Re: [Haskell-cafe] Question about Newtype op() function arguments.
Ah, so is the idea, then, to use *op()* when `n` wasn't actually constructed formally, but rather assembled by the user, so as to match the type of the accessor function normally supplied as the argument to the constructor? On 6/7/2013 4:51 PM, Tom Ellis wrote: On Fri, Jun 07, 2013 at 04:05:09PM -0400, Joe Q wrote: The phantom parameter solves the same problem as scoped type variables. Granted, if you find yourself in that kind of polymorphic soup you have deeper problems... I don't understand this. Scoped type variables are used when you want to use a type variable from the top level within the body of a function. If you use op and specify a particular constructor then you don't have a variable but a concrete instance of a type. But maybe I'm missing some more powerful way this can be used ... Tom You can use scoped type variables to correct an ambiguous type error. You can think of op as a variation on asTypeOf, as documented here on http://www.haskell.org/haskellwiki/Scoped_type_variables#Avoiding_Scoped_Type_Variables . If I tried to come up with an example that's specific to op, it would only be horribly contrived. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Question about Newtype op() function arguments.
Hi All, Referring to the following, which is taken from the *Control.Newtype *documentation page: op :: Newtypehttp://hackage.haskell.org/packages/archive/newtype/0.2/doc/html/Control-Newtype.html#t:Newtype n o = (o - n) - n - oSourcehttp://hackage.haskell.org/packages/archive/newtype/0.2/doc/html/src/Control-Newtype.html#op This function serves two purposes: 1. Giving you the unpack of a newtype without you needing to remember the name. 2. Showing that the first parameter is *completely ignored* on the value level, meaning the only reason you pass in the constructor is to provide type information. Typeclasses sure are neat. As point #2, above, emphasizes, the only purpose for the first argument to the function (i.e. - the constructor (o - n)) is to specify the type of 'n'. However, being a *newtype*, 'n' can have only one constructor. So, why is it necessary to pass in the constructor to this function, when we're already passing in 'n'? Thanks, -db ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Question about Newtype op() function arguments.
* David Banas capn.fre...@gmail.com [2013-06-07 07:08:19-0700] Hi All, Referring to the following, which is taken from the *Control.Newtype *documentation page: op :: Newtypehttp://hackage.haskell.org/packages/archive/newtype/0.2/doc/html/Control-Newtype.html#t:Newtype n o = (o - n) - n - oSourcehttp://hackage.haskell.org/packages/archive/newtype/0.2/doc/html/src/Control-Newtype.html#op This function serves two purposes: 1. Giving you the unpack of a newtype without you needing to remember the name. 2. Showing that the first parameter is *completely ignored* on the value level, meaning the only reason you pass in the constructor is to provide type information. Typeclasses sure are neat. As point #2, above, emphasizes, the only purpose for the first argument to the function (i.e. - the constructor (o - n)) is to specify the type of 'n'. However, being a *newtype*, 'n' can have only one constructor. So, why is it necessary to pass in the constructor to this function, when we're already passing in 'n'? Hi David, Note that that argument is in fact not necessary — 'op' without its first argument is exactly 'unpack'. I imagine that 'op' may be useful if you need to disambiguate the instance in an otherwise ambiguous context. One way to do this is, of course, to provide an explicit type signature, but passing a constructor is another, and rather elegant, way to achieve that. I am not a heavy user of the newtype package, so I can't say how often this situation occurs in practice. Roman ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Question about Newtype op() function arguments.
On Fri, Jun 07, 2013 at 07:08:19AM -0700, David Banas wrote: op :: Newtypehttp://hackage.haskell.org/packages/archive/newtype/0.2/doc/html/Control-Newtype.html#t:Newtype n o = (o - n) - n - oSourcehttp://hackage.haskell.org/packages/archive/newtype/0.2/doc/html/src/Control-Newtype.html#op This function serves two purposes: 1. Giving you the unpack of a newtype without you needing to remember the name. 2. Showing that the first parameter is *completely ignored* on the value level, meaning the only reason you pass in the constructor is to provide type information. Typeclasses sure are neat. As point #2, above, emphasizes, the only purpose for the first argument to the function (i.e. - the constructor (o - n)) is to specify the type of 'n'. However, being a *newtype*, 'n' can have only one constructor. So, why is it necessary to pass in the constructor to this function, when we're already passing in 'n'? I am puzzled by this too. What does op stand for? I hypothesis opposite in the sense of inverse, since as Roman points out op Constructor :: n - o is the inverse of Constructor :: o - n. But I admit I do not see how this provides value over unpack itself. Tom ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Question about Newtype op() function arguments.
