Why does GHC infer a monomorphic type here, even with MonomorphismRestriction disabled? The Next CEO of Stack OverflowResolving the type of `f = f (<*>) pure`NoMonomorphismRestriction helps preserve sharing?How can eta-reduction of a well typed function result in a type error?Can I write such polymorphic function? What language extensions do I need?GHC rewrite rule specialising a function for a type classType Inference in PatternsHow to type check recursive definitions using Algorithm W?What is the monomorphism restriction?Why are higher rank types so fragile in HaskellWhy can't GHC typecheck this function involving polymorphism and existential types?Problems With Type Inference on (^)

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Why does GHC infer a monomorphic type here, even with MonomorphismRestriction disabled?



The Next CEO of Stack OverflowResolving the type of `f = f (<*>) pure`NoMonomorphismRestriction helps preserve sharing?How can eta-reduction of a well typed function result in a type error?Can I write such polymorphic function? What language extensions do I need?GHC rewrite rule specialising a function for a type classType Inference in PatternsHow to type check recursive definitions using Algorithm W?What is the monomorphism restriction?Why are higher rank types so fragile in HaskellWhy can't GHC typecheck this function involving polymorphism and existential types?Problems With Type Inference on (^)










13















This was prompted by Resolving the type of `f = f (<*>) pure`, which discusses a more complicated example, but this one works too.



The following definition compiles without problem:



w :: Integral a => a
w = fromInteger w


...Of course it doesn't work runtime-wise, but that's beside the question. The point is that the definition of w itself uses a specialised version of w :: Integer. Clearly that is a suitable instantiation, and therefore typechecks.



However, if we remove the signature, then GHC infers not the above type, but only the concrete one:



w' = fromInteger w'


GHCi> :t w
w :: Integral a => a
GHCi> :t w'
w' :: Integer


Well, when I saw this, I was fairly sure this was the monomorphism restriction at work. It's well known that also e.g.



i = 3


GHCi> :t i
i :: Integer


although i :: Num p => p would be perfectly possible. And indeed, i :: Num p => p is inferred if -XNoMonomorphismRestriction is active, i.e. if the monomorphism restriction is disabled.



However, in case of w' only the type Integer is inferred even when the monomorphism restriction is disabled.



To count out that this has something to do with defaulting:



fromFloat :: RealFrac a => Float -> a
q :: RealFrac a => a
q = fromFloat q
q' = fromFloat q'


GHCi> :t q
q :: RealFrac a => a
GHCi> :t q'
q' :: Float


Why is the polymorphic type not inferred?










share|improve this question
























  • Doesn't the monomorphism restriction apply only to simple bindings anyways (and w' = fromInteger w', being recursive, is not simple)?

    – Alec
    4 hours ago











  • @Alec possible, but still – why does something like the monomorphism restriction seem to kick in here?

    – leftaroundabout
    4 hours ago











  • I'm probably being dense and missing something here, but fromInteger has type (Num a) => Integer -> a, and since w' is used as the input to fromInteger, doesn't that mean Integer is the only possible type for it? Indeed I'm rather surprised that the version with the polymorphic type signature compiles. (So as I said, probably missing something.)

    – Robin Zigmond
    4 hours ago











  • @RobinZigmond Integer is certainly the only possible monomorphic type for w', but as w demonstrates a polymorphic type is perfectly fine as well. After all, a polymorphic type can be instantiated to a monomorphic one, provided it fulfills the constraints.

    – leftaroundabout
    4 hours ago
















13















This was prompted by Resolving the type of `f = f (<*>) pure`, which discusses a more complicated example, but this one works too.



The following definition compiles without problem:



w :: Integral a => a
w = fromInteger w


...Of course it doesn't work runtime-wise, but that's beside the question. The point is that the definition of w itself uses a specialised version of w :: Integer. Clearly that is a suitable instantiation, and therefore typechecks.



However, if we remove the signature, then GHC infers not the above type, but only the concrete one:



w' = fromInteger w'


GHCi> :t w
w :: Integral a => a
GHCi> :t w'
w' :: Integer


Well, when I saw this, I was fairly sure this was the monomorphism restriction at work. It's well known that also e.g.



i = 3


GHCi> :t i
i :: Integer


although i :: Num p => p would be perfectly possible. And indeed, i :: Num p => p is inferred if -XNoMonomorphismRestriction is active, i.e. if the monomorphism restriction is disabled.



