The Haskell definition says:
An expression is in weak head normal form (WHNF), if it is either:
a constructor (eventually applied to arguments) like True, Just (square 42) or (:) 1
a built-in function applied to too few arguments (perhaps none) like (+) 2 or sqrt.
or a lambda abstraction \x -> expression.
Why do built-in functions receive special treatment? According to lambda calculus, there is no difference between a partially applied function and any other function, because at the end we have only one argument functions.
A normal function applied to an argument, like the following:
(\x y -> x + 1 : y) 1
Can be reduced, to give:
\y -> 1 + 1 : y
In the first expression, the "outermost" thing was an application, so it was not in WHNF. In the second, the outermost thing is a lambda abstraction, so it is in WHNF (even though we could do more reductions inside the function body).
Now lets consider the application of a built-in (primitive) function:
(+) 1
Because this is a built-in, there's no function body in which we can substitute 1 for the first parameter. The evaluator "just knows" how to evaluate fully "saturated" applications of (+), like (+) 1 2. But there's nothing that can be done with a partially applied built-in; all we can do is produce a data structure describing "apply (+) to 1 and wait for one more argument", but that's exactly what the thing we're trying to reduce is. So we do nothing.
Built-ins are special because they're not defined by lambda calculus expressions, so the reduction process can't "see inside" their definition. Thus, unlike normal functions applications, built-in function applications have to be "reduced" by just accumulating arguments until they are fully "saturated" (in which case reduction to WHNF is by running whatever the magic implementation of the built-in is). Unsaturated built-in applications cannot be reduced any further, and so are already in WHNF.
Consider
Prelude> let f n = [(+x) | x <- [1..]] !! n
Prelude> let g = f 20000000 :: Int -> Int
g is at this point not in WHNF! You can see this by evaluating, say, g 3, which takes a noticable lag because you need WHNF before you can apply an argument. That's when the list is traversed in search for the right built-in function. But afterwards, this choice is then fixed, g is WHNF (and indeed NF: that's the same for lambdas, perhaps what you meant with your question) and thus any subsequent calls will give a result immediately.
Related
Say I have a general recursive definition in haskell like this:
foo a0 a1 ... = base_case
foo b0 b1 ...
| cond1 = recursive_case_1
| cond2 = recursive_case_2
...
Can it always rewritten using foldr? Can it be proved?
If we interpret your question literally, we can write const value foldr to achieve any value, as #DanielWagner pointed out in a comment.
A more interesting question is whether we can instead forbid general recursion from Haskell, and "recurse" only through the eliminators/catamorphisms associated to each user-defined data type, which are the natural generalization of foldr to inductively defined data types. This is, essentially, (higher-order) primitive recursion.
When this restriction is performed, we can only compose terminating functions (the eliminators) together. This means that we can no longer define non terminating functions.
As a first example, we lose the trivial recursion
f x = f x
-- or even
a = a
since, as said, the language becomes total.
More interestingly, the general fixed point operator is lost.
fix :: (a -> a) -> a
fix f = f (fix f)
A more intriguing question is: what about the total functions we can express in Haskell? We do lose all the non-total functions, but do we lose any of the total ones?
Computability theory states that, since the language becomes total (no more non termination), we lose expressiveness even on the total fragment.
The proof is a standard diagonalization argument. Fix any enumeration of programs in the total fragment so that we can speak of "the i-th program".
Then, let eval i x be the result of running the i-th program on the natural x as input (for simplicity, assume this is well typed, and that the result is a natural). Note that, since the language is total, then a result must exist. Moreover, eval can be implemented in the unrestricted Haskell language, since we can write an interpreter of Haskell in Haskell (left as an exercise :-P), and that would work as fine for the fragment. Then, we simply take
f n = succ $ eval n n
The above is a total function (a composition of total functions) which can be expressed in Haskell, but not in the fragment. Indeed, otherwise there would be a program to compute it, say the i-th program. In such case we would have
eval i x = f x
for all x. But then,
eval i i = f i = succ $ eval i i
which is impossible -- contradiction. QED.