The phantom parameter solves the same problem as scoped type variables. Granted, if you find yourself in that kind of polymorphic soup you have deeper problems... On Jun 7, 2013 2:53 PM, Tom Ellis tom-lists-haskell-cafe-2...@jaguarpaw.co.uk wrote: On Fri, Jun 07, 2013 at 07:08:19AM -0700, David Banas wrote: op :: Newtype http://hackage.haskell.org/packages/archive/newtype/0.2/doc/html/Control-Newtype.html#t:Newtype n o = (o - n) - n - oSource http://hackage.haskell.org/packages/archive/newtype/0.2/doc/html/src/Control-Newtype.html#op This function serves two purposes: 1. Giving you the unpack of a newtype without you needing to remember the name. 2. Showing that the first parameter is *completely ignored* on the value level, meaning the only reason you pass in the constructor is to provide type information. Typeclasses sure are neat. As point #2, above, emphasizes, the only purpose for the first argument to the function (i.e. - the constructor (o - n)) is to specify the type of 'n'. However, being a *newtype*, 'n' can have only one constructor. So, why is it necessary to pass in the constructor to this function, when we're already passing in 'n'? I am puzzled by this too. What does op stand for? I hypothesis opposite in the sense of inverse, since as Roman points out op Constructor :: n - o is the inverse of Constructor :: o - n. But I admit I do not see how this provides value over unpack itself. Tom ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Question about Newtype op() function arguments.
On Fri, Jun 07, 2013 at 04:05:09PM -0400, Joe Q wrote: The phantom parameter solves the same problem as scoped type variables. Granted, if you find yourself in that kind of polymorphic soup you have deeper problems... I don't understand this. Scoped type variables are used when you want to use a type variable from the top level within the body of a function. If you use op and specify a particular constructor then you don't have a variable but a concrete instance of a type. But maybe I'm missing some more powerful way this can be used ... Tom On Jun 7, 2013 2:53 PM, Tom Ellis tom-lists-haskell-cafe-2...@jaguarpaw.co.uk wrote: On Fri, Jun 07, 2013 at 07:08:19AM -0700, David Banas wrote: op :: Newtype http://hackage.haskell.org/packages/archive/newtype/0.2/doc/html/Control-Newtype.html#t:Newtype n o = (o - n) - n - oSource http://hackage.haskell.org/packages/archive/newtype/0.2/doc/html/src/Control-Newtype.html#op This function serves two purposes: 1. Giving you the unpack of a newtype without you needing to remember the name. 2. Showing that the first parameter is *completely ignored* on the value level, meaning the only reason you pass in the constructor is to provide type information. Typeclasses sure are neat. As point #2, above, emphasizes, the only purpose for the first argument to the function (i.e. - the constructor (o - n)) is to specify the type of 'n'. However, being a *newtype*, 'n' can have only one constructor. So, why is it necessary to pass in the constructor to this function, when we're already passing in 'n'? I am puzzled by this too. What does op stand for? I hypothesis opposite in the sense of inverse, since as Roman points out op Constructor :: n - o is the inverse of Constructor :: o - n. But I admit I do not see how this provides value over unpack itself. Tom ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Question about Newtype op() function arguments.
On 6/7/2013 4:51 PM, Tom Ellis wrote: On Fri, Jun 07, 2013 at 04:05:09PM -0400, Joe Q wrote: The phantom parameter solves the same problem as scoped type variables. Granted, if you find yourself in that kind of polymorphic soup you have deeper problems... I don't understand this. Scoped type variables are used when you want to use a type variable from the top level within the body of a function. If you use op and specify a particular constructor then you don't have a variable but a concrete instance of a type. But maybe I'm missing some more powerful way this can be used ... Tom You can use scoped type variables to correct an ambiguous type error. You can think of op as a variation on asTypeOf, as documented here on http://www.haskell.org/haskellwiki/Scoped_type_variables#Avoiding_Scoped_Type_Variables. If I tried to come up with an example that's specific to op, it would only be horribly contrived. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
