问题
I was thinking about how I could save myself from undefinition, and one idea I had was to enumerate all possible sources of partiality. At least I would know what of to beware. I found three yet:
- Incomplete pattern matches or guards.
- Recursion. (Optionally excluding structural recursion on algebraic types.)
- If a function is unsafe, any use of that function infects the user code. (Should I be saying "partiality is transitive"?)
I have heard of other ways to obtain a logical contradiction, for instance by using negative types, but I am not sure if anything of that sort applies to Haskell. There are many logical paradoxes out there, and some of them can be encoded in Haskell, but may it be true that any logical paradox requires the use of recursion, and is therefore covered by the point 2 above?
For instance, if it were proven that a Haskell expression free of recursion can always be evaluated to normal form, then the three points I give would be a complete list. I fuzzily remember seeing something like a proof of this in one of Simon Peyton Jones's books, but that was written like 30 years ago, so even if I remember correctly and it used to apply to a prototype Haskell back then, it may be false today, seeing how many a language extension we have. Possibly some of them enable other ways to undefine a program?
And then, if it were so easy to detect expressions that cannot be partial, why do we not do that? How easier would life be!
回答1:
This is a partial answer (pun intended), where I'll only list a few arguably non obvious ways one can achieve non termination.
First, I'll confirm that negative-recursive types can indeed cause non termination. Indeed, it is known that allowing a recursive type such as
data R a = R (R a -> a)
allows one to define fix, and obtain non termination from there.
{-# LANGUAGE ScopedTypeVariables #-}
{-# OPTIONS -Wall #-}
data R a = R (R a -> a)
selfApply :: R a -> a
selfApply t@(R x) = x t
-- Church's fixed point combinator Y
-- fix f = (\x. f (x x))(\x. f (x x))
fix :: forall a. (a -> a) -> a
fix f = selfApply (R (\x -> f (selfApply x)))
Total languages like Coq or Agda prohibit this by requiring recursive types to use only strictly-positive recursion.
Another potential source of non-termination is that Haskell allows Type :: Type. As far as I can see, that makes it possible to encode System U in Haskell, where Girard's paradox can be used to cause a logical inconsistency, constructing a term of type Void. That term (as far as I understand) would be non terminating.
Girard's paradox is unfortunately rather complex to fully describe, and I have not completely studied it yet. I only know it is related to the so-called hypergame, a game where the first move is to choose a finite game to play. A finite game is one which causes every match to terminate after finitely many moves. The next moves after that would correspond to a match according to the chosen finite game at step one. Here's the paradox: since the chosen game must be finite, no matter what it is, the whole hypergame match will always terminate after a finite amount of moves. This makes hypergame itself a finite game, making the infinite sequence of moves "I choose hypergame, I choose hypergame, ..." a valid play, in turn proving that hypergame is not finite.
Apparently, this argument can be encoded in a rich enough pure type system like System U, and Type :: Type allows to embed the same argument.
来源：https://stackoverflow.com/questions/58355074/how-to-make-a-partial-function