use*_*928 20 haskell typeclass ghc
当条件类型类实例运行得很深时,可能很难弄清楚为什么ghc抱怨缺少类型类实例.例如:
class MyClass1 c
class MyClass2 c
class MyClass3 c
data MyType1 a
data MyType2 a
instance MyClass1 a => MyClass2 (MyType1 a)
instance MyClass2 a => MyClass3 (MyType2 a)
foo :: (MyClass3 c) => c
foo = undefined
bar :: MyType2 (MyType1 Int)
bar = foo
GHC给出以下错误:
Example.hs:149:7-9: error:
• No instance for (MyClass1 Int) arising from a use of ‘foo’
• In the expression: foo
In an equation for ‘bar’: bar = foo
|
149 | bar = foo
| ^^^
假设我只写了foo和的定义bar,而其他一切都是我没写过的导入代码,我可能会为ghc为什么试图找到一个MyClass1实例而感到困惑Int.这可能是我第一次听说过这个MyClass1课程,如果到目前为止我一直依赖导入的实例.如果ghc可以给我一个类型类实例链的"堆栈跟踪",那将是很好的,例如
Sought (MyClass2 (MyType1 Int)) to satisfy (MyClass3 (MyType2 (MyType1 Int))) from conditional type class instance OtherModule.hs:37:1-18
Sought (MyClass1 Int) to satisfy (MyClass2 (MyType1 Int)) from conditional type class instance OtherModule.hs:36:1-18
ghc有这个命令行选项吗？如果没有,我该如何调试呢？
请记住,我的真正问题比这个简单的例子复杂得多.例如
Search.hs:110:31-36: error:
• Could not deduce (Ord
(Vars (DedupingMap (Rep (Index gc)) (IndexedProblem ac))))
arising from a use of ‘search’
from the context: (PP gc (IndexedProblem ac),
Show (Vars (DedupingMap (Rep (Index gc)) (IndexedProblem ac))),
Foldable f, MonadNotify m)
bound by the type signature for:
searchIndexedReplicaProblem :: forall gc ac (f :: * -> *) (m :: *
-> *).
(PP gc (IndexedProblem ac),
Show
(Vars
(DedupingMap
(Rep (Index gc)) (IndexedProblem ac))),
Foldable f, MonadNotify m) =>
f (Index
(Clzs
(PartitionedProblem
gc (IndexedProblem ac))))
-> m (Maybe
(Vars
(PartitionedProblem
gc (IndexedProblem ac))))
at Search.hs:(103,1)-(109,131)
• In the expression: search
In an equation for ‘searchIndexedReplicaProblem’:
searchIndexedReplicaProblem = search
|
110 | searchIndexedReplicaProblem = search
| ^^^^^^
PP有五种覆盖条件,我使用的是类型族和不可判定的实例,因此ghc给我错误的原因非常不明显.我可以使用哪些工具来追踪问题？
您可以尝试-ddump-cs-trace选项,虽然它旨在帮助GHC开发人员在调试约束解决代码时,但它也可能对您有用.以下是您的示例的输出:
Step 1[l:2,d:0] Kept as inert:
[G] $dMyClass3_a1rt {0}:: MyClass3 c_a1rs[sk:2]
Step 2[l:2,d:0] Dict equal (keep):
[WD] $dMyClass3_a1rv {0}:: MyClass3 c_a1rs[sk:2]
Constraint solver steps = 2
Step 1[l:1,d:0] Top react: Dict/Top (solved wanted):
[WD] $dMyClass3_a2uc {0}:: MyClass3 (MyType2 (MyType1 Int))
Step 2[l:1,d:1] Top react: Dict/Top (solved wanted):
[WD] $dMyClass2_a2up {1}:: MyClass2 (MyType1 Int)
Step 3[l:1,d:2] Kept as inert:
[WD] $dMyClass1_a2uq {2}:: MyClass1 Int
Step 4[l:2,d:0] Kept as inert:
[G] $dMyClass3_a1rB {0}:: MyClass3 c_a1rz[sk:2]
Step 5[l:2,d:0] Wanted CallStack IP:
[WD] $dIP_a2u8 {0}:: ?callStack::GHC.Stack.Types.CallStack
Step 6[l:2,d:0] Kept as inert:
[WD] $dIP_a2uA {0}:: ?callStack::GHC.Stack.Types.CallStack
Step 7[l:2,d:0] Kept as inert:
[G] $dMyClass2_a2uh {0}:: MyClass2 a_a2ug[ssk:2]
Step 8[l:2,d:0] Kept as inert:
[G] $dMyClass1_a2ul {0}:: MyClass1 a_a2uk[ssk:2]
Constraint solver steps = 8
从这个转储中提取有用信息并不容易,但AFAIK现在是唯一可用的选项.一些相关的门票:13443,15044
补充:我将尝试解释转储的含义.我实际上并不熟悉GHC内部,所以这只是我(可能是错误的)理解.
相关位是下一个:
Step 1[l:1,d:0] Top react: Dict/Top (solved wanted):
[WD] $dMyClass3_a2uc {0}:: MyClass3 (MyType2 (MyType1 Int))
Step 2[l:1,d:1] Top react: Dict/Top (solved wanted):
[WD] $dMyClass2_a2up {1}:: MyClass2 (MyType1 Int)
Step 3[l:1,d:2] Kept as inert:
[WD] $dMyClass1_a2uq {2}:: MyClass1 Int
这里d代表"深度",WD是"想要派生"的约束.所以我们有一些像所需约束的堆栈跟踪:最初我们想要MyClass3 (MyType2 (MyType1 Int)),然后我们找到了MyClass3实例MyType2,现在我们想要MyClass2 (MyType1 Int)满足它.然后我们找到了MyClass2实例MyType1,现在我们想要MyClass1 Int满足它.我知道,解释是模糊的,但这就是我对你的全部,抱歉.