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Glasgow Haskell Compiler
GHC
Commits
fcb4fe57
Commit
fcb4fe57
authored
Oct 31, 2012
by
Simon Peyton Jones
Browse files
Wibbles to error messages
parent
f28972a1
Changes
6
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6 changed files
with
127 additions
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90 deletions
+127
-90
testsuite/tests/indexed-types/should_fail/T2239.stderr
testsuite/tests/indexed-types/should_fail/T2239.stderr
+28
-30
testsuite/tests/indexed-types/should_fail/T5934.stderr
testsuite/tests/indexed-types/should_fail/T5934.stderr
+7
-6
testsuite/tests/typecheck/should_fail/T5570.stderr
testsuite/tests/typecheck/should_fail/T5570.stderr
+9
-9
testsuite/tests/typecheck/should_fail/T5689.stderr
testsuite/tests/typecheck/should_fail/T5689.stderr
+39
-10
testsuite/tests/typecheck/should_fail/T5691.stderr
testsuite/tests/typecheck/should_fail/T5691.stderr
+17
-8
testsuite/tests/typecheck/should_fail/tcfail133.stderr
testsuite/tests/typecheck/should_fail/tcfail133.stderr
+27
-27
No files found.
testsuite/tests/indexed-types/should_fail/T2239.stderr
View file @
fcb4fe57
T2239.hs:47:13:
Couldn't match type `forall b1. MyEq b1 Bool => b1 -> b1'
with `b -> b'
Expected type: (forall b1. MyEq b1 Bool => b1 -> b1) -> b -> b
Actual type: (forall b. MyEq b Bool => b -> b)
-> forall b. MyEq b Bool => b -> b
In the expression:
id ::
(forall b. MyEq b Bool => b -> b)
-> (forall b. MyEq b Bool => b -> b)
In an equation for `complexFD':
complexFD
= id ::
(forall b. MyEq b Bool => b -> b)
-> (forall b. MyEq b Bool => b -> b)
T2239.hs:50:13:
Couldn't match type `forall b1. b1 ~ Bool => b1 -> b1'
with `Bool -> Bool'
Expected type: (forall b1. b1 ~ Bool => b1 -> b1) -> b -> b
Actual type: (forall b. b ~ Bool => b -> b)
-> forall b. b ~ Bool => b -> b
In the expression:
id ::
(forall b. b ~ Bool => b -> b) -> (forall b. b ~ Bool => b -> b)
In an equation for `complexTF':
complexTF
= id ::
(forall b. b ~ Bool => b -> b) -> (forall b. b ~ Bool => b -> b)
T2239.hs:47:13:
Couldn't match type `b -> b'
with `forall b1. MyEq b1 Bool => b1 -> b1'
Expected type: (forall b1. MyEq b1 Bool => b1 -> b1) -> b -> b
Actual type: (b -> b) -> b -> b
In the expression:
id ::
(forall b. MyEq b Bool => b -> b)
-> (forall b. MyEq b Bool => b -> b)
In an equation for `complexFD':
complexFD
= id ::
(forall b. MyEq b Bool => b -> b)
-> (forall b. MyEq b Bool => b -> b)
T2239.hs:50:13:
Couldn't match type `Bool -> Bool'
with `forall b1. b1 ~ Bool => b1 -> b1'
Expected type: (forall b1. b1 ~ Bool => b1 -> b1) -> b -> b
Actual type: (b -> b) -> b -> b
In the expression:
id ::
(forall b. b ~ Bool => b -> b) -> (forall b. b ~ Bool => b -> b)
In an equation for `complexTF':
complexTF
= id ::
(forall b. b ~ Bool => b -> b) -> (forall b. b ~ Bool => b -> b)
testsuite/tests/indexed-types/should_fail/T5934.stderr
View file @
fcb4fe57
T5934.hs:10:7:
No instance for (Num ((forall s. GenST s) -> Int))
arising from the literal `0'
In the expression: 0
In an equation for `run': run = 0
T5934.hs:10:7:
Cannot instantiate unification variable `a0'
with a type involving foralls: (forall s. GenST s) -> Int
Perhaps you want -XImpredicativeTypes
In the expression: 0
In an equation for `run': run = 0
testsuite/tests/typecheck/should_fail/T5570.stderr
View file @
fcb4fe57
T5570.hs:7:16:
Couldn't match kind `*' with `#'
When matching types
t
0 :: *
Double# :: #
In the second argument of `($)', namely `D# $ 3.0##'
In the expression: print $ D# $ 3.0##
In an equation for `main': main = print $ D# $ 3.0##
T5570.hs:7:16:
Couldn't match kind `*' with `#'
When matching types
s
0 :: *
Double# :: #
In the second argument of `($)', namely `D# $ 3.0##'
In the expression: print $ D# $ 3.0##
In an equation for `main': main = print $ D# $ 3.0##
testsuite/tests/typecheck/should_fail/T5689.stderr
View file @
fcb4fe57
T5689.hs:10:36:
Couldn't match expected type `Bool' with actual type `t'
Relevant bindings include
r :: IORef (t -> t) (bound at T5689.hs:7:14)
v :: t (bound at T5689.hs:10:28)
In the expression: v
In the expression: if v then False else True
In the second argument of `writeIORef', namely
`(\ v -> if v then False else True)'
T5689.hs:10:36:
Couldn't match expected type `Bool' with actual type `t'
Relevant bindings include
