Ross Paterson <ross@soi.city.ac.uk> wrote,
The new wording:
\code{unsafePerformIO} may compromise typing; to avoid this, the programmer should ensure that the result of \code{unsafePerformIO} has a monomorphic type.
rules out the following:
my_hash :: Storable a => a -> Int my_hash a = fromIntegral $ unsafePerformIO $ allocaBytes (sizeof a) $ \p -> do let size = fromIntegral (sizeOf a) c_memset p 0 size poke p a hash_bytes p size
foreign import ccall unsafe "memset" c_memset :: Ptr a -> CInt -> CSize -> IO () foreign import ccall unsafe hash_bytes :: Ptr a -> CSize -> IO CInt
Why is this ruled out? hash_bytes returns a `CInt', which is a monomorphic type.
Manuel writes:
However, it is possible to construct examples that are deterministic, but still dubious from a typing perspective. Let's assume a C routine
void *foo();
that *always returns the same pointer* to a buffer area. To bind this in Haskell as
foreign import ccall foo :: Ptr a
is problematic[1].
(a) It's constant across a run of the program, but its value still depends on the environment, and
Yes, and that's nothing that we want to rule out. A standard idiom for obtaining constant values from C is -= In C land =- int my_const () { ... return ...; } -= In Haskell land =- const :: Int const = unsafePerformIO my_const foreign import ccall my_const :: IO Int All that's required here is that my_const() is constant within a program run. Of course, I could have given a pure type in the foreign import in this simple example, but that's not different from an explicit `unsafePerformIO' and it is easy enough to construct an example where this is not possible.
(b) the declaration contains incorrect type information.
I guess, that's open to debate; ie, depends on how you interpret C types in Haskell. Manuel