However, in case of w' only the type Integer is inferred even when the monomorphism restriction is disabled.



To count out that this has something to do with defaulting:



fromFloat :: RealFrac a => Float -> a
q :: RealFrac a => a
q = fromFloat q
q' = fromFloat q'


GHCi> :t q
q :: RealFrac a => a
GHCi> :t q'
q' :: Float


Why is the polymorphic type not inferred?










share|improve this question
























  • Doesn't the monomorphism restriction apply only to simple bindings anyways (and w' = fromInteger w', being recursive, is not simple)?

    – Alec
    4 hours ago











  • @Alec possible, but still – why does something like the monomorphism restriction seem to kick in here?

    – leftaroundabout
    4 hours ago











  • I'm probably being dense and missing something here, but fromInteger has type (Num a) => Integer -> a, and since w' is used as the input to fromInteger, doesn't that mean Integer is the only possible type for it? Indeed I'm rather surprised that the version with the polymorphic type signature compiles. (So as I said, probably missing something.)

    – Robin Zigmond
    4 hours ago











  • @RobinZigmond Integer is certainly the only possible monomorphic type for w', but as w demonstrates a polymorphic type is perfectly fine as well. After all, a polymorphic type can be instantiated to a monomorphic one, provided it fulfills the constraints.

    – leftaroundabout
    4 hours ago














13












13








13


2






This was prompted by Resolving the type of `f = f (<*>) pure`, which discusses a more complicated example, but this one works too.



The following definition compiles without problem:



w :: Integral a => a
w = fromInteger w


...Of course it doesn't work runtime-wise, but that's beside the question. The point is that the definition of w itself uses a specialised version of w :: Integer. Clearly that is a suitable instantiation, and therefore typechecks.



However, if we remove the signature, then GHC infers not the above type, but only the concrete one:



w' = fromInteger w'


GHCi> :t w
w :: Integral a => a
GHCi> :t w'
w' :: Integer


Well, when I saw this, I was fairly sure this was the monomorphism restriction at work. It's well known that also e.g.



i = 3


GHCi> :t i
i :: Integer


although i :: Num p => p would be perfectly possible. And indeed, i :: Num p => p is inferred if -XNoMonomorphismRestriction is active, i.e. if the monomorphism restriction is disabled.



However, in case of w' only the type Integer is inferred even when the monomorphism restriction is disabled.



To count out that this has something to do with defaulting:



fromFloat :: RealFrac a => Float -> a
q :: RealFrac a => a
q = fromFloat q
q' = fromFloat q'


GHCi> :t q
q :: RealFrac a => a
GHCi> :t q'
q' :: Float


Why is the polymorphic type not inferred?










share|improve this question
















This was prompted by Resolving the type of `f = f (<*>) pure`, which discusses a more complicated example, but this one works too.



The following definition compiles without problem:



w :: Integral a => a
w = fromInteger w


...Of course it doesn't work runtime-wise, but that's beside the question. The point is that the definition of w itself uses a specialised version of w :: Integer. Clearly that is a suitable instantiation, and therefore typechecks.



However, if we remove the signature, then GHC infers not the above type, but only the concrete one:



w' = fromInteger w'


GHCi> :t w
w :: Integral a => a
GHCi> :t w'
w' :: Integer


Well, when I saw this, I was fairly sure this was the monomorphism restriction at work. It's well known that also e.g.



i = 3


GHCi> :t i
i :: Integer


although i :: Num p => p would be perfectly possible. And indeed, i :: Num p => p is inferred if -XNoMonomorphismRestriction is active, i.e. if the monomorphism restriction is disabled.



However, in case of w' only the type Integer is inferred even when the monomorphism restriction is disabled.



To count out that this has something to do with defaulting:



fromFloat :: RealFrac a => Float -> a
q :: RealFrac a => a
q = fromFloat q
q' = fromFloat q'


GHCi> :t q
q :: RealFrac a => a
GHCi> :t q'
q' :: Float


Why is the polymorphic type not inferred?







haskell recursion type-inference parametric-polymorphism






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago







leftaroundabout

















asked 4 hours ago









leftaroundaboutleftaroundabout

80.1k3119237




80.1k3119237












  • Doesn't the monomorphism restriction apply only to simple bindings anyways (and w' = fromInteger w', being recursive, is not simple)?