In type theory, it is indeed the case that you can elaborate all definitions by dependent pattern-matching into ones only using eliminators (a more strongly-typed version of folds, the generalisation of lists' foldr).
See e.g. Eliminating Dependent Pattern Matching (pdf)
I wanted to flip a list constructor usage, to have type:
[a] -> a -> [a]
(for use in a fold), so tried:
(flip :)
but it gives the type:
Prelude> :t (flip :)
(flip :) :: [(a -> b -> c) -> b -> a -> c] -> [(a -> b -> c) -> b -> a -> c]
This surprised me, but it appears that this was parsed as a left section of (:), instead of a partial application of flip. Rewriting it using flip as infix seems to overcome this,
Prelude> :t ((:) `flip`)
((:) `flip`) :: [a] -> a -> [a]
But I couldn't find the rule defining this behavior, and I thought that function application was the highest precedence, and was evaluated left->right, so I would have expected these two forms to be equivalent.
What you want to do is this:
λ> :t (flip (:))
(flip (:)) :: [a] -> a -> [a]
Operators in Haskell are infix. So when you do flip : it operates in an infix fashion i.e. flip is applied to : function. By putting parenthesis explicitly in flip (:), you tell that : should be applied to flip. You can also use the backtick operator in flip for making that infix which you have tried already.
It was putting : in parentheses that made your second example work, not using backticks around flip.
We often say that "function application has highest precedence" to emphasise that e.g. f x + 1 should be read as (f x) + 1, and not as f (x + 1). But this isn't really wholly accurate. If it was, and (flip :) parsed as you expected, then the highest precedence after (f x) + 1 would be the application of (f x) to +; the whole expression f x + 1 would end up being parsed as f applied to 3 arguments: x, +, and 1. But this would happen with all expressions involving infix operators! Even a simple 1 + 1 would be recognised as 1 applied to + and 1 (and then complain about the missing Num instance that would allow 1 to be a function).
Essentially this strict understanding of "function application has highest precedence" would mean that function application would be all that ever happens; infix operators would always end up as arguments to some function, never actually working as infix operators.
Actually precedence (and associativity) are mechanisms for resolving the ambiguity of expressions involving multiple infix operators. Function application is not an infix operator, and simply doesn't take part in the precedence/associativity system. Chains of terms that don't involve operators are resolved as function application before precedence is invoked to resolve the operator applications (hence "highest precedence"), but it's not really precedence that causes it.
Here's how it works. You start with a linear sequence of terms and operators; there's no structure, they were simply written next to each other.
What I'm calling a "term" here can be a non-operator identifier like flip; or a string, character, or numeric literal; or a list expression; or a parenthesised subexpression; etc. They're all opaque as far as this process is concerned; we only know (and only need to know) that they're not infix operators. We can always tell an operator because it will either be a "symbolic" identifier like ++!#>, or an alphanumeric identifier in backticks.
So, sequence of terms and operators. You find all chains of one or more terms in a row that contain no operators. Each such chain is a chain of function applications, and becomes a single term.1
Now if you have two operators directly next to each other you've got an error. If your sequence starts or ends in an operator, that's also an error (unless this is an operator section).
At this point you're guaranteed to have a strictly alternating sequence like term operator term operator term operator term, etc. So you pick the operator with the highest precedence together with the terms to its left and right, call that an operator application, and those three items become a single term. Associativity acts as a tie break when you have multiple operators with the same precedence. Rinse and repeat until the whole expression has become a single term (or associativity fails to break a tie, which is also an error). This means that in an expression involving operators, the "top level application" is always one of the operators, never ordinary function application.
A consequence of this is that there are no circumstances under which an operator can end up passed as the argument to a function. It's simply impossible. This is why we need the (:) syntax to disable the "operator-ness" of operators, and get at their identity as values.
For flip : the only chain of non-operator terms is just flip, so there's no ordinary function application to resolve "at highest precedence". : then goes looking for its left and right arguments (but this is a section, so there's no right argument), and finds flipon its left.
To make flip receive : as an argument instead of the other way around, you must write flip (:). (:) is not an operator (it's in parentheses, so it doesn't matter what's inside), and so we have a chain of two terms with no operators, so that gets resolved to a single expression by applying flip to (:).
