f x zero = Nothing
f x y = Just $ x / y
where zero = 0
The literal-bound identifier zero simply matches all after the warning Pattern match(es) are overlapped.
That's how Haskell's syntax works; every lowercase-initial variable name in a pattern (re)binds that name. Any existing binding will be shadowed.
But even if that weren't the case, the binding for zero would not be visible to the first alternative, because of how Haskell's syntax works. A similar thing happens in the following version:
f = \v1 v2 -> case (v1, v2) of
(x, zero) -> Nothing
(x, y) -> Just $ x / y
where zero = 0
The where clause only applies to the one alternative that it's part of, not to the whole list of alternatives. That code is pretty much the same thing as
f = \v1 v2 -> case (v1, v2) of
(x, zero) -> Nothing
(x, y) -> let zero = 0 in Just $ x / y
If bound identifiers had different semantics than unbound identifiers in a pattern match, that could be quite error prone as binding a new identifier could mess up pattern matches anywhere that identifier is in scope.
For example let's say you're importing some module Foo (unqualified). And now the module Foo is changed to add the binding x = 42 for some reason. Now in your pattern match you'd suddenly be comparing the first argument against 42 rather than binding it to x. That's a pretty hard to find bug.
So to avoid this kind of scenario, identifier patterns have the same semantics regardless of whether they're already bound somewhere.
Because they are very fragile. What does this compute?
f x y z = 2*x + 3*y + z
Would you expect this to be equal to
f x 3 z = 2*x + 9 + z
f _ _ _ = error "non-exhaustive patterns!"
only because there's a y = 3 defined somewhere in the same 1000+ line module?
Also consider this:
import SomeLibrary
f x y z = 2*x + 3*y + z
What if in a future release SomeLibrary defines y? We don't want that to suddenly stop working.
Finally, what if there is no Eq instance for y?
y :: a -> a
y = id
f :: a -> (a -> a) -> a
f x y = y x
f x w = w (w x)
Sure, it is a contrived example, but there's no way the runtime can compare the input function to check whether it is equal to y or not.
To disambiguate this, some new languages like Swift uses two different syntaxes. E.g. (pseudo-code)
switch someValue {
case .a(x) : ... // compare by equality using the outer x
case .b(let x) : ... // redefine x as a new local variable, shadowing the outer one
}
zero is just a variable that occurs inside a pattern, just like y does in the second line. There is no difference between the two. When a variable that occurs inside a pattern, this introduces a new variable. If there was a binding for that variable already, the new variable shadows the old one.
So you cannot use an already bound variable inside a pattern. Instead, you should do something like that:
f x y | y == zero = Nothing
where zero = 0
f x y = Just $ x / y
Notice that I also moved the where clause to bring it in scope for the first line.
Related
So I have a function that returns a function. The returned function is supposed to return 0 for all inputs except x. For x it should return 10.
g x = do
let
f x = 10
f _ = 0
f
But instead the function always returns 10:
(g 3) 4
10
It seems that the parameter x is no the same x that gets used to create the function f. So how can I achieve that?
Your approach seems to be trying to use pattern matching like it's unification - Haskell doesn't perform unification, and pattern matching only works for things like data constructors (eg. Just), not for values.
The name x you're using in the inner definition shadows the name x used in the outer definition - so your inner function is equivalent to:
f x = 10
You should use a different variable name, and make your inner function compare its argument to the outer function's argument:
g x = let f y = if x == y then 10 else 0
in f
Or using currying, which is arguably better style:
g x y = if x == y then 10 else 0
This is semantically equivalent to the above version, and is also semantically equivalent to a function returning a lambda. We can partially apply g to a value to produce a function taking one parameter, eg:
> (g 3) 4
0
> (g "hi") "hi"
10
Pattern-matching "f X" compares argument with X only if X is a literal or data constructor (like "True"). When X is a variable it binds a new variable - in this case another "x" which shadows the outer one. You have to use good old comparison. For example, with guards:
f y | x == y = ...
f _ = ...
Haskell does not allow unification of variables that share the same name.
For example, f x x = ... is a syntax error rather than to say "match this case when the first and the second arguments are equal".
In your case, you have the same conceptual problem but no syntax error because Haskell simply shadows x from g once you try to match it using an argument in f. As matching a variable x without a constructor is unconditional it always succeeds and returns 10.
You want a function like this:
g x = \y -> if x == y then 10 else 0
which uses the (==) operator explicitly.
Reading this answer, I'm getting puzzled by the very first code fragment:
data Pair a = P a a
instance Functor Pair where
fmap f (P x y) = P (f x) (f y)
instance Monad Pair where
return x = P x x
P a b >>= f = P x y
where P x _ = f a
P _ y = f b
What I see is the author redefining a data constructor two times and applying it to undefined variables.
First off, how does the second of the two definitions of P (those two that are found in the where clause of the instance Monad definition) matter if, as I believe, the first one (whichever we put first) always matches?
Second, according to what syntax rules could the expression P x y get evaluated when there are no expressions for x and y in scope, but rather some kind of a redefinition of a data constructor that happens to mention these variables' names?
