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Within LTS Haskell 24.15 (ghc-9.10.3)

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  1. deriveRepresentable0 :: Name -> Q [Dec]

    generic-deriving Generics.Deriving.TH

    Given the type and the name (as string) for the Representable0 type synonym to derive, generate the Representable0 instance.

  2. deriveRepresentable0Options :: Options -> Name -> Q [Dec]

    generic-deriving Generics.Deriving.TH

    Like deriveRepresentable0, but takes an Options argument.

  3. deriveRepresentable1 :: Name -> Q [Dec]

    generic-deriving Generics.Deriving.TH

    Given the type and the name (as string) for the Representable1 type synonym to derive, generate the Representable1 instance.

  4. deriveRepresentable1Options :: Options -> Name -> Q [Dec]

    generic-deriving Generics.Deriving.TH

    Like deriveRepresentable1, but takes an Options argument.

  5. deriveSelectors :: Name -> Q [Dec]

    generic-deriving Generics.Deriving.TH

    Given a datatype name, derive datatypes and instances of class Selector. On GHC 7.11 and up, this functionality is no longer used in GHC generics, so this function generates no declarations.

  6. deriveLift :: Name -> Q [Dec]

    th-lift Language.Haskell.TH.Lift

    No documentation available.

  7. deriveLift' :: [Role] -> Info -> Q [Dec]

    th-lift Language.Haskell.TH.Lift

    Obtain Info values through a custom reification function. This is useful when generating instances for datatypes that have not yet been declared.

  8. deriveLiftMany :: [Name] -> Q [Dec]

    th-lift Language.Haskell.TH.Lift

    Derive Lift instances for many datatypes.

  9. deriveLiftMany' :: [([Role], Info)] -> Q [Dec]

    th-lift Language.Haskell.TH.Lift

    No documentation available.

  10. deriveListable :: Name -> DecsQ

    leancheck Test.LeanCheck

    Derives a Listable instance for a given type Name. Consider the following Stack datatype:

    data Stack a  =  Stack a (Stack a) | Empty
    
    Writing
    deriveListable ''Stack
    
    will automatically derive the following Listable instance:
    instance Listable a => Listable (Stack a) where
    tiers  =  cons2 Stack \/ cons0 Empty
    
    Warning: if the values in your type need to follow a data invariant, the derived instance won't respect it. Use this only on "free" datatypes. Needs the TemplateHaskell extension.

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