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# Seq magic

The innocent-looking `seq` operator causes all manner of mayhem in GHC. This page summarises the issues. See also discussion in Trac #5129.

## The baseline position

Our initial story was that `(seq e1 e2)` meant precisely

case e1 of { _ -> e2 }

Indeed this was `seq`'s inlining. This translation validates some important rules

* `seq` is strict in both its arguments * (e1 `seq` e2) e3 ===> e1 `seq` (e2 e3) * case (e1 `seq` e2) of alts ===> e1 `seq` (case e2 of alts) * value `seq` e ===> e

But this approach has problems; see `Note [Deguaring seq]` in `DsUtils`.

### Problem 1 (Trac #1031)

Consider

f x y = x `seq` (y `seq` (# x,y #))

The `[CoreSyn let/app invariant]` (see `CoreSyn`) means that, other things being equal, because
the argument to the outer `seq` has an unlifted type, we'll use call-by-value thus:

f x y = case (y `seq` (# x,y #)) of v -> x `seq` v

But that is bad for two reasons:

- we now evaluate
`y`before`x`, and - we can't bind
`v`to an unboxed pair

Seq is very, very special! Treating it as a two-argument function, strict in
both arguments, doesn't work. We "fixed" this by treating `seq` as a language
construct, desugared by the desugarer, rather than as a function that may (or
may not) be inlined by the simplifier. So the above term is desugared to:

case x of _ -> case y of _ -> (# x,y #)

### Problem 2 (Trac #2273)

Consider

let chp = case b of { True -> fst x; False -> 0 } in chp `seq` ...chp...

Here the `seq` is designed to plug the space leak of retaining `(snd x)`
for too long.

If we rely on the ordinary inlining of `seq`, we'll get

let chp = case b of { True -> fst x; False -> 0 } case chp of _ { I# -> ...chp... }

But since `chp` is cheap, and the case is an alluring contet, we'll
inline `chp` into the case scrutinee. Now there is only one use of `chp`,
so we'll inline a second copy. Alas, we've now ruined the purpose of
the seq, by re-introducing the space leak:

case (case b of {True -> fst x; False -> 0}) of I# _ -> ...case b of {True -> fst x; False -> 0}...

We can try to avoid doing this by ensuring that the binder-swap in the case happens, so we get his at an early stage:

case chp of chp2 { I# -> ...chp2... }

But this is fragile. The real culprit is the source program. Perhaps we should have said explicitly

let !chp2 = chp in ...chp2...

But that's painful. So the desugarer does a little hack to make `seq`
more robust: a saturated application of `seq` is turned **directly** into
the case expression, thus:

x `seq` e2 ==> case x of x -> e2 -- Note shadowing! e1 `seq` e2 ==> case x of _ -> e2

So we desugar our example to:

let chp = case b of { True -> fst x; False -> 0 } case chp of chp { I# -> ...chp... }

And now all is well.

Be careful not to desugar

True `seq` e ==> case True of True { ... }

which stupidly tries to bind the datacon 'True'. This is easily avoided.

The whole thing is a hack though; if you define `mySeq=seq`, the hack
won't work on `mySeq`.

### Problem 3 (Trac #5262)

Consider

f x = x `seq` (\y.y)

With the above desugaring we get

f x = case x of x { _ -> \y.y }

and now ete expansion gives

f x y = case x of x { _ -> y }

Now suppose that we have

f (length xs) `seq` 3

Plainly `(length xs)` should be evaluated... but it isn't because `f` has arity 2.
(Without -O this doesn't happen.)

### Problem 4: seq in the IO monad ==

See the extensive discussion in Trac #5129.

### Problem 5: the need for special rules

Roman found situations where he had

case (f n) of _ -> e

where he knew that `f` (which was strict in `n`) would terminate if n did.
Notice that the result of `(f n)` is discarded. So it makes sense to
transform to

case n of _ -> e

Rather than attempt some general analysis to support this, I've added enough support that you can do this using a rewrite rule:

RULE "f/seq" forall n e. seq (f n) e = seq n e

You write that rule. When GHC sees a case expression that discards
its result, it mentally transforms it to a call to `seq` and looks for
a RULE. (This is done in `Simplify.rebuildCase`.) As usual, the
correctness of the rule is up to you.

To make this work, we need to be careful that `seq` is **not** desguared
into a case expression on the LHS of a rule.

To increase applicability of these user-defined rules, we also
have the following built-in rule for `seq`

seq (x |> co) y = seq x y

This eliminates unnecessary casts and also allows other seq rules to match more often. Notably,

seq (f x |> co) y --> seq (f x) y

and now a user-defined rule for `seq` may fire.

# A better way

Here's our new plan.

- Introduce a new primop
`seq# :: a -> State# s -> (# a, State# s #)` - An application of the primop is not considered cheap.
- Desugar
`seq`thus:x `seq` e2 ==> case seq# x RW of (# x, _ #) -> e2 -- Note shadowing! e1 `seq` e2 ==> case seq# x RW of (# _, _ #) -> e2

- Define
`evaluate`thusevaluate :: a -> IO () evaluate x = IO (\s -> case seq# x s of (# _, s' #) -> (# (), s' #)

All the same equations hold as with the old defn for `seq`, but the problems
go away:

- Problem 1: (seq x y) is elaborated in the desugarer
- Problem 2: problem largely unaffected
- Problem 3: if we regard
`(seq# a b)`as expensive, we won't eta expand. - Problem 4: unchanged