#7220 closed bug (fixed)
Confusing error message in type checking related to type family, fundep, and higher-rank type
Reported by: | tsuyoshi | Owned by: | simonpj |
---|---|---|---|
Priority: | normal | Milestone: | 7.8.1 |
Component: | Compiler (Type checker) | Version: | 7.6.1-rc1 |
Keywords: | Cc: | ||
Operating System: | Unknown/Multiple | Architecture: | Unknown/Multiple |
Type of failure: | None/Unknown | Test Case: | typecheck/should_fail/T7220 |
Blocked By: | Blocking: | ||
Related Tickets: | Differential Rev(s): |
Description
(This is related to, but different from, the message which I posted to glasgow-haskell-users mailing list: http://www.haskell.org/pipermail/glasgow-haskell-users/2012-September/022815.html.)
GHC 7.6.1-rc1 (7.6.0.20120810; 64-bit Windows) rejects the attached code (Test2.hs) with the following error message:
Test2.hs:24:52: Couldn't match expected type `Y' with actual type `TF (forall b. (C A b, TF b ~ Y) => b)' In the first argument of `f :: (forall b. (C A b, TF b ~ Y) => b) -> X', namely `u' In the expression: (f :: (forall b. (C A b, TF b ~ Y) => b) -> X) u In an equation for `v': v = (f :: (forall b. (C A b, TF b ~ Y) => b) -> X) u
I am not sure whether the code is supposed to be accepted or rejected, but even if it is correct to reject the code, this error message does not look right to me. If I am not mistaken, the error message is saying that the type checker expects the argument of (f :: (forall b. (C A b, TF b ~ Y) => b) -> X) to have type Y, but I cannot think of any reason why it should relate the type (forall b. (C A b, TF b ~ Y) => b) with Y.
The same code is available also at https://gist.github.com/3606856.
Attachments (1)
Change History (6)
Changed 3 years ago by tsuyoshi
comment:1 Changed 3 years ago by tsuyoshi
comment:2 Changed 3 years ago by igloo
- difficulty set to Unknown
- Milestone set to 7.8.1
- Owner set to simonpj
Simon, could you take a look at this please?
comment:3 Changed 3 years ago by simonpj
- Resolution set to fixed
- Status changed from new to closed
- Test Case set to typecheck/should_fail/T7220
"I cannot think of any reason why it should relate the type (forall b. (C A b, TF b ~ Y) => b) with Y". Here's why:
u :: (C A b, TF b ~ Y) => b u = undefined v :: X v = (f :: (forall b. (C A b, TF b ~ Y) => b) -> X) u
- GHC instanatiates the occurence of u, given type just beta, where beta is a fresh unification variable. Plus constraints (C A beta, TF beta ~ Y).
- Now it tries to unify beta with the expected arugment type of the (f :: sig) expression. This argument type is (forall b. (C A b, TF b ~ Y) => b)
- So GHC (until today) would unify beta := forall b. (C A b, TF b ~ Y) => b.
- So the constraint (TF beta ~ Y) turns into (TF (forall b. (C A b, TF b ~ Y) => b) ~ Y)`, and that's the error you saw.
But really GHC should not have taken the third step above (at least not without -XImpredicativeTypes). With the patch that fixes #6069 and #7264 it no longer does so, yielding instead
T7220.hs:24:6: Cannot instantiate unification variable `b0' with a type involving foralls: forall b. (C A b, TF b ~ Y) => b Perhaps you want -XImpredicativeTypes In the expression: f :: (forall b. (C A b, TF b ~ Y) => b) -> X In the expression: (f :: (forall b. (C A b, TF b ~ Y) => b) -> X) u In an equation for `v': v = (f :: (forall b. (C A b, TF b ~ Y) => b) -> X) u
which isn't fantastic but is perhaps less confusing than before.
comment:4 Changed 11 months ago by simonpj
Richard's fix for #9404 makes this compile. He says:
The reason this went wrong before is that the old method of inferring a type would get caught when unifying a TauTv (the metavariable given as the guess for f's type) with a polytype, deferring to the solver. If we defer, then we don't have enough information to type-check u successfully, and we fail. With the new inferring types with PolyTv, no deferring happens, and u type-checks fine.
But with the #9404 fix, f is inferred to have the type as given. Then, u is checked (by tcPolyExprNC) to have the argument type of f, which it does. No problems.
I actually think that success is the right behavior here, so I chalk this up to a success of this patch.
When I posted this report, I had been overlooking TF in the “actual type” part of the error message. Now the error message at least makes some sense.
(Although I do not fully understand why the code is rejected in the first place, I think that I have to understand the algorithm used by the type checker to understand this.)
I am not sure whether this bug report still makes sense or not. If not, please close it. Sorry about this.