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-Exception.html#3
"... 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-Exception-Base.html#try:
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
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