1 The other way to look at this is that you identify all sequences of terms not otherwise separated by operators and insert the "function application operator" between them. This "operator" has higher precedence than it's possible to assign to other operators and is left-associative. Then the operator-resolution logic will automatically treat function application the way I've been describing.
I'm trying to better understand Haskell's laziness, such as when it evaluates an argument to a function.
From this source:
But when a call to const is evaluated (that’s the situation we are interested in, here, after all), its return value is evaluated too ... This is a good general principle: a function obviously is strict in its return value, because when a function application needs to be evaluated, it needs to evaluate, in the body of the function, what gets returned. Starting from there, you can know what must be evaluated by looking at what the return value depends on invariably. Your function will be strict in these arguments, and lazy in the others.
So a function in Haskell always evaluates its own return value? If I have:
foo :: Num a => [a] -> [a]
foo [] = []
foo (_:xs) = map (* 2) xs
head (foo [1..]) -- = 4
According to the above paragraph, map (* 2) xs, must be evaluated. Intuitively, I would think that means applying the map to the entire list- resulting in an infinite loop.
But, I can successfully take the head of the result. I know that : is lazy in Haskell, so does this mean that evaluating map (* 2) xs just means constructing something else that isn't fully evaluated yet?
What does it mean to evaluate a function applied to an infinite list? If the return value of a function is always evaluated when the function is evaluated, can a function ever actually return a thunk?
Edit:
bar x y = x
var = bar (product [1..]) 1
This code doesn't hang. When I create var, does it not evaluate its body? Or does it set bar to product [1..] and not evaluate that? If the latter, bar is not returning its body in WHNF, right, so did it really 'evaluate' x? How could bar be strict in x if it doesn't hang on computing product [1..]?
First of all, Haskell does not specify when evaluation happens so the question can only be given a definite answer for specific implementations.
The following is true for all non-parallel implementations that I know of, like ghc, hbc, nhc, hugs, etc (all G-machine based, btw).
BTW, something to remember is that when you hear "evaluate" for Haskell it normally means "evaluate to WHNF".
Unlike strict languages you have to distinguish between two "callers" of a function, the first is where the call occurs lexically, and the second is where the value is demanded. For a strict language these two always coincide, but not for a lazy language.
Let's take your example and complicate it a little:
foo [] = []
foo (_:xs) = map (* 2) xs
bar x = (foo [1..], x)
main = print (head (fst (bar 42)))
The foo function occurs in bar. Evaluating bar will return a pair, and the first component of the pair is a thunk corresponding to foo [1..]. So bar is what would be the caller in a strict language, but in the case of a lazy language it doesn't call foo at all, instead it just builds the closure.
Now, in the main function we actually need the value of head (fst (bar 42)) since we have to print it. So the head function will actually be called. The head function is defined by pattern matching, so it needs the value of the argument. So fst is called. It too is defined by pattern matching and needs its argument so bar is called, and bar will return a pair, and fst will evaluate and return its first component. And now finally foo is "called"; and by called I mean that the thunk is evaluated (entered as it's sometimes called in TIM terminology), because the value is needed. The only reason the actual code for foo is called is that we want a value. So foo had better return a value (i.e., a WHNF). The foo function will evaluate its argument and end up in the second branch. Here it will tail call into the code for map. The map function is defined by pattern match and it will evaluate its argument, which is a cons. So map will return the following {(*2) y} : {map (*2) ys}, where I have used {} to indicate a closure being built. So as you can see map just returns a cons cell with the head being a closure and the tail being a closure.
To understand the operational semantics of Haskell better I suggest you look at some paper describing how to translate Haskell to some abstract machine, like the G-machine.
I always found that the term "evaluate," which I had learned in other contexts (e.g., Scheme programming), always got me all confused when I tried to apply it to Haskell, and that I made a breakthrough when I started to think of Haskell in terms of forcing expressions instead of "evaluating" them. Some key differences:
"Evaluation," as I learned the term before, strongly connotes mapping expressions to values that are themselves not expressions. (One common technical term here is "denotations.")
