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Why is foldl signature is a pipeline instead of a tuple like type in SML?

I do not understand why the signature of foldl is like that.

Standard ML of New Jersey v110.79 [built: Tue Aug  8 23:21:20 2017]
- foldl;
val it = fn : ('a * 'b -> 'b) -> 'b -> 'a list -> 'b

In particular, the arrows -> should've been asterisk * like the following:

val it = fn : ('a * 'b -> 'b) * 'b * 'a list -> 'b

And the reason is because foldl takes the following:

  1. Function ('a * 'b -> 'b) that compares & returns one of two strings; assuming acc is 'b, thus, returning acc
  2. acc & 'a list to compare them.
  3. Finally, returning the base 'b which is acc.
Solution

  • * entails it to be a tuple, and which necessitates that all the components of the tuple will have to be provided... -> entails currying, which allows partial application. Both are 2 different semantics.

    Also, it is interesting that you write ...foldl takes a function that compares two strings....

    foldl is generic and quite polymorphic, so why will it work only with string. A particular function application may work over string data, but not the function definition itself.