Where indentation doesn't seem to use configuration value

[owst]

I like to indent the "where" keyword by 2 spaces from the expression it is contained within, and then indent the bindings in the where (on a newline) by 2 more spaces, so the bindings are indented 4 spaces from the expression containing them, thus:

foop = doop 1
  where 
    doop x = x + woop x
    woop x = 0

However, when I hit enter after doop 1 and type where, it correctly gets indented by 2 spaces (I don't think this is configurable, looking at the README?) but then after hitting enter and typing doop x and woop x, the indentation is as follows:

foop = doop 1
  where 
      doop x = x + woop x
      woop x = 0

I have set let g:haskell_indent_where = 2 in my .vimrc, but it doesn't appear to have any effect.

[owst]

A further comment, it appears that the indentation is applied if there is a binding on the where keyword line (and not otherwise); typing the doop x binding on the same line as the where keyword gives:

foop = doop 1
  where doop x = x + woop x
    woop x = 0

I would expect the indentation not to rely on doop being bound on the same line as the where keyword, as per my original desired output.

[raichoo]

I'm not able to reproduce this with the current version of haskell-vim. Is this still happening for you?

[owst]

Yes, it still happens for me. I have just updated using vundle and still have the same behaviour.

I typed the following to reproduce (CR = carriage return):

"f = g 1" CR "where" CR "g x = x" CR "h x = x"

which gave:

f = g 1
  where
      g x = x
      h x = x

However, "f = g 1" CR "where g x = x" CR "h x = x" gives, which seems to be correct.

f = g 1
  where g x = x
    h x = x

Ideally, I'd like the first example to look like:

f = g 1
  where
    g x = x
    h x = x

Furthermore, it seems the indentation is taken from the RHS expression rather than the LHS; without any manual indentation I got this:

thisIsALongFunctionName withSome longArguments =
    theRHSFunction withSome
      where
          theRHSFunction x = x

ideally I'd want it to look like:

thisIsALongFunctionName withSome longArguments =
    theRHSFunction withSome
  where
    theRHSFunction x = x

[raichoo]

That would be pretty hard to realize with the tools vim gives you. It would certainly work in some cases but would break others. If you really want that I'd say "patches welcome".

[johntyree]

@owst I'd also point out that those are different things. In the example you gave, the where clause is defining things at the top level of the function, whereas I think the plugin is defining them within the scope of theRHSFunction withSome.