[Haskell-cafe] evaluation semantics of bind
Richard O'Keefe
ok at cs.otago.ac.nz
Mon Feb 9 21:37:58 EST 2009
On 10 Feb 2009, at 1:35 am, Gregg Reynolds wrote:
> My original question was motivated by the observation that a human
> reader of an expression of the form "e >>= f" , on seeing that f is
> constant, may pull the constant value out of f, disregard e and
> dispense with the application f e.
There isn't any "application f e".
Any human reader who does that is simply WRONG to do so.
> So can a compiler, unless IO expressions are involved, in which
> case such optimizations are forbidden.
Remember that (e >>= f) means (>>=) e f.
When/whether it is sound to replace (>>=) e f by (f undefined)
depends on the semantics of >>=.
But >>= is an operation of a type class (the Monad type class).
If we come across
g :: Monad m => m a -> b -> m b
g e r = e >>= \x -> return r
then the compiler doesn't know what m is, so it doesn't know
which, if any, of its arguments >>= depends on.
data Trivial x = Trivial x
instance Monad Trivial
where return = Trivial
(>>=) (Trivial a) f = f a
If we now try
h :: Trivial a -> b -> Trivial b
h e r = g e r
then the compiler can inline the call to g
h e r = e >>= \x -> return r
and inline the calls to >>= and return
h (Trivial a) r = (\x -> Trivial r) a
then simplify to
h (Trivial _) r = Trivial r
You might have thought that the result doesn't depend on
the first argument to h, but as written, it does. This
version of h is strict in its first argument, so even
though there are NO side effects whatever, the first
argument will be evaluated to weak head normal form.
Of course, since there is only one constructor for Trivial,
it could be a newtype. If I change that declaration to
newtype Trivial x = Trivial x
then the pattern match is implicitly irrefutable,
h ~(Trivial _) r = Trivial r
and if the compiler doesn't simplify ~(NTC _) to _ when NTC
is a newtype constructor, I'll be surprised, so it's like
h _ r = Trivial r
and the function will *NOT* be strict in its first argument.
Here you see that the soundness or otherwise of eliminating
the first argument of >>= when the second argument doesn't
depend on the eventual result has nothing to do with IO as
such and certainly nothing to do with side effects. It's
really all about whether the version of >>= used is strict
in its first argument or not.
> I wondered if that was due to the semantics of >>= or the
> semantics of IO.
It's about the semantics of IO's implementation of >>=.
>
>
> To summarize what I've concluded thanks to the helpful people on
> haskell-cafe:
>
> The compiler can optimize e >>= f except for any IO expressions in e
> and f.
False. There is nothing unusual about IO here.
> IO expressions must be evaluated, due to the semantics of IO.
False.
> The may not be disregarded, memoized, or substituted.
False. In Haskell there is nothing whatever unusual about
expressions of type IO something. They *MAY* be disregarded:
let n = length [getChar, putChar 'f']
can be optimised to let n = 2. They MAY be memoised. They
MAY be substituted.
> IO semantics may be implemented in different ways by different
> compilers;
True. But then, so may Int semantics.
> these implementation techniques are not part of the formal semantics
> of the language, which may be expressed as above: IO expressions
> must be evaluated wherever and whenever they occur.
False. Utterly false.
>
>
> The bind operator >>= enforces sequencing on arguments containing IO
> expressions, but does not force evaluation.
False. For the special case of IO (and for other similar monads)
>>= enforced sequence BY forcing evaluation (more precisely, by
being strict in its first argument).
> Even bind expressions involving IO may be optimized. For example:
>
> getChar >>= \x -> ...<monster computation>... putChar 'c'
>
> The compiler may discard <monster computation> (assuming it contains
> no IO expressions), but it must evaluate getChar and putChar (due to
> IO semantics) in the correct order (due to bind semantics).
You are conflating *evaluating* (by the Haskell evaluator) with
*performance* (by the Haskell environment). IO *values* are
determined by the Haskell evaluator exactly like any other values;
IO *actions* are performed by the Haskell environment as part of
the process of forcing the lazy evaluator to do some work.
In your fragmentary example, <monster computation> may be discarded
EVEN IF it contains IO expressions, it's only if they are linked into
the IO chain using >> and/or >>= that the environment will perform
their values.
>
>
> Thanks all,
>
> gregg
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