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The test was failing with: T7861: T7861.hs:15:13: Cannot instantiate unification variable ‘t0’ with a type involving foralls: A a0 -> a0 GHC doesn't yet support impredicative polymorphism In the first argument of ‘seq’, namely ‘f’ In a stmt of a 'do' block: f `seq` print "Hello 2" It requires ImpredicativeTypes, at least since 7.8, because we instantiate seq's type (c->d->d) with f's type (c:= (forall b. a) -> a), which is polymorphic (it has foralls). I simplified the test a bit by removing the type synonym, and verified that ghc-7.6.3 still panics on this test. Reviewers: simonpj, austin, bgamari Reviewed By: bgamari Subscribers: thomie Differential Revision: https://phabricator.haskell.org/D1080 GHC Trac Issues: #7861
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