Why does failwithf expect an unit at the end?

[Maverick Woo]

Hi,

Out of curiosity, may I know the reason why failwithf (and invalid_argf) wants an unit at the end of an invocation? What advantage does this bring over a version that does not need an ending unit?

Thank you!

Maverick

[Ben Millwood]

It enables better format string checking. The problem is that failwith-like functions have fully polymorphic return types (they don't return, so they can claim to return whatever type they like), and in particular [failwith "oh no"] can act as a function and take further parameters.

Thus, if you have a failwith-like function that takes a variable number of parameters, you can easily lose the static checking that you aren't giving too many. For example, you can define a unitless failwithf:

utop # let f fmt = ksprintf (fun s -> failwith s) fmt;;

val f : ('a, unit, string, 'b) format4 -> 'a = <fun>

Now [f "%d" 4] has type ['a]. Unfortunately, this type can be unified with, say, [int -> 'b], so [f "%d" 4 5] is no longer a type error, it just raises [Failure "4"].

If, on the other hand, you define failwithf:

utop # let f fmt = ksprintf (fun s -> fun () -> failwith s) fmt;;

val f : ('a, unit, string, unit -> 'b) format4 -> 'a = <fun>

Now [f "%d" 4 5] complains that 5 doesn't match type unit. The () parameter is serving as a signpost for "end of format arguments".

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