    – Alec
    4 hours ago











  • @Alec possible, but still – why does something like the monomorphism restriction seem to kick in here?

    – leftaroundabout
    4 hours ago











  • I'm probably being dense and missing something here, but fromInteger has type (Num a) => Integer -> a, and since w' is used as the input to fromInteger, doesn't that mean Integer is the only possible type for it? Indeed I'm rather surprised that the version with the polymorphic type signature compiles. (So as I said, probably missing something.)

    – Robin Zigmond
    4 hours ago











  • @RobinZigmond Integer is certainly the only possible monomorphic type for w', but as w demonstrates a polymorphic type is perfectly fine as well. After all, a polymorphic type can be instantiated to a monomorphic one, provided it fulfills the constraints.

    – leftaroundabout
    4 hours ago


















  • Doesn't the monomorphism restriction apply only to simple bindings anyways (and w' = fromInteger w', being recursive, is not simple)?

    – Alec
    4 hours ago











  • @Alec possible, but still – why does something like the monomorphism restriction seem to kick in here?

    – leftaroundabout
    4 hours ago











  • I'm probably being dense and missing something here, but fromInteger has type (Num a) => Integer -> a, and since w' is used as the input to fromInteger, doesn't that mean Integer is the only possible type for it? Indeed I'm rather surprised that the version with the polymorphic type signature compiles. (So as I said, probably missing something.)

    – Robin Zigmond
    4 hours ago











  • @RobinZigmond Integer is certainly the only possible monomorphic type for w', but as w demonstrates a polymorphic type is perfectly fine as well. After all, a polymorphic type can be instantiated to a monomorphic one, provided it fulfills the constraints.

    – leftaroundabout
    4 hours ago

















Doesn't the monomorphism restriction apply only to simple bindings anyways (and w' = fromInteger w', being recursive, is not simple)?

– Alec
4 hours ago





Doesn't the monomorphism restriction apply only to simple bindings anyways (and w' = fromInteger w', being recursive, is not simple)?

– Alec
4 hours ago













@Alec possible, but still – why does something like the monomorphism restriction seem to kick in here?

– leftaroundabout
4 hours ago





@Alec possible, but still – why does something like the monomorphism restriction seem to kick in here?

– leftaroundabout
4 hours ago













I'm probably being dense and missing something here, but fromInteger has type (Num a) => Integer -> a, and since w' is used as the input to fromInteger, doesn't that mean Integer is the only possible type for it? Indeed I'm rather surprised that the version with the polymorphic type signature compiles. (So as I said, probably missing something.)

– Robin Zigmond
4 hours ago





I'm probably being dense and missing something here, but fromInteger has type (Num a) => Integer -> a, and since w' is used as the input to fromInteger, doesn't that mean Integer is the only possible type for it? Indeed I'm rather surprised that the version with the polymorphic type signature compiles. (So as I said, probably missing something.)

– Robin Zigmond
4 hours ago













@RobinZigmond Integer is certainly the only possible monomorphic type for w', but as w demonstrates a polymorphic type is perfectly fine as well. After all, a polymorphic type can be instantiated to a monomorphic one, provided it fulfills the constraints.

– leftaroundabout
4 hours ago






@RobinZigmond Integer is certainly the only possible monomorphic type for w', but as w demonstrates a polymorphic type is perfectly fine as well. After all, a polymorphic type can be instantiated to a monomorphic one, provided it fulfills the constraints.

– leftaroundabout
4 hours ago













1 Answer
1






active

oldest

votes


















13














Polymorphic recursion (where a function calls itself at a different type than the one at which it was called) always requires a type signature. The full explanation is in Section 4.4.1 of the Haskell 2010 Report:




If a variable f is defined without providing a corresponding type signature declaration, then each use of f outside its own declaration group (see Section 4.5) is treated as having the corresponding inferred, or principal type. However, to ensure that type inference is still possible, the defining occurrence, and all uses of f within its declaration group must have the same monomorphic type (from which the principal type is obtained by generalization, as described in Section 4.5.2).




The same section later presents an example of polymorphic recursion supported by a type signature.



My understanding is that unaided type inference is generally undecidable in the presence of polymorphic recursion, so Haskell doesn't even try.