r :: IORef (t -> t) (bound at T5689.hs:7:14)
v :: t (bound at T5689.hs:10:28)
In the expression: v
In the expression: if v then False else True
In the second argument of `writeIORef', namely
`(\ v -> if v then False else True)'
T5689.hs:10:43:
Couldn't match expected type `t' with actual type `Bool'
Relevant bindings include
r :: IORef (t -> t) (bound at T5689.hs:7:14)
v :: t (bound at T5689.hs:10:28)
In the expression: False
In the expression: if v then False else True
In the second argument of `writeIORef', namely
`(\ v -> if v then False else True)'
T5689.hs:10:54:
Couldn't match expected type `t' with actual type `Bool'
Relevant bindings include
r :: IORef (t -> t) (bound at T5689.hs:7:14)
v :: t (bound at T5689.hs:10:28)
In the expression: True
In the expression: if v then False else True
In the second argument of `writeIORef', namely
`(\ v -> if v then False else True)'
T5689.hs:14:23:
Couldn't match expected type `t' with actual type `Bool'
Relevant bindings include
r :: IORef (t -> t) (bound at T5689.hs:7:14)
c :: t -> t (bound at T5689.hs:12:13)
In the first argument of `c', namely `True'
In the second argument of `($)', namely `c True'
In a stmt of a 'do' block: print $ c True
testsuite/tests/typecheck/should_fail/T5691.stderr
View file @
fcb4fe57
T5691.hs:14:9:
Couldn't match type `p' with `PrintRuleInterp'
Expected type: p a
Actual type: PrintRuleInterp a
In the pattern: f :: p a
In an equation for `test': test (f :: p a) = MkPRI $ printRule_ f
In the instance declaration for `Test PrintRuleInterp'
T5691.hs:14:9:
Couldn't match type `p' with `PrintRuleInterp'
Expected type: PrintRuleInterp a
Actual type: p a
In the pattern: f :: p a
In an equation for `test': test (f :: p a) = MkPRI $ printRule_ f
In the instance declaration for `Test PrintRuleInterp'
T5691.hs:15:24:
Couldn't match type `p' with `PrintRuleInterp'
Expected type: PrintRuleInterp a
Actual type: p a
Relevant bindings include f :: p a (bound at T5691.hs:14:9)
In the first argument of `printRule_', namely `f'
In the second argument of `($)', namely `printRule_ f'
In the expression: MkPRI $ printRule_ f
testsuite/tests/typecheck/should_fail/tcfail133.stderr
View file @
fcb4fe57
tcfail133.hs:2:61: Warning:
-XDatatypeContexts is deprecated: It was widely considered a misfeature, and has been removed from the Haskell language.
tcfail133.hs:68:7:
No instance for (Show
t
0) arising from a use of `show'
The type variable `
t
0' is ambiguous
Possible fix: add a type signature that fixes these type variable(s)
Note: there are several potential instances:
instance Show Zero -- Defined at tcfail133.hs:8:29
instance Show One -- Defined at tcfail133.hs:9:28
instance (Show a, Show b, Number a, Digit b) => Show (a :@ b)
-- Defined at tcfail133.hs:11:54
...plus 26 others
In the expression: show
In the expression: show $ add (One :@ Zero) (One :@ One)
In an equation for `foo':
foo = show $ add (One :@ Zero) (One :@ One)
tcfail133.hs:68:14:
No instance for (AddDigit (Zero :@ (One :@ One)) One
t
0)
arising from a use of `add'
In the second argument of `($)', namely
`add (One :@ Zero) (One :@ One)'
In the expression: show $ add (One :@ Zero) (One :@ One)
In an equation for `foo':
foo = show $ add (One :@ Zero) (One :@ One)
tcfail133.hs:2:61: Warning:
-XDatatypeContexts is deprecated: It was widely considered a misfeature, and has been removed from the Haskell language.
tcfail133.hs:68:7:
No instance for (Show
s
0) arising from a use of `show'
The type variable `
s
0' is ambiguous
Possible fix: add a type signature that fixes these type variable(s)
Note: there are several potential instances:
instance Show Zero -- Defined at tcfail133.hs:8:29
instance Show One -- Defined at tcfail133.hs:9:28
instance (Show a, Show b, Number a, Digit b) => Show (a :@ b)
-- Defined at tcfail133.hs:11:54
...plus 26 others
In the expression: show
In the expression: show $ add (One :@ Zero) (One :@ One)
In an equation for `foo':
foo = show $ add (One :@ Zero) (One :@ One)
tcfail133.hs:68:14:
No instance for (AddDigit (Zero :@ (One :@ One)) One
s
0)
arising from a use of `add'
In the second argument of `($)', namely
`add (One :@ Zero) (One :@ One)'
In the expression: show $ add (One :@ Zero) (One :@ One)
In an equation for `foo':
foo = show $ add (One :@ Zero) (One :@ One)
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