package base

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type 'a arg

the underlying applicative

type ('f, 'r) t

'f is the type of a function that consumes the list of arguments and returns an 'r.

val nil : ('r, 'r) t

the empty argument list *

val cons : 'a arg -> ('f, 'r) t -> ('a -> 'f, 'r) t

prepend an argument

val (@>) : 'a arg -> ('f, 'r) t -> ('a -> 'f, 'r) t

infix operator for cons

val step : ('f1, 'r) t -> f:('f2 -> 'f1) -> ('f2, 'r) t

Transform argument values in some way. For example, one can label a function argument like so:

step ~f:(fun f x -> f ~foo:x) : ('a -> 'r1, 'r2) t -> (foo:'a -> 'r1, 'r2) t

The preferred way to factor out an Args sub-sequence:

let args =
  Foo.Args.(
    bar "A"
    (* TODO: factor out the common baz qux sub-sequence *)
    @> baz "B"
    @> qux "C"
    @> zap "D"
    @> nil
  )

is to write a function that prepends the sub-sequence:

let baz_qux remaining_args =
  Foo.Args.(
    baz "B"
    @> qux "C"
    @> remaining_args
  )

and splice it back into the original sequence using @@ so that things line up nicely:

let args =
  Foo.Args.(
    bar "A"
    @> baz_qux
    @@ zap "D"
    @> nil
  )
val mapN : f:'f -> ('f, 'r) t -> 'r arg
val applyN : 'f arg -> ('f, 'r) t -> 'r arg