I think about StableName like the "&" operator in C, that returns you the
memory address of a variable. It's not the same for many reasons, but by
analogy, if &x == &y then x == y, but &x != &y does not imply x != y.
So, values that are semantically equal, may be stored in different memory
locations and have different StableNames.
The fact that changing the order of the lines also changes the result of
the computation is obviously stated in the type signature of
makeStableName, which lives in the IO monad. On the other hand
hashStableNAme is a pure function.
L.
On Tue, Jun 26, 2012 at 3:26 PM, Ismael Figueroa Palet wrote: 2012/6/26 Lorenzo Bolla The point I was making is that StableName might be what you want. You are
using it to check if two functions are the same by comparing their
"stablehash". But from StableName documentation: The reverse is not necessarily true: if two stable names are not equal, then the objects they name may still be equal. The `eq` you implemented means this, I reckon: if `eq` returns True then
the 2 functions are equal, if `eq` returns False then you can't tell! Does it make sense?
L. Yes it does make sense, and I'm wondering why the hash are equal in one
case but are not equal on the other case (i.e. using let/where vs not using
it) because I'd like it to behave the same in both situations Thanks again On Tue, Jun 26, 2012 at 1:54 PM, Ismael Figueroa Palet <
ifigueroap@gmail.com> wrote: Thanks Lorenzo, I'm cc'ing the list with your response also: As you point out, when you do some kind of "let-binding", using the
where clause, or explicit let as in: main :: IO ()
main = do
let f1 = (successor :: Int -> State Int Int)
let f2 = (successor :: Int -> Maybe Int)
b2 <- eq f2 f2
b1 <- eq f1 f1
print (show b1 ++ " " ++ show b2) The behavior is as expected. I guess the binding triggers some internal
optimization or gives more information to the type checker; but I'm still
not clear why it is required to be done this way -- having to let-bind
every function is kind of awkward. I know the details of StableNames are probably implementation-dependent,
but I'm still wondering about how to detect / restrict this situation. Thanks 2012/6/26 Lorenzo Bolla From StableName docs: The reverse is not necessarily true: if two stable names are not
equal, then the objects they name may still be equal. This version works as expected: import System.Mem.StableName
import Control.Monad.State eq :: a -> b -> IO Bool
eq a b = do
pa <- makeStableName a
pb <- makeStableName b
return (hashStableName pa == hashStableName pb) successor :: (Num a, Monad m) => a -> m a
successor n = return (n+1) -- main :: IO ()
-- main = do
-- b2 <- eq (successor :: Int -> State Int Int) (successor ::
Int -> State Int Int)
-- b1 <- eq (successor :: Int -> Maybe Int) (successor :: Int
-> Maybe Int)
-- print (show b1 ++ " " ++ show b2) main :: IO ()
main = do
b2 <- eq f2 f2
b1 <- eq f1 f1
print (show b1 ++ " " ++ show b2)
where f1 = (successor :: Int -> Maybe Int)
f2 = (successor :: Int -> State Int Int) hth,
L. On Tue, Jun 26, 2012 at 1:15 PM, Ismael Figueroa Palet <
ifigueroap@gmail.com> wrote: I'm using StableNames to have a notion of function equality, and I'm
running into problems when using monadic functions. Consider the code below, file Test.hs import System.Mem.StableName
import Control.Monad.State eq :: a -> b -> IO Bool
eq a b = do
pa <- makeStableName a
pb <- makeStableName b
return (hashStableName pa == hashStableName pb) successor :: (Num a, Monad m) => a -> m a
successor n = return (n+1) main :: IO ()
main = do
b1 <- eq (successor :: Int -> Maybe Int) (successor :: Int ->
Maybe Int)
b2 <- eq (successor :: Int -> State Int Int) (successor :: Int
-> State Int Int)
print (show b1 ++ " " ++ show b2) Running the code into ghci the result is "False False". There is some
old post saying that this is due to the dictionary-passing style for
typeclasses, and compiling with optimizations improves the situation. Compiling with ghc --make -O Tests.hs and running the program, the
result is "True True", which is what I expect.
However, if I change main to be like the following: main :: IO ()
main = do
b2 <- eq (successor :: Int -> State Int Int) (successor :: Int
-> State Int Int)
b1 <- eq (successor :: Int -> Maybe Int) (successor :: Int ->
Maybe Int)
print (show b1 ++ " " ++ show b2) i.e. just changing the sequential order, and then compiling again with
the same command, I get "True False", which is very confusing for me.
Similar situations happens when using the state monad transformer, and
manually built variations of it. It sounds the problem is with hidden closures created somewhere that
do not point to the same memory locations, so StableNames yields false for
that cases, but it is not clear to me under what circumstances this
situation happens. Is there other way to get some approximation of function
equality? or a way to "configure" the behavior of StableNames in presence
of class constraints? I'm using the latests Haskell Platform on OS X Lion, btw. Thanks --
Ismael _______________________________________________
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