Re: [Haskell-cafe] Re: Adding a field to a data record
With the RecordWildCard extension you should be able to write
newFoo Old.Foo{..} = New.Foo { .., z=1 }
On Tue, Jul 28, 2009 at 3:47 PM, Henry Laxennadine.and.he...@pobox.com wrote:
Malcolm Wallace Malcolm.Wallace at cs.york.ac.uk writes:
and perhaps use emacs to
query-replace all the Foo1's back to Foo's
At least this bit can be avoided easily enough, by using
module qualification during the conversion process.
module Original (Foo(..)) where
data Foo = Foo { ... y :: Int } deriving ...
module New (Foo(..)) where
data Foo = Foo { ... y, z :: Int } deriving ...
module Convert where
import Original as Old
import New as New
newFoo :: Old.Foo - New.Foo
newFoo old{..} = New.Foo { a=a, b=b, ... z=1 }
Finally rename module New.
Regards,
Malcolm
Thanks Malcolm, yes, that keeps me out of emacs, but the part I would really
like to avoid is writing the New.Foo { a=a, b=b, ... z=1 } part, where the
field
names are many, long, and varied. Yes, I could cut and paste, but I'm hoping
for a better way. Thanks.
Best wishes,
Henry Laxen
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Re: [Haskell-cafe] Re: Adding a field to a data record
On Tue, Jul 28, 2009 at 7:47 AM, Henry Laxen nadine.and.he...@pobox.comwrote:
Malcolm Wallace Malcolm.Wallace at cs.york.ac.uk writes:
and perhaps use emacs to
query-replace all the Foo1's back to Foo's
At least this bit can be avoided easily enough, by using
module qualification during the conversion process.
module Original (Foo(..)) where
data Foo = Foo { ... y :: Int } deriving ...
module New (Foo(..)) where
data Foo = Foo { ... y, z :: Int } deriving ...
module Convert where
import Original as Old
import New as New
newFoo :: Old.Foo - New.Foo
newFoo old{..} = New.Foo { a=a, b=b, ... z=1 }
Finally rename module New.
Regards,
Malcolm
Thanks Malcolm, yes, that keeps me out of emacs, but the part I would
really
like to avoid is writing the New.Foo { a=a, b=b, ... z=1 } part, where the
field
names are many, long, and varied. Yes, I could cut and paste, but I'm
hoping
for a better way. Thanks.
I guess you could define:
type UpgradeFoo = (Foo, Int)
And then write the conversion code as a zip. upgradeFoo foos = zip foos
[1..]
instance Show UpgradeFoo where ...
And then use the module trick to switch the code around?
Jason
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Re: [Haskell-cafe] Re: Adding a field to a data record
the part I would really like to avoid is writing the
New.Foo { a=a, b=b, ... z=1 } part, where the field
names are many, long, and varied.
OK, here is another hack-ish trick, since I notice your data is stored
on disk as text, using show. I assume you are using something like
Read to retrieve it. Well, how about using a real parser instead?
The parser during conversion can be slightly more lax, automatically
adding in the extra field.
For instance, using polyparse's Text.Parse, and DrIFT to derive the
appropriate Parse instance for your datatype:
module Foo where
data Foo = Foo { a :: Int
, b :: Bool
, c :: Maybe Foo }
{-! derive : Parse !-}
DrIFT gives you this instance:
{-* Generated by DrIFT : Look, but Don't Touch. *-}
instance Parse Foo where
parse = constructors
[ ( Foo
, return Foo `discard` isWord { `apply` field a
`discard` isWord , `apply` field b
`discard` isWord , `apply` field c
`discard` isWord }
)
]
Let's say the field 'b' is new, and your existing data does not have
it. So just take the parser generated by DrIFT and make a small
modification:
{-* Generated by DrIFT but modified by hand for conversion
purposes *-}
instance Parse Foo where
parse = constructors
[ ( Foo
, return Foo `discard` isWord { `apply` field a
`apply` return True -- this field does not yet exist in
data
`discard` isWord , `apply` field c
`discard` isWord }
)
]
Then do the obvious thing: parse the old data, immediately write it
out again, and then throw away the modified parser in favour of the
pure generated one.
Regards,
Malcolm
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Re: [Haskell-cafe] Re: Adding a field to a data record
Hello,
you may also find the package pretty-show
(http://hackage.haskell.org/package/pretty-show) useful. It contains
code to convert automatically derived instances of Show into an
explicit data structure, which you can then manipulate (e.g., by
adding the extra field), and then render back to text.
-Iavor
On Tue, Jul 28, 2009 at 6:07 PM, Malcolm
Wallacemalcolm.wall...@cs.york.ac.uk wrote:
the part I would really like to avoid is writing the
New.Foo { a=a, b=b, ... z=1 } part, where the field
names are many, long, and varied.
OK, here is another hack-ish trick, since I notice your data is stored on
disk as text, using show. I assume you are using something like Read to
retrieve it. Well, how about using a real parser instead? The parser
during conversion can be slightly more lax, automatically adding in the
extra field.
For instance, using polyparse's Text.Parse, and DrIFT to derive the
appropriate Parse instance for your datatype:
module Foo where
data Foo = Foo { a :: Int
, b :: Bool
, c :: Maybe Foo }
{-! derive : Parse !-}
DrIFT gives you this instance:
{-* Generated by DrIFT : Look, but Don't Touch. *-}
instance Parse Foo where
parse = constructors
[ ( Foo
, return Foo `discard` isWord { `apply` field a
`discard` isWord , `apply` field b
`discard` isWord , `apply` field c
`discard` isWord }
)
]
Let's say the field 'b' is new, and your existing data does not have it. So
just take the parser generated by DrIFT and make a small modification:
{-* Generated by DrIFT but modified by hand for conversion purposes *-}
instance Parse Foo where
parse = constructors
[ ( Foo
, return Foo `discard` isWord { `apply` field a
`apply` return True -- this field does not yet exist
in data
`discard` isWord , `apply` field c
`discard` isWord }
)
]
Then do the obvious thing: parse the old data, immediately write it out
again, and then throw away the modified parser in favour of the pure
generated one.
Regards,
Malcolm
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