What are the semantics of "strict returns"?

Question

In the Haskell Performance Resource wiki-section, the not further explained recommendation is given to

• Use strict returns (return $! ...) unless you absolutely need them lazy.

Why is that a good thing to do? When exactly is the ...-expression (Whnf-)forced?

Given the "Left identity" monad-law and the definition

f $! x = x `seq` f x

I can rewrite (in do-notation`):

do x' <- return $! x
   f x'

to

do x' <- x `seq` return x
   f x'

But it seems I can't get to

do f $! x

PS: If the BangPatterns-extension is available, is

do !x' <- return x
   f x'

semantically the same as the first do-expression given above?

Answer 1

There's a reason you can't get from

do x' <- x `seq` return x
   f x'

to

f $! x

It's because they are not the same. Just expand the do notation:

(x `seq` return x) >>= (\ x' -> f x')

The seq will only be evaluated if the (>>=) is strict in its first argument. That's not necessarily true.

Answer 2

For IO there is also the useful Control.Exception.evaluate:

Forces its argument to be evaluated to weak head normal form when the resultant IO action is executed. It can be used to order evaluation with respect to other IO operations; its semantics are given by

evaluate :: a -> IO a
evaluate x = (return $! x) >>= return