Commit 852fe754 authored by panne's avatar panne

[project @ 2004-11-06 10:01:41 by panne]

Updated expected output
parent a4b3ba35
Loading package base ... linking ... done.
Loading package haskell98 ... linking ... done.
Loading package template-haskell ... linking ... done.
foo :: GHC.Base.Int -> GHC.Base.Int
foo x_0 | x_0 GHC.Base.== 5 = 6
foo x_1 = 7
......@@ -8,3 +5,6 @@ bar :: Data.Maybe.Maybe GHC.Base.Int -> GHC.Base.Int
bar x_0 | Data.Maybe.Just y_1 <- x_0
= y_1
bar _ = 9
Loading package base ... linking ... done.
Loading package haskell98 ... linking ... done.
Loading package template-haskell ... linking ... done.
tcfail103.hs:15:0:
Inferred type is less polymorphic than expected
Quantified type variable `t' escapes
It is mentioned in the environment:
s = t is bound by the pattern type signature at tcfail103.hs:15:7
Quantified type variable `t' escapes
It is mentioned in the environment:
s = t bound at: tcfail103.hs:15:7
When trying to generalise the type inferred for `f'
Signature type: forall t. ST t Int
Type to generalise: ST t Int
Signature type: forall t. ST t Int
Type to generalise: ST t Int
In the type signature for `f'
When generalising the type(s) for `f'
tcfail104.hs:9:14:
Inferred type is less polymorphic than expected
Quantified type variable `a' escapes
Expected type: forall a1. a1 -> a1
Inferred type: a -> a
When checking the pattern: x :: forall a. a -> a
Quantified type variable `a' escapes
Expected type: forall a1. a1 -> a1
Inferred type: a -> a
In a lambda abstraction: \ (x :: forall a. a -> a) -> x
In the definition of `f':
f v = (if v then (\ (x :: forall a. a -> a) -> x) else (\ x -> x)) id 'c'
tcfail104.hs:15:14:
Inferred type is less polymorphic than expected
Quantified type variable `a' escapes
Expected type: forall a1. a1 -> a1
Inferred type: a -> a
When checking the pattern: x :: forall a. a -> a
Quantified type variable `a' escapes
Expected type: forall a1. a1 -> a1
Inferred type: a -> a
In a lambda abstraction: \ (x :: forall a. a -> a) -> x
In the definition of `g':
g v = (if v then (\ (x :: forall a. a -> a) -> x) else (\ x -> x)) id 'c'
Markdown is supported
0%
or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment