[Haskell-cafe] What is the role of $!?

Thomas Davie tom.davie at gmail.com
Thu Nov 29 02:29:24 EST 2007

On 29 Nov 2007, at 06:32, PR Stanley wrote:

> Hi
> Thanks for the response.
> 	JCC: In most languages, if you have some expression E, and when the  
> computer attempts to evaluate E it goes in to an infinite loop, then  
> when the computer attempts to evaluate the expression f(E), it also
> goes into an infinite loop, regardless of what f is.  That's the  
> definition of a strict language.
> PRS: Does that mean that a strict language is also imperative?

Nope, not at all.  Just a strict language has slightly fewer programs  
it can evaluate correctly, as more will loop infinitely.

> Either e or f(e) could result in an infinite loop.
> 	JCC: In Haskell, this isn't the case ---we can write functions f  
> such that the computation f(E)  terminates,
> even when E does not.  (:) is one such function, as are some  
> functions built from it, such as (++); xn ++ ys terminates whenever  
> xn does, even if ys is an infinite loop.  This is what makes it easy
> and convenient to build infinite loops in Haskell; in most strict  
> languages, if you said
> 	let fibs = 0 : 1 : zipWith (+) fibs (tail fibs)
> the language would insist on evaluating fibs before it actually  
> assigned anything to the memory cell for fibs, giving rise to an  
> infinite loop.  (For this reason, most strict languages make such  
> definitions compile-time errors).
> Unfortunately, non-strictness turns out to be a pain in the ass to  
> implement, since it means when the code generator sees an  
> expression, it can't just generate code to evaluate it --- it has to  
> hide the code somewhere else, and then substitute a pointer to that  
> code for the value of the expression.
> 	PRS: Is there a kind of strictness applied when the compiler/ 
> interpreter sorts the various sub-expressions into little memory  
> compartments indexed with pointers for later evaluation? To put it  
> another way, does lazy evaluation begin with the outer-most  
> expression, the most abstract, and determine what sshould go where  
> in relation to the subsequent inner expressions?  For example:
> 		takeWhile (<20) [0..9] ++ [10..]
> The compiler determiens at the outset that the result of takeWhile  
> is a list followed by the calculation of the length of that list  
> based on the predicate (<20), and then calls ++ which is for all  
> intents and purposes on its own an infinite loop. Is this what  
> happens?

Not really.  For lazy evaluation the compiler doesn't "decide" the  
order statically -- it merely gives the program rules to follow for  
what the next expression to be evaluated should be.  Lets look at a  
slightly simpler example:

takeWhile (< 2) (map (+1) [0..])

We will always attempt to evaluate the outermost left most  
expression.  We do this by matching against the rules given in the  
program, to make this clearer, here are the rules for takeWhile and map:

takeWhile _ []                =  []
takeWhile p (x:xs) | p x      =  x : takeWhile p xs
                   | otherwise =  []

map _ [] = []
map f (x:xs) = f x : map f xs

    takeWhile (< 2) (map (+1) [0..])
    -- We start by evaluating the leftmost outermost expression.  We  
attempt to match on the first rule of takeWhile, and discover that we  
can't because we don't know whether the result of (map (+1) [0..]) is  
the empty list or not.  Therefore we demand the evaluation of (map +1)  
-> takeWhile (< 2) ((+1) 0 : map (+1) [1..])
    -- We now know that we don't have the empty list, so we must use  
the second rule of takeWhile.  We must evaluate the guard first though:
-> (<2) ((+1) 0) |
    -- To do this, we must evaluate ((+1) 0)
-> (<2) 1 |
    -- This evaluates to True, so we may insert the right hand side --  
note that x remains evaluated
-> True | 1 : takeWhile (<2) (map (+1) [1..])
    -- We can drop the guard now, but lets carry on.  We have already  
evaluated the outermost expression, so lets evaluate the next in.   
Again pattern matching on takeWhile demands the evaluation of map:
-> 1 : takeWhile (<2) ((+1) 1 : map (+1) [2..])
    -- We again, can pattern match on takeWhile, and must evaluate the  
guard again:
-> 1 : ((<2) ((+1) 1) |)
    -- Again, we must evaluate the result of the addition
-> 1 : ((<2) 2 |)
    -- This time we get False, so we must evaluate the next guard
-> 1 : (otherwise |)
    -- otherwise is a synonym for True, so we use this right hand side.
-> 1 : (True | [])
    -- and we can get rid of the guard, and prettify the result,  
giving us:
-> [1]

Note that we followed a set of rules that gave us non-strict  
semantics.  The set of rules is called lazy evaluation.  We may come  
up with several other sets of rules that give us different evaluation  
orders, but still non-strict semantics (e.g. Optimistic Evaluation).

> This is a very simple example, that's to say, I am aware that the  
> compiler may be faced with a much more complex job of applying lazy  
> evaluation. Nevertheless, I wonder if there are a set of fundamental  
> rules to which the compiler must always adhere in lazy evaluation.
> 	JCC: There are a number of clever optimizations you can use here  
> (indeed, most of the history of Haskell compilation techniques is a  
> list of clever techniques to get around the limitations of compiling  
> non-strict languages), but most of them rely on the compiler knowing  
> that, in this case, if a sub- expression is an infinite loop, the  
> entire expression is an infinite loop.  This is actually pretty easy  
> to figure out (most of the time), but sometimes the compiler needs a  
> little help.
> That's where $! (usually) comes in.  When the compiler sees (f $ x),  
> it has to look at f to see whether, if x is an infinite loop, f $ x  
> is one as well.  When the compiler sees (f $! x), it doesn't need to  
> look at f --- if x is an infinite loop, (f $! x) always is one as  
> well.  So, where in (f $ x) the compiler sometimes needs to put the  
> code for x in a separate top-level block, to be called later when  
> it's needed, in (f $! x) the compiler can always generate code for x  
> inline, like a compiler for a normal language would.  Since most CPU  
> architectures are optimized for normal languages that compile f(E)  
> by generating code for E inline, this is frequently a big speed-up.
> PrS: Your description of $! reminds me of the difference between  
> inline functions and "ordinary" functions in C++ with the former  
> being faster. Am I on the right track? In either case, (f $ x) and  
> (f $! x), lazy evaluation must be applied at a higher level
> otherwise either instruction could result in an infinite loop.  
> Therefore, is efficiency the only consideration here?
> If Haskell is a lazy language and $ merely implies lazy evaluation  
> then what's the difference between (f $ x \oplus y) and (f (x \oplus  
> y))?

$ does not mean "do lazy evaluation" it means "apply".  It's a  
function, like any other:

f ($) x = f x

All it does is takes the function, and the argument and applies one to  
the other, it can be used for eliminating ugly bracketing, or is  
useful in creating sections, e.g.

 > map ($ 5) [(1 +), (2 *), (3 ^)]
[6, 10, 243]

$! is the special case, which means strictly apply.  It evaluates its  
argument first, *then* does the application.  This may or may not be  
faster (and usually isn't, due to evaluating more of the argument):

f ($!) x = seq x (f x)

seq is a special function that says "first fully evaluate my first  
argument, then return my second argument", it breaks non-strict  
semantics.  Personally, my only use for such functions is a little bit  
of debugging work, seq for example can be used to force something to  
be printed whenever an expression is evaluated:

seq (unsafePerformIO $ putStrLn "At the nasty evaluation") (some  
problematic expression)

There is however a nicer version of this in the libraries, that masks  
it nicely for me:

trace "At the nasty evaluation" (some problematic expression)

I hope this helped somewhat.

Tom Davie

More information about the Haskell-Cafe mailing list