seq -parallel +template-haskell

sequenceQ :: [Q a] -> Q [a]
template-haskell Language.Haskell.TH.Syntax
ArithSeqE :: Range -> Exp
template-haskell Language.Haskell.TH.Syntax, template-haskell Language.Haskell.TH
> { [ 1 ,2 .. 10 ] }
arithSeqE :: RangeQ -> ExpQ
template-haskell Language.Haskell.TH.Lib, template-haskell Language.Haskell.TH
type ClauseQ = Q Clause
template-haskell Language.Haskell.TH.Lib, template-haskell Language.Haskell.TH