In Haskell, the process of forcing is IMHO most easily understood as expression rewriting. You start with an expression, and you repeatedly rewrite it according to certain rules until you get an equivalent expression that satisfies a certain property.
In Haskell the "certain property" has the unfriendly name weak head normal form ("WHNF"), which really just means that the expression is either a nullary data constructor or an application of a data constructor.
Let's translate that to a very rough set of informal rules. To force an expression expr:
If expr is a nullary constructor or a constructor application, the result of forcing it is expr itself. (It's already in WHNF.)
If expr is a function application f arg, then the result of forcing it is obtained this way:
Find the definition of f.
Can you pattern match this definition against the expression arg? If not, then force arg and try again with the result of that.
Substitute the pattern match variables in the body of f with the parts of (the possibly rewritten) arg that correspond to them, and force the resulting expression.
One way of thinking of this is that when you force an expression, you're trying to rewrite it minimally to reduce it to an equivalent expression in WHNF.
Let's apply this to your example:
foo :: Num a => [a] -> [a]
foo [] = []
foo (_:xs) = map (* 2) xs
-- We want to force this expression:
head (foo [1..])
We will need definitions for head and `map:
head [] = undefined
head (x:_) = x
map _ [] = []
map f (x:xs) = f x : map f x
-- Not real code, but a rule we'll be using for forcing infinite ranges.
[n..] ==> n : [(n+1)..]
So now:
head (foo [1..]) ==> head (map (*2) [1..]) -- using the definition of foo
==> head (map (*2) (1 : [2..])) -- using the forcing rule for [n..]
==> head (1*2 : map (*2) [2..]) -- using the definition of map
==> 1*2 -- using the definition of head
==> 2 -- using the definition of *
I believe the idea must be that in a lazy language if you're evaluating a function application, it must be because you need the result of the application for something. So whatever reason caused the function application to be reduced in the first place is going to continue to need to reduce the returned result. If we didn't need the function's result we wouldn't be evaluating the call in the first place, the whole application would be left as a thunk.
A key point is that the standard "lazy evaluation" order is demand-driven. You only evaluate what you need. Evaluating more risks violating the language spec's definition of "non-strict semantics" and looping or failing for some programs that should be able to terminate; lazy evaluation has the interesting property that if any evaluation order can cause a particular program to terminate, so can lazy evaluation.1
But if we only evaluate what we need, what does "need" mean? Generally it means either
a pattern match needs to know what constructor a particular value is (e.g. I can't know what branch to take in your definition of foo without knowing whether the argument is [] or _:xs)
a primitive operation needs to know the entire value (e.g. the arithmetic circuits in the CPU can't add or compare thunks; I need to fully evaluate two Int values to call such operations)
the outer driver that executes the main IO action needs to know what the next thing to execute is
So say we've got this program:
foo :: Num a => [a] -> [a]
foo [] = []
foo (_:xs) = map (* 2) xs
main :: IO ()
main = print (head (foo [1..]))
To execute main, the IO driver has to evaluate the thunk print (head (foo [1..])) to work out that it's print applied to the thunk head (foo [1..]). print needs to evaluate its argument on order to print it, so now we need to evaluate that thunk.
head starts by pattern matching its argument, so now we need to evaluate foo [1..], but only to WHNF - just enough to tell whether the outermost list constructor is [] or :.
foo starts by pattern matching on its argument. So we need to evaluate [1..], also only to WHNF. That's basically 1 : [2..], which is enough to see which branch to take in foo.2
The : case of foo (with xs bound to the thunk [2..]) evaluates to the thunk map (*2) [2..].
So foo is evaluated, and didn't evaluate its body. However, we only did that because head was pattern matching to see if we had [] or x : _. We still don't know that, so we must immediately continue to evaluate the result of foo.
This is what the article means when it says functions are strict in their result. Given that a call to foo is evaluated at all, its result will also be evaluated (and so, anything needed to evaluate the result will also be evaluated).
