strict bits of datatypes

Ian Lynagh igloo at
Tue Mar 20 10:55:26 EDT 2007

On Tue, Mar 20, 2007 at 01:53:47PM +0000, Malcolm Wallace wrote:
> Now, in the definition
>     x = x `seq` foo
> one can also make the argument that, if the value of x (on the lhs of
> the defn) is demanded, then of course the x on the rhs of the defn is
> also demanded.  There is no need for the `seq` here either.
> Semantically, the definition is equivalent to
>     x = foo
> I am arguing that, as a general rule, eliding the `seq` in such a case
> is an entirely valid and correct transformation.

So does nhc98 print "Foo" for this program?

    main = putStrLn $ let x = x `seq` "Foo" in x

(yhc tells me my program has deadlocked, but my recent attempt to
compile nhc98 failed so I can't check it).

I don't fully understand what your interpretation is; is it also true
    y = x
    x = y `seq` foo
is equivalent to
    y = x
    x = foo

And is it true that
    y = if True then x else undefined
    x = y `seq` foo
is equivalent to
    y = x
    x = foo

> The objection to this point of view is that if you have a definition
>     x = x `seq` foo
> then, operationally, you have a loop, because to evaluate x, one must
> first evaluate x before evaluating foo.  But as I said at the beginning,
> `seq` does _not_ imply order of evaluation, so the objection is not
> well-founded.

I'm having trouble finding a non-operational description of the
behaviour I think seq should have. (Nor, for that matter, can I think of
a description that makes it clear that it has the semantics that you
think it should have). Anyone?

I think you could make a similar argument that
    let x = x in x :: ()
is () rather than _|_, and similarly
    let x = x in x :: Int
is 3, or is there some key difference I'm missing?


More information about the Haskell-prime mailing list