In this case, the type checker starts with



w :: a


where a is a meta-variable. Since fromInteger is called with w as an argument within its own declaration (and therefore within its declaration group), the type checker unifies a with Integer. There are no variables left to generalize.






share|improve this answer




















  • 1





    My bachelor was about a simple type inferencer for a subset of Haskell including typeclasses & polymorphic recursion. A very simple approach is to limit the depth of the polymorphic recursion up to k depth. Most useful cases of polymorphic recursion can be inferred with a very low depth bound (like k=1 or k=2). Anyway Haskell type inference is already undecidable so that's not the only reason why it's not allowed. An other reason is probably performance, it surely makes type inference O(k·f(n)) instead of O(f(n)) since you may need to do all over again for k times.

    – Bakuriu
    2 hours ago






  • 1





    @Bakuriu, I am pretty sure that Haskell 2010 without polymorphic recursion has full type inference--it's basically Hindley-Milner at that point, plus type classes and defaulting. Do you have a reference saying otherwise? As for some limited recursion depth: that sounds like a potentially useful extension, but it has a very different flavor from what the Haskell Report tends to do. I would find such a feature most useful for discovering the right type signatures for polymorphic recursive code.

    – dfeuer
    2 hours ago












  • Yes, but I think all extensions to the type systems on top of Haskell2010 make type inference undecidable. Note that for example Type families are "artificially" limited to avoid undecidable instances by forbidding certain well-formed programs by default, so allowing a "k-recursive" polymorphic recursion would not be very different from that case, IMHO.

    – Bakuriu
    1 hour ago











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









13














Polymorphic recursion (where a function calls itself at a different type than the one at which it was called) always requires a type signature. The full explanation is in Section 4.4.1 of the Haskell 2010 Report:




If a variable f is defined without providing a corresponding type signature declaration, then each use of f outside its own declaration group (see Section 4.5) is treated as having the corresponding inferred, or principal type. However, to ensure that type inference is still possible, the defining occurrence, and all uses of f within its declaration group must have the same monomorphic type (from which the principal type is obtained by generalization, as described in Section 4.5.2).




The same section later presents an example of polymorphic recursion supported by a type signature.



My understanding is that unaided type inference is generally undecidable in the presence of polymorphic recursion, so Haskell doesn't even try.



In this case, the type checker starts with



w :: a


where a is a meta-variable. Since fromInteger is called with w as an argument within its own declaration (and therefore within its declaration group), the type checker unifies a with Integer. There are no variables left to generalize.






share|improve this answer




















  • 1





    My bachelor was about a simple type inferencer for a subset of Haskell including typeclasses & polymorphic recursion. A very simple approach is to limit the depth of the polymorphic recursion up to k depth. Most useful cases of polymorphic recursion can be inferred with a very low depth bound (like k=1 or k=2). Anyway Haskell type inference is already undecidable so that's not the only reason why it's not allowed. An other reason is probably performance, it surely makes type inference O(k·f(n)) instead of O(f(n)) since you may need to do all over again for k times.

    – Bakuriu
    2 hours ago






  • 1





    @Bakuriu, I am pretty sure that Haskell 2010 without polymorphic recursion has full type inference--it's basically Hindley-Milner at that point, plus type classes and defaulting. Do you have a reference saying otherwise? As for some limited recursion depth: that sounds like a potentially useful extension, but it has a very different flavor from what the Haskell Report tends to do. I would find such a feature most useful for discovering the right type signatures for polymorphic recursive code.

    – dfeuer
    2 hours ago












  • Yes, but I think all extensions to the type systems on top of Haskell2010 make type inference undecidable. Note that for example Type families are "artificially" limited to avoid undecidable instances by forbidding certain well-formed programs by default, so allowing a "k-recursive" polymorphic recursion would not be very different from that case, IMHO.

    – Bakuriu
    1 hour ago















13














Polymorphic recursion (where a function calls itself at a different type than the one at which it was called) always requires a type signature. The full explanation is in Section 4.4.1 of the Haskell 2010 Report:




If a variable f is defined without providing a corresponding type signature declaration, then each use of f outside its own declaration group (see Section 4.5) is treated as having the corresponding inferred, or principal type. However, to ensure that type inference is still possible, the defining occurrence, and all uses of f within its declaration group must have the same monomorphic type (from which the principal type is obtained by generalization, as described in Section 4.5.2).




The same section later presents an example of polymorphic recursion supported by a type signature.



My understanding is that unaided type inference is generally undecidable in the presence of polymorphic recursion, so Haskell doesn't even try.