But how far it needs to be evaluated depends on the calling context. head is only pattern matching on the result of foo, so it only needs a result to WHNF. We can get an infinite list to WHNF (we already did so, with 1 : [2..]), so we don't necessarily get in an infinite loop when evaluating a call to foo. But if head were some sort of primitive operation implemented outside of Haskell that needed to be passed a completely evaluated list, then we'd be evaluating foo [1..] completely, and thus would never finish in order to come back to head.
So, just to complete my example, we're evaluating map (2 *) [2..].
map pattern matches its second argument, so we need to evaluate [2..] as far as 2 : [3..]. That's enough for map to return the thunk (2 *) 2 : map (2 *) [3..], which is in WHNF. And so it's done, we can finally return to head.
head ((2 *) 2 : map (2 *) [3..]) doesn't need to inspect either side of the :, it just needs to know that there is one so it can return the left side. So it just returns the unevaluated thunk (2 *) 2.
Again though, we only evaluated the call to head this far because print needed to know what its result is, so although head doesn't evaluate its result, its result is always evaluated whenever the call to head is.
(2 *) 2 evaluates to 4, print converts that into the string "4" (via show), and the line gets printed to the output. That was the entire main IO action, so the program is done.
1 Implementations of Haskell, such as GHC, do not always use "standard lazy evaluation", and the language spec does not require it. If the compiler can prove that something will always be needed, or cannot loop/error, then it's safe to evaluate it even when lazy evaluation wouldn't (yet) do so. This can often be faster so GHC optimizations do actually do this.
2 I'm skipping over a few details here, like that print does have some non-primitive implementation we could step inside and lazily evaluate, and that [1..] could be further expanded to the functions that actually implement that syntax.
Not necessarily. Haskell is lazy, meaning that it only evaluates when it needs to. This has some interesting effects. If we take the below code, for example:
-- File: lazinessTest.hs
(>?) :: a -> b -> b
a >? b = b
main = (putStrLn "Something") >? (putStrLn "Something else")
This is the output of the program:
$ ./lazinessTest
Something else
This indicates that putStrLn "Something" is never evaluated. But it's still being passed to the function, in the form of a 'thunk'. These 'thunks' are unevaluated values that, rather than being concrete values, are like a breadcrumb-trail of how to compute the value. This is how Haskell laziness works.
In our case, two 'thunks' are passed to >?, but only one is passed out, meaning that only one is evaluated in the end. This also applies in const, where the second argument can be safely ignored, and therefore is never computed. As for map, GHC is smart enough to realise that we don't care about the end of the array, and only bothers to compute what it needs to, in your case the second element of the original list.
However, it's best to leave the thinking about laziness to the compiler and keep coding, unless you're dealing with IO, in which case you really, really should think about laziness, because you can easily go wrong, as I've just demonstrated.
There are lots and lots of online articles on the Haskell wiki to look at, if you want more detail.
Function could evaluate either return type:
head (x:_) = x
or exception/error:
head _ = error "Head: List is empty!"
or bottom (⊥)
a = a
b = last [1 ..]
On ZVON, one of the definitions provided for the takeWhile function is
Input: takeWhile (\x -> 6*x < 100) [1..20]
Output: [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
Can someone explain what the portion (\x -> 6*x < 100) means?
It's an anonymous function definition, otherwise known as a lambda-expression. (\x -> 6*x < 100) is a function which takes a number, and returns the boolean result of the inequality.
Since functional languages like Haskell frequently take functions as arguments, it is convenient to be able to define simple functions in-line, without needing to assign them a name.
Originally, the story goes, Alonzo Church wanted to mark variables in functional expressions with a circumflex, like e.g. (ŷ.x(yz)) but the Princeton printing press just couldn't do that at the time. He then wanted at least to print carets before the vars, like this: (^y.x(yz)), but they couldn't do that either.
The next best option was to use the Greek letter lambda instead, and so they ended up writing (λy.x(yz)) etc., hence the "lambda" in lambda-expression. It was all just a typographical accident.
Today on ASCII terminals we can't even use the letter λ, and so in Haskell we use a backslash instead (and an arrow in place of a dot in original lambda-expressions notation):
(\y -> x (y z))
stands for a function g such that
g y = x (y z)
Source: read it somewhere, don't remember where.
