问题
I am using this type to reason about strings on which decidable parsing can be performed:
data Every : (a -> Type) -> List a -> Type where
Nil : {P : a -> Type} -> Every P []
(::) : {P : a -> Type} -> P x -> Every P xs -> Every P (x::xs)
For example, defining the digits [0-9] like this:
data Digit : Char -> Type where
Zero : Digit '0'
One : Digit '1'
Two : Digit '2'
Three : Digit '3'
Four : Digit '4'
Five : Digit '5'
Six : Digit '6'
Seven : Digit '7'
Eight : Digit '8'
Nine : Digit '9'
digitToNat : Digit a -> Nat
digitToNat Zero = 0
digitToNat One = 1
digitToNat Two = 2
digitToNat Three = 3
digitToNat Four = 4
digitToNat Five = 5
digitToNat Six = 6
digitToNat Seven = 7
digitToNat Eight = 8
digitToNat Nine = 9
then we can have the following functions:
fromDigits : Every Digit xs -> Nat -> Nat
fromDigits [] k = 0
fromDigits (x :: xs) k = (digitToNat x) * (pow 10 k) + fromDigits xs (k-1)
s2n : (s : String) -> {auto p : Every Digit (unpack s)} -> Nat
s2n {p} s = fromDigits p (length s - 1)
This s2n function will now work fine at compile time, but then that isn't very useful in of itself. To use it at runtime we must construct the proof Every Digit (unpack s) before we can use the function.
So I think I now want to write something like this function:
every : (p : a -> Type) -> (xs : List a) -> Maybe $ Every p xs
That or we want to return either a proof of membership or a proof of non-membership, but I'm not entirely sure how to do either of these things in a general way. So instead I tried doing the Maybe version for just characters:
every : (p : Char -> Type) -> (xs : List Char) -> Maybe $ Every p xs
every p [] = Just []
every p (x :: xs) with (decEq x '0')
every p ('0' :: xs) | (Yes Refl) = Just $ p '0' :: !(every p xs)
every p (x :: xs) | (No contra) = Nothing
But then I get this unification error:
Can't unify
Type
with
p '0'
Specifically:
Can't unify
Type
with
p '0'
But p is of type Char -> Type. I'm not sure what is causing this unification failure, but think the problem may be related to my previous question.
Is this a sensible approach to what I am trying to do? I feel like it is a little to much work at the moment, and more general versions of these functions should be possible. It would be nice if the auto keyword could be used to write a function gives you a Maybe proof or an Either proof proofThatItIsNot, in a similar way to how the DecEq class works.
回答1:
The error message is correct: you provided a value of type Type, but you need a value of type p '0'. You are also correct that p is of type Char -> Type, and therefore that p '0' is of type Type. However, p '0' is not of type p '0'.
Perhaps the issue would be easier to see with simpler types: 3 has type Int, and Int has type Type, but Int does not have type Int.
Now, how do we fix the problem? Well, p is a predicate, meaning that it constructs types whose inhabitants are proofs of this predicate. So the value of type p '0' we need to provide would be a proof, in this case a proof that '0' is a digit. Zero happens to be such a proof. But in the signature of every, the p variable isn't talking about digits: it's an abstract predicate, about which we know nothing. For this reason, there are no values we could use instead of p '0'. We must instead change the type of every.
One possibility would be to write a more specialized version version of every, one which would only work for the particular predicate Digit instead of working for an arbitrary p:
everyDigit : (xs : List Char) -> Maybe $ Every Digit xs
everyDigit [] = Just []
everyDigit (x :: xs) with (decEq x '0')
everyDigit ('0' :: xs) | (Yes Refl) = Just $ Zero :: !(everyDigit xs)
everyDigit (x :: xs) | (No contra) = Nothing
Instead of incorrectly using the value p '0' in a spot which needs a value of type p '0', I have used the value Zero in a spot which now needs a value of type Digit '0'.
Another possibility would be to modify every so that in addition to the predicate p which gives a proof type for every Char, we would also receive a proof-making function mkPrf which would give the corresponding proof value for every Char, when possible.
every : (p : Char -> Type)
-> (mkPrf : (c : Char) -> Maybe $ p c)
-> (xs : List Char)
-> Maybe $ Every p xs
every p mkPrf [] = Just []
every p mkPrf (x :: xs) with (mkPrf x)
every p mkPrf (x :: xs) | Just prf = Just $ prf :: !(every p mkPrf xs)
every p mkPrf (x :: xs) | Nothing = Nothing
I am no longer pattern-matching on the Char, instead I am asking mkPrf to examine the Char. I then pattern-match on the result, to see if it found a proof. It is the implementation of mkPrf which pattern-matches on the Char.
everyDigit' : (xs : List Char) -> Maybe $ Every Digit xs
everyDigit' = every Digit mkPrf
where
mkPrf : (c : Char) -> Maybe $ Digit c
mkPrf '0' = Just Zero
mkPrf _ = Nothing
In the implementation of mkPrf, we are again constructing a proof for the concrete type Digit '0' instead of the abstract type p '0', so Zero is an acceptable proof.
来源：https://stackoverflow.com/questions/27095563/generating-run-time-proofs-with-type-predicates-in-idris