In this case, the type checker starts with



w :: a


where a is a meta-variable. Since fromInteger is called with w as an argument within its own declaration (and therefore within its declaration group), the type checker unifies a with Integer. There are no variables left to generalize.






share|improve this answer




















  • 1





    My bachelor was about a simple type inferencer for a subset of Haskell including typeclasses & polymorphic recursion. A very simple approach is to limit the depth of the polymorphic recursion up to k depth. Most useful cases of polymorphic recursion can be inferred with a very low depth bound (like k=1 or k=2). Anyway Haskell type inference is already undecidable so that's not the only reason why it's not allowed. An other reason is probably performance, it surely makes type inference O(k·f(n)) instead of O(f(n)) since you may need to do all over again for k times.

    – Bakuriu
    2 hours ago






  • 1





    @Bakuriu, I am pretty sure that Haskell 2010 without polymorphic recursion has full type inference--it's basically Hindley-Milner at that point, plus type classes and defaulting. Do you have a reference saying otherwise? As for some limited recursion depth: that sounds like a potentially useful extension, but it has a very different flavor from what the Haskell Report tends to do. I would find such a feature most useful for discovering the right type signatures for polymorphic recursive code.

    – dfeuer
    2 hours ago












  • Yes, but I think all extensions to the type systems on top of Haskell2010 make type inference undecidable. Note that for example Type families are "artificially" limited to avoid undecidable instances by forbidding certain well-formed programs by default, so allowing a "k-recursive" polymorphic recursion would not be very different from that case, IMHO.

    – Bakuriu
    1 hour ago













13












13








13







Polymorphic recursion (where a function calls itself at a different type than the one at which it was called) always requires a type signature. The full explanation is in Section 4.4.1 of the Haskell 2010 Report:




If a variable f is defined without providing a corresponding type signature declaration, then each use of f outside its own declaration group (see Section 4.5) is treated as having the corresponding inferred, or principal type. However, to ensure that type inference is still possible, the defining occurrence, and all uses of f within its declaration group must have the same monomorphic type (from which the principal type is obtained by generalization, as described in Section 4.5.2).




The same section later presents an example of polymorphic recursion supported by a type signature.



My understanding is that unaided type inference is generally undecidable in the presence of polymorphic recursion, so Haskell doesn't even try.



In this case, the type checker starts with



w :: a


where a is a meta-variable. Since fromInteger is called with w as an argument within its own declaration (and therefore within its declaration group), the type checker unifies a with Integer. There are no variables left to generalize.






share|improve this answer















Polymorphic recursion (where a function calls itself at a different type than the one at which it was called) always requires a type signature. The full explanation is in Section 4.4.1 of the Haskell 2010 Report:




If a variable f is defined without providing a corresponding type signature declaration, then each use of f outside its own declaration group (see Section 4.5) is treated as having the corresponding inferred, or principal type. However, to ensure that type inference is still possible, the defining occurrence, and all uses of f within its declaration group must have the same monomorphic type (from which the principal type is obtained by generalization, as described in Section 4.5.2).




The same section later presents an example of polymorphic recursion supported by a type signature.



My understanding is that unaided type inference is generally undecidable in the presence of polymorphic recursion, so Haskell doesn't even try.



In this case, the type checker starts with



w :: a


where a is a meta-variable. Since fromInteger is called with w as an argument within its own declaration (and therefore within its declaration group), the type checker unifies a with Integer. There are no variables left to generalize.







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 hours ago

























answered 4 hours ago









dfeuerdfeuer

33.5k349133




33.5k349133







  • 1





    My bachelor was about a simple type inferencer for a subset of Haskell including typeclasses & polymorphic recursion. A very simple approach is to limit the depth of the polymorphic recursion up to k depth. Most useful cases of polymorphic recursion can be inferred with a very low depth bound (like k=1 or k=2). Anyway Haskell type inference is already undecidable so that's not the only reason why it's not allowed. An other reason is probably performance, it surely makes type inference O(k·f(n)) instead of O(f(n)) since you may need to do all over again for k times.

    – Bakuriu
    2 hours ago






  • 1





    @Bakuriu, I am pretty sure that Haskell 2010 without polymorphic recursion has full type inference--it's basically Hindley-Milner at that point, plus type classes and defaulting. Do you have a reference saying otherwise? As for some limited recursion depth: that sounds like a potentially useful extension, but it has a very different flavor from what the Haskell Report tends to do. I would find such a feature most useful for discovering the right type signatures for polymorphic recursive code.