(\x -> 6*x < 100) is a lambda, an anonymous function that takes one argument (here called x) and computes & returns 6*x < 100, i.e., tests whether that number multiplied by 6 is less than 100.
It is a lambda function, that is, a function that you define in the spot mostly for convenience. You read it as "take x as your input, multiply it by 6 and see if it is less than 100". There are some other amenities related, though. For example, in Haskell Lambda functions and ordinary functions have a lexical environment associated and are properly speaking closures, so that they can perform computations using the environment as input.
I have been wondering about this a lot, but I haven't been able to find anything about it.
When using the seq function, how does it then really work? Everywhere, it is just explained saying that seq a b evaluates a, discards the result and returns b.
But what does that really mean? Would the following result in strict evaluation:
foo s t = seq q (bar q t) where
q = s*t
What I mean is, is q strictly evaluated before being used in bar? And would the following be equivalent:
foo s t = seq (s*t) (bar (s*t) t)
I find it a little hard getting specifics on the functionality of this function.
You're not alone. seq is probably one of the most difficult Haskell functions to use properly, for a few different reasons. In your first example:
foo s t = seq q (bar q t) where
q = s*t
q is evaluated before bar q t is evaluated. If bar q t is never evaluated, q won't be either. So if you have
main = do
let val = foo 10 20
return ()
as val is never used, it won't be evaluated. So q won't be evaluated either. If you instead have
main = print (foo 10 20)
the result of foo 10 20 is evaluated (by print), so within foo q is evaluated before the result of bar.
This is also why this doesn't work:
myseq x = seq x x
Semantically, this means the first x will be evaluated before the second x is evaluated. But if the second x is never evaluated, the first one doesn't need to be either. So seq x x is exactly equivalent to x.
Your second example may or may not be the same thing. Here, the expression s*t will be evaluated before bar's output, but it may not be the same s*t as the first parameter to bar. If the compiler performs common sub-expression elimination, it may common-up the two identical expressions. GHC can be fairly conservative about where it does CSE though, so you can't rely on this. If I define bar q t = q*t it does perform the CSE and evaluate s*t before using that value in bar. It may not do so for more complex expressions.
You might also want to know what is meant by strict evaluation. seq evaluates the first argument to weak head normal form (WHNF), which for data types means unpacking the outermost constructor. Consider this:
baz xs y = seq xs (map (*y) xs)
xs must be a list, because of map. When seq evaluates it, it will essentially transform the code into
case xs of
[] -> map (*y) xs
(_:_) -> map (*y) xs
This means it will determine if the list is empty or not, then return the second argument. Note that none of the list values are evaluated. So you can do this:
Prelude> seq [undefined] 4
4
but not this
Prelude> seq undefined 5
*** Exception: Prelude.undefined
Whatever data type you use for seqs first argument, evaluating to WHNF will go far enough to figure out the constructor and no further. Unless the data type has components that are marked as strict with a bang pattern. Then all the strict fields will also be evaluated to WHNF.
Edit: (thanks to Daniel Wagner for suggestion in comments)
For functions, seq will evaluate the expression until the function "has a lambda showing", which means that it's ready for application. Here are some examples that might demonstrate what this means:
-- ok, lambda is outermost
Prelude> seq (\x -> undefined) 'a'
'a'
-- not ok. Because of the inner seq, `undefined` must be evaluated before
-- the lambda is showing
Prelude> seq (seq undefined (\x -> x)) 'b'
*** Exception: Prelude.undefined
If you think of a lambda binding as a (built-in) data constructor, seq on functions is perfectly consistent with using it on data.
Also, "lambda binding" subsumes all types of function definitions, whether defined by lambda notation or as a normal function.
The Controversy section of the HaskellWiki's seq page has a little about some of the consequences of seq in relation to functions.
You can think of seq as:
seq a b = case a of
_ -> b
This will evaluate a to head-normal form (WHNF) and then continue with evaluating b.
Edit after augustss comment: this case ... of is the strict, GHC Core one, which always forces its argument.