Interesting to note that, if I instead write like:
P a b >>= f = P x y
where P u _ = f a
P _ v = f b
— substituting u & v for x & y
— I will observe an error: Variable not in scope for each of x & y, even though by all sane intuition renaming a bound variable makes no difference.
The equation
P x y = z
does not define P, only x and y. This is the case in general: only the data and newtype keywords can introduce new constructors. All other equations define only term variables. To disambiguate between the two, constructors always begin with upper case (for the special case of operators, : counts as "upper case punctuation") and variables always begin with lower case.
So, the meaning of
P a b >>= f = P x y
where P x _ = f a
P _ y = f b
reads like this:
Define a new function named (>>=). It is defined when the first argument matches the pattern P a b (binding a and b to the actual values P is applied to in the first argument), and when the second argument matches the pattern f (that is, always, and binding f to the value of that second argument).
Apply f to a. Make sure the result of this matches the pattern P x _ (binding x to the first value P is applied to and throwing away the second).
Apply f to b. Make sure the result of this matches the pattern P _ y (binding y to the second value P is applied to and throwing away the first).
Return the value P x y.
All:
How can I write the following in the "Curry" syntax:
let y = 2 in
let f x = x + y in
let f x = let y = 3 in f y in
f 5
I at first tried something like this:
(y -> (f -> ((f x -> f 5) (y -> f y) 3)) x + y) 2
However this does not seem evaluate properly.
Even better yet would be a Lambda-expression to see binding.
Thanks!
let v = e1 in e2 translates to lambda calculus as (\v.e2)(e1) (where I use the backslash to denote a lambda). So, your example would be
(\y1.(\f1.(\f2.f2 5)(\x2.(\y2.f1(y2))(3)))(\x1.x1+y1))(2)
I used alpha conversion to differentiate between variables that otherwise would have the same name. Observe that the f in the middle has become f1, that is the f in f y in the third line of your example uses the f defined in the second line, not the one which is about to be defined in the third line. In other words, your definition is not recursive; you have used let, not let rec.
Digression: Translating let rec into the lambda calculus requires a fixed point combinator Y (or some like technique). Y is characterized by the property that Y(f) reduces to f(Y(f)). Then, let rec v = e1 in e2 roughly translates to (\v.e2)(Y(\v.e1)).
I am trying to understand lambda functions (i.e. anonymous functions) in Haskell by writing a few simple functions that use them.
In the following example, I am simply trying to take in 3 parameters and add two of the three using an anonymous function and adding the third parameter after that.
I am getting an error saying that I must declare an instance first.
specialAdd x y z = (\x y -> x + y) + z
I appreciate any explanation of why my example is not working and/or any explanation that would help me better understand lambda functions.
specialAdd x y z = (\x y -> x + y) + z
In this example, what you are trying to do is add a function to a number, which is not going to work. Look at (\x y -> x + y) + z: it has the form a + b. In order for such an expression to work, the a part and the b part must be numbers of the same type.
Haskell is a bit of an unusual language, so its error messages are rarely of the form "you can't do that". So what's going on here is that Haskell sees that (\x y -> x + y) is a function, and since in an expression like a + b, b must be the same type as a, it concludes that b must also be a function. Haskell also allows you to define your own rules for adding non-built-in types; so it can't just give you an error saying "you can't add two functions," but instead the error is "you have not defined a rule that allows me to add two functions."
The following would do what you want:
specialAdd x y z = ((\x y -> x + y) x y) + z
Here you are applying the function (\x y -> x + y) to arguments x and y, then adding the result to z.
A good way to practice anonymous function is to use them with high order function as fold or map.
Using map as an entry point,
Basic definition of map,
map f [] = []
map f (x:xs) = f x : f xs
Built up an example,
>>> let list = [0..4]
>>> let f x = x + 1
Applying map we obtain,
>>> map f list
[1,2,3,4,5]
Now, we can omit the declaration of f and replace it using anonymous function,
>>> map (\x->x+1) list
[1,2,3,4,5]
then we deduce, map f list == map (\x->x+1) list, thus
f = \x-> x + 1 --- The same as f x = x + 1, but this is the equivalent lambda notation.
then starting with a simple function we see how to translate it into an anonymous function and then how an anonymous function can be rely to a lambda abstraction.
As an exercise try to translate f x = 2*x.
Now more complex, an anonymous function which take two arguments,
Again an working example,
>>> let add x y = x + y
>>> foldl' add 0 [0..4]
10
Can be rewrite using anonymous function as,
>>> foldl' (\x y -> x + y) 0 [0..4]
Again using equality we deduce that add = \x y -> x + y
Moreover as in hakell all function are function of one argument, and we can partial apply it, we can rewrite our previous anonymous function as, add = \x -> (\y -> x + y).
Then where is the trick ?? Because, I just show the use of anonymous function into high order one, and starting from that, showing how this can be exploited to rewrite function using lambda notation. I mean how can it help you to learn how to write down anonymous function ?
Simply cause I've give you (show you) an existing framework using high order function.