    – dfeuer
    2 hours ago












  • Yes, but I think all extensions to the type systems on top of Haskell2010 make type inference undecidable. Note that for example Type families are "artificially" limited to avoid undecidable instances by forbidding certain well-formed programs by default, so allowing a "k-recursive" polymorphic recursion would not be very different from that case, IMHO.

    – Bakuriu
    1 hour ago












  • 1





    My bachelor was about a simple type inferencer for a subset of Haskell including typeclasses & polymorphic recursion. A very simple approach is to limit the depth of the polymorphic recursion up to k depth. Most useful cases of polymorphic recursion can be inferred with a very low depth bound (like k=1 or k=2). Anyway Haskell type inference is already undecidable so that's not the only reason why it's not allowed. An other reason is probably performance, it surely makes type inference O(k·f(n)) instead of O(f(n)) since you may need to do all over again for k times.

    – Bakuriu
    2 hours ago






  • 1





    @Bakuriu, I am pretty sure that Haskell 2010 without polymorphic recursion has full type inference--it's basically Hindley-Milner at that point, plus type classes and defaulting. Do you have a reference saying otherwise? As for some limited recursion depth: that sounds like a potentially useful extension, but it has a very different flavor from what the Haskell Report tends to do. I would find such a feature most useful for discovering the right type signatures for polymorphic recursive code.

    – dfeuer
    2 hours ago












  • Yes, but I think all extensions to the type systems on top of Haskell2010 make type inference undecidable. Note that for example Type families are "artificially" limited to avoid undecidable instances by forbidding certain well-formed programs by default, so allowing a "k-recursive" polymorphic recursion would not be very different from that case, IMHO.

    – Bakuriu
    1 hour ago







1




1





My bachelor was about a simple type inferencer for a subset of Haskell including typeclasses & polymorphic recursion. A very simple approach is to limit the depth of the polymorphic recursion up to k depth. Most useful cases of polymorphic recursion can be inferred with a very low depth bound (like k=1 or k=2). Anyway Haskell type inference is already undecidable so that's not the only reason why it's not allowed. An other reason is probably performance, it surely makes type inference O(k·f(n)) instead of O(f(n)) since you may need to do all over again for k times.

– Bakuriu
2 hours ago





My bachelor was about a simple type inferencer for a subset of Haskell including typeclasses & polymorphic recursion. A very simple approach is to limit the depth of the polymorphic recursion up to k depth. Most useful cases of polymorphic recursion can be inferred with a very low depth bound (like k=1 or k=2). Anyway Haskell type inference is already undecidable so that's not the only reason why it's not allowed. An other reason is probably performance, it surely makes type inference O(k·f(n)) instead of O(f(n)) since you may need to do all over again for k times.

– Bakuriu
2 hours ago




1




1





@Bakuriu, I am pretty sure that Haskell 2010 without polymorphic recursion has full type inference--it's basically Hindley-Milner at that point, plus type classes and defaulting. Do you have a reference saying otherwise? As for some limited recursion depth: that sounds like a potentially useful extension, but it has a very different flavor from what the Haskell Report tends to do. I would find such a feature most useful for discovering the right type signatures for polymorphic recursive code.

– dfeuer
2 hours ago






@Bakuriu, I am pretty sure that Haskell 2010 without polymorphic recursion has full type inference--it's basically Hindley-Milner at that point, plus type classes and defaulting. Do you have a reference saying otherwise? As for some limited recursion depth: that sounds like a potentially useful extension, but it has a very different flavor from what the Haskell Report tends to do. I would find such a feature most useful for discovering the right type signatures for polymorphic recursive code.

– dfeuer
2 hours ago














Yes, but I think all extensions to the type systems on top of Haskell2010 make type inference undecidable. Note that for example Type families are "artificially" limited to avoid undecidable instances by forbidding certain well-formed programs by default, so allowing a "k-recursive" polymorphic recursion would not be very different from that case, IMHO.

– Bakuriu
1 hour ago





Yes, but I think all extensions to the type systems on top of Haskell2010 make type inference undecidable. Note that for example Type families are "artificially" limited to avoid undecidable instances by forbidding certain well-formed programs by default, so allowing a "k-recursive" polymorphic recursion would not be very different from that case, IMHO.

– Bakuriu
1 hour ago



















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