Typed holes are a feature GHC borrowed from Agda. They are incredibly useful for incremental development, making it really easy to go from correct (but incomplete) state to a more complete version.

Unfortunately, Austin Seipp’s tutorial assumes knowledge of type classes and to some extent free monads. This isn’t necessary, and can be off-putting to beginners, which is a shame: I’ve found them one of the easiest ways for newbie Haskellers to break down problems iteratively. One of the worst feelings you can get learning a language is getting stuck and not knowing how to unstick yourself: following hole-driven dev when you aren’t quite sure what you’re doing is a great way to avoid getting lost in the weeds.

In aid of that, I’ve taken a very easy exercise and shown how you might solve it in microscopic detail. I deliberately don’t use pattern matching, user datatypes or anything beyond the absolute basics.

Here, we set up the problem - in practice, we would not name g,g2… but just change in-place. I’ve named them all to make it clear.

In our first version, we’ll just name the arguments to the function and insert a hole for the result.

The problem we are trying to solve: given a function that takes an ‘a’ to a ‘b’, and a pair with an ‘a’ and a ‘c’, return a ‘b’ and a ‘c’.

g,g2,g3,g4,g5,g6,g7 :: (a -> b) -> (a, c) -> (b, c) g x y = _foo

src/Text/FastEdit.hs:56:9: Found hole ‘_foo’ with type: (b, c) Where: ‘b’ is a rigid type variable bound by the type signature for g :: (a -> b) -> (a, c) -> (b, c) at src/Text/FastEdit.hs:55:24 ‘c’ is a rigid type variable bound by the type signature for g :: (a -> b) -> (a, c) -> (b, c) at src/Text/FastEdit.hs:55:24 Relevant bindings include y :: (a, c) (bound at src/Text/FastEdit.hs:56:5) x :: a -> b (bound at src/Text/FastEdit.hs:56:3) g :: (a -> b) -> (a, c) -> (b, c) (bound at src/Text/FastEdit.hs:56:1) In the expression: _foo In an equation for ‘g’: g x y = _foo

If we look at the type GHC has found for _foo, clearly our result is going to be a pair: let’s split our hole up.

g2 x y = (_foo,_bar)

src/Text/FastEdit.hs:57:11: Found hole ‘_foo’ with type: b Where: ‘b’ is a rigid type variable bound by the type signature for g2 :: (a -> b) -> (a, c) -> (b, c) at src/Text/FastEdit.hs:55:24 Relevant bindings include y :: (a, c) (bound at src/Text/FastEdit.hs:57:6) x :: a -> b (bound at src/Text/FastEdit.hs:57:4) g2 :: (a -> b) -> (a, c) -> (b, c) (bound at src/Text/FastEdit.hs:57:1) In the expression: _foo In the expression: (_foo, _bar) In an equation for ‘g2’: g2 x y = (_foo, _bar) src/Text/FastEdit.hs:57:16: Found hole ‘_bar’ with type: c Where: ‘c’ is a rigid type variable bound by the type signature for g2 :: (a -> b) -> (a, c) -> (b, c) at src/Text/FastEdit.hs:55:24 Relevant bindings include y :: (a, c) (bound at src/Text/FastEdit.hs:57:6) x :: a -> b (bound at src/Text/FastEdit.hs:57:4) g2 :: (a -> b) -> (a, c) -> (b, c) (bound at src/Text/FastEdit.hs:57:1) In the expression: _bar In the expression: (_foo, _bar) In an equation for ‘g2’: g2 x y = (_foo, _bar)

Focusing on _bar first, we know that it’s got type ‘c’. Now we have to look at the relevant bindings: we need to find some way of getting a ‘c’ out of them. To do this, we look at the final return type of each. x doesn’t mention c, so we can ignore it. g2 does, but it also needs to have a c passed in, so it can’t be the ultimate source. That leaves y. It’s not quite the right type, but let’s pick that out as an argument and update the hole.

