On 28/04/10 14:45, Mads Lindstrøm wrote:
Hi
From http://haskell.org/ghc/docs/6.12.1/html/libraries/base-4.2.0.0/Control-Excep...
"... The difference between using try and catch for recovery is that in catch the handler is inside an implicit block (see "Asynchronous Exceptions") which is important when catching asynchronous exceptions ..."
However, 'try' is implemented by calling catch http://haskell.org/ghc/docs/6.12.1/html/libraries/base-4.2.0.0/src/Control-E...:
try :: Exception e => IO a -> IO (Either e a) try a = catch (a>>= \ v -> return (Right v)) (\e -> return (Left e))
Thus, I wonder, why do 'try' not "inherit" the implicit block mentioned above?
There's nothing magic going on - the "handler" in the case of try is just (return . Left), and that does indeed get executed with an implicit block.
Looking at catch:
catch :: Exception e => IO a -- ^ The computation to run -> (e -> IO a) -- ^ Handler to invoke if an exception is raised -> IO a catch io h = H'98.catch io (h . fromJust . fromException . toException)
I see no call to 'block'. But maybe it is hidden in H'98.catch? And is H'98.catch == Prelude.catch ?
The block is implicit (it's built into the implementation of catch, in fact). Cheers, Simon