Find the type error in the following Haskell expression:

if null xs then tail xs else xs

You can’t, of course: this program is obviously nonsense unless you?re a typechecker. The trouble is that only certain computations make sense if the `null xs`

test is `True`

, whilst others make sense if it is `False`

. However, as far as the type system is concerned, the type of the then branch is the type of the else branch is the type of the entire conditional. Statically, the test is irrelevant. Which is odd, because if the test really were irrelevant, we wouldn?t do it. Of course, `tail []`

doesn?t go wrong – well-typed programs don?t go wrong – so we?d better pick a different word for the way they do go.

The above quote is an opening paragraph of Conor McBride’s “Epigram: Practical Programming with Dependent Types” paper. As always, Conor makes a good point – this test is completely irrelevant for the typechecker although it is very relevant at run time. Clearly the type system fails to accurately approximate runtime behaviour of our program. In this short post I will show how to fix this in Haskell using dependent types.

The problem is that the types used in this short program carry no information about the manipulated data. This is true both for `Bool`

returned by `null xs`

, which contains no evidence of the result, as well as lists, that store no information about their length. As some of you probably realize the latter is easily fixed by using vectors, ie. length-indexed lists:

data N = Z | S N -- natural numbers
data Vec a (n :: N) where
Nil :: Vec a Z
Cons :: a -> Vec a n -> Vec a (S n) |

The type of vector encodes its length, which means that the type checker can now be aware whether it is dealing with an empty vector. Now let’s write `null`

and `tail`

functions that work on vectors:

vecNull :: Vec a n -> Bool
vecNull Nil = True
vecNull (Cons _ _) = False
vecTail :: Vec a (S n) -> Vec a n
vecTail (Cons _ tl) = tl |

`vecNull`

is nothing surprising – it returns `True`

for empty vector and `False`

for non-empty one. But the tail function for vectors differs from its implementation for lists. `tail`

from Haskell’s standard prelude is not defined for an empty list so calling `tail []`

results in an exception (that would be the case in Conor’s example). But the type signature of `vecTail`

requires that input vector is non-empty. As a result we can rule out the `Nil`

case. That also means that Conor’s example will no longer typecheck^{1}. But how can we write a correct version of this example, one that removes first element of a vector only when it is non-empty? Here’s an attempt:

shorten :: Vec a n -> Vec a m
shorten xs = case vecNull xs of
True -> xs
False -> vecTail xs |

That however won’t compile: now that we written type-safe tail function typechecker requires a proof that vector passed to it as an argument is non empty. The weak link in this code is the `vecNull`

function. It tests whether a vector is empty but delivers no type-level proof of the result. In other words we need:

vecNull` :: Vec a n -> IsNull n |

ie. a function with result type carrying the information about the length of the list. This data type will have the runtime representation isomorphic to `Bool`

, ie. it will be an enumeration with two constructors, and the type index will correspond to length of a vector:

data IsNull (n :: N) where
Null :: IsNull Z
NotNull :: IsNull (S n) |

`Null`

represents empty vectors, `NotNull`

represents non-empty ones. We can now implement a version of `vecNull`

that carries proof of the result at the type level:

vecNull` :: Vec a n -> IsNull n
vecNull` Nil = Null
vecNull` (Cons _ _) = NotNull |

The type signature of `vecNull``

says that the return type must have the same index as the input vector. Pattern matching on the `Nil`

case provides the type checker with the information that the `n`

index of `Vec`

is `Z`

. This means that the return value in this case must be `Null`

– the `NotNull`

constructor is indexed with `S`

and that obviously does not match `Z`

. Similarly in the `Cons`

case the return value must be `NotNull`

. However, replacing `vecNull`

in the definition of `shorten`

with our new `vecNull``

will again result in a type error. The problem comes from the type signature of `shorten`

:

shorten :: Vec a n -> Vec a m |

By indexing input and output vectors with different length indices – `n`

and `m`

– we tell the typechecker that these are completely unrelated. But that is not true! Knowing the input length `n`

we know exactly what the result should be: if the input vector is empty the result vector is also empty; if the input vector is not empty it should be shortened by one. Since we need to express this at the type level we will use a type family:

type family Pred (n :: N) :: N where
Pred Z = Z
Pred (S n) = n |

(In a fully-fledged dependently-typed language we would write normal function and then apply it at the type level.) Now we can finally write:

shorten :: Vec a n -> Vec a (Pred n)
shorten xs = case vecNull` xs of
Null -> xs
NotNull -> vecTail xs |

This definition should not go wrong. Trying to swap expression in the branches will result in a type error.