g3 x y = (_foo,_bar y)

src/Text/FastEdit.hs:58:11: Found hole ‘_foo’ with type: b Where: ‘b’ is a rigid type variable bound by the type signature for g3 :: (a -> b) -> (a, c) -> (b, c) at src/Text/FastEdit.hs:55:24 Relevant bindings include y :: (a, c) (bound at src/Text/FastEdit.hs:58:6) x :: a -> b (bound at src/Text/FastEdit.hs:58:4) g3 :: (a -> b) -> (a, c) -> (b, c) (bound at src/Text/FastEdit.hs:58:1) In the expression: _foo In the expression: (_foo, _bar y) In an equation for ‘g3’: g3 x y = (_foo, _bar y) src/Text/FastEdit.hs:58:16: Found hole ‘_bar’ with type: (a, c) -> c Where: ‘a’ is a rigid type variable bound by the type signature for g3 :: (a -> b) -> (a, c) -> (b, c) at src/Text/FastEdit.hs:55:24 ‘c’ is a rigid type variable bound by the type signature for g3 :: (a -> b) -> (a, c) -> (b, c) at src/Text/FastEdit.hs:55:24 Relevant bindings include y :: (a, c) (bound at src/Text/FastEdit.hs:58:6) x :: a -> b (bound at src/Text/FastEdit.hs:58:4) g3 :: (a -> b) -> (a, c) -> (b, c) (bound at src/Text/FastEdit.hs:58:1) In the expression: _bar In the expression: _bar y In the expression: (_foo, _bar y)

Progress! Again focusing on the second error message, we now need a function from (a,c) to c. The relevant bindings don’t seem to help us here, so we look up (a,c) -> c on hoogle - the first result is snd, which seems about right.

g4 x y = (_foo,snd y)

src/Text/FastEdit.hs:59:11: Found hole ‘_foo’ with type: b Where: ‘b’ is a rigid type variable bound by the type signature for g4 :: (a -> b) -> (a, c) -> (b, c) at src/Text/FastEdit.hs:55:24 Relevant bindings include y :: (a, c) (bound at src/Text/FastEdit.hs:59:6) x :: a -> b (bound at src/Text/FastEdit.hs:59:4) g4 :: (a -> b) -> (a, c) -> (b, c) (bound at src/Text/FastEdit.hs:59:1) In the expression: _foo In the expression: (_foo, snd y) In an equation for ‘g4’: g4 x y = (_foo, snd y)

We’ve finished the second hole, so we focus on _foo now. Looking at the relevant bindings, we can dismiss y because it doesn’t return a ‘b’. x looks like the most relevant piece, but it is clear we need to pass something into it, because it’s a function. We don’t know what yet, so we let the hole do the work.

g5 x y = ( x _foo,snd y)

src/Text/FastEdit.hs:60:14: Found hole ‘_foo’ with type: a Where: ‘a’ is a rigid type variable bound by the type signature for g5 :: (a -> b) -> (a, c) -> (b, c) at src/Text/FastEdit.hs:55:24 Relevant bindings include y :: (a, c) (bound at src/Text/FastEdit.hs:60:6) x :: a -> b (bound at src/Text/FastEdit.hs:60:4) g5 :: (a -> b) -> (a, c) -> (b, c) (bound at src/Text/FastEdit.hs:60:1) In the first argument of ‘x’, namely ‘_foo’ In the expression: x _foo In the expression: (x _foo, snd y)

On the home straight now! We have an ‘a’ hidden inside y:

g6 x y = ( x (_foo y) ,snd y)

src/Text/FastEdit.hs:61:15: Found hole ‘_foo’ with type: (a, c) -> a Where: ‘a’ is a rigid type variable bound by the type signature for g6 :: (a -> b) -> (a, c) -> (b, c) at src/Text/FastEdit.hs:55:24 ‘c’ is a rigid type variable bound by the type signature for g6 :: (a -> b) -> (a, c) -> (b, c) at src/Text/FastEdit.hs:55:24 Relevant bindings include y :: (a, c) (bound at src/Text/FastEdit.hs:61:6) x :: a -> b (bound at src/Text/FastEdit.hs:61:4) g6 :: (a -> b) -> (a, c) -> (b, c) (bound at src/Text/FastEdit.hs:61:1) In the expression: _foo In the first argument of ‘x’, namely ‘(_foo y)’ In the expression: x (_foo y)

We could use hoogle again, but we happen to remember that ‘fst’ can be used to pick out the first element of a pair:

g7 x y = (x (fst y) ,snd y)

and we’re done.

I realise this may seem like a huge amount of work for a fairly trivial function, but the important point is that we never got off-track: if GHC gives us an error rather than information about a hole, we don’t even try to fix it: we just roll back to our last good state and try again. We were never lost in the weeds; we moved relentlessly forward. In practice, experienced Haskellers will have something like ghc-mod running so they can get faster feedback in-editor, but we can still do it with just an editor and an open GHCi.

# ADDENDUM

As @benno37 pointed out, it’s the underscore that makes it a typed
hole. This is quite clever really - it is *only* considered a hole if
it would otherwise be a scope error, so if for some misbegotten reason
you decide to name a parameter to a function “_foo”, it will remain a
normal variable.