.
As such they are seen as standalone expressions, rather than being nested
under the one introduced by bar.
See http://en.wikibooks.org/wiki/Haskell/Indentation
On Thu, Oct 3, 2013 at 8:31 PM, Corentin Dupont
wrote:
>