This framework is a huge opportunity to accommodate you with this notation.
Starting from that an infinity range of exercise can be deduce, for example try to do the following.
A - Find the corresponding anonymous function ?
1 - let f (x,y) = x + y in map f [(0,1),(2,3),(-1,1)]
2 - let f x y = x * y in foldl' f 1 [1..5]
B - Rewrite all of them using lambda notation into a declarative form (f = \x-> (\y-> ...)
And so on ....
To summarize,
A function as
(F0) f x1 x2 ... xn = {BODY of f}
can always be rewrite as,
(F1) f = \x1 x2 ... xn -> {BODY of f}
where
(F2) (\x1 x2 ... xn -> {BODY of f})
F2 form are just anonymous function, a pure translation of the function into lambda calculus form. F1 is a declarative lambda notation (because we declare f, as we define it, binding it to the anonymous F2). F0 being the usual notation of Haskeller.
A last note focusing on the fact we can put parenthesis between the argument, this create a closure. Doing that mean that a subset of the function's code can be fully evaluated using a subset of the function's argument, (mean converting to a form where no more free variable occurs), but that's another story.
Here is correct form:
specialAdd a b c = ((\x y -> x + y) a b) + c
Example from Learn You a Haskell...:
zipWith (\a b -> (a * 30 + 3) / b) [5,4,3,2,1] [1,2,3,4,5]
Great explanation:
http://learnyouahaskell.com/higher-order-functions#lambdas
From what I understand Labmbda/Anonymous functions help you declare a function "inline" without the need to give it a name. The "\" (ASCII for the Greek, λ) precedes the variable names for the expression that follows the "->". For example,
(\x y -> x + y)
is an anonymous (lambda) function similar to (+). It takes two parameters of type Num and returns their sum:
Prelude> :type (+)
(+) :: Num a => a -> a -> a
Prelude> :type (\x y -> x + y)
(\x y -> x + y) :: Num a => a -> a -> a
Your example is not working because, as others have pointed out, the right hand side of it is using a lambda function, (\x y -> x + y), as a parameter for the (+) operator, which is defined by default only for parameters of type Num. Some of the beauty of the lambda function can be in its "anonymous" use. Vladimir showed how you can use the lambda function in your declaration by passing it the variables from the left side. A more "anonymous" use could be simply calling it, with variables, without giving the function a name (hence anonymous). For example,
Prelude> (\x y z -> x + y + z) 1 2 3
6
and if you like writing parentheses:
Prelude> (((+).) . (+)) 1 2 3
6
Or in a longer expression (as in your example declaration), e.g.,
Prelude> filter (\x -> length x < 3) [[1],[1,2],[1,2,3]]
[[1],[1,2]]
You are trying to use (+) as something like (Num a) => (a -> a -> a) -> a -> ?? which is not correct.
(+) is defined in the class Num and (a -> a -> a) is not an instance of this class.
What exactly are you trying to achieve ?
I am new to Haskell and I am very confused by Where vs. Let. They both seem to provide a similar purpose. I have read a few comparisons between Where vs. Let but I am having trouble discerning when to use each. Could someone please provide some context or perhaps a few examples that demonstrate when to use one over the other?
Where vs. Let
A where clause can only be defined at the level of a function definition. Usually, that is identical to the scope of let definition. The only difference is when guards are being used. The scope of the where clause extends over all guards. In contrast, the scope of a let expression is only the current function clause and guard, if any.
Haskell Cheat Sheet
The Haskell Wiki is very detailed and provides various cases but it uses hypothetical examples. I find its explanations too brief for a beginner.
Advantages of Let:
f :: State s a
f = State $ \x -> y
where y = ... x ...
Control.Monad.State
will not work, because where refers to
the pattern matching f =, where no x
is in scope. In contrast, if you had
started with let, then you wouldn't
have trouble.
Haskell Wiki on Advantages of Let
f :: State s a
f = State $ \x ->
let y = ... x ...
in y
Advantages of Where:
f x
| cond1 x = a
| cond2 x = g a
| otherwise = f (h x a)
where
a = w x
f x
= let a = w x
in case () of
_ | cond1 x = a
| cond2 x = g a
| otherwise = f (h x a)
Declaration vs. Expression
The Haskell wiki mentions that the Where clause is declarative while the Let expression is expressive. Aside from style how do they perform differently?
Declaration style | Expression-style
--------------------------------------+---------------------------------------------
where clause | let expression
arguments LHS: f x = x*x | Lambda abstraction: f = \x -> x*x
Pattern matching: f [] = 0 | case expression: f xs = case xs of [] -> 0
Guards: f [x] | x>0 = 'a' | if expression: f [x] = if x>0 then 'a' else ...
In the first example why is the Let in scope but Where is not?
Is it possible to apply Where to the first example?
Can some apply this to real examples where the variables represent actual expressions?
Is there a general rule of thumb to follow when to use each?
Update
For those that come by this thread later on I found the best explanation to be found here: "A Gentle Introduction to Haskell".
Let Expressions.
Haskell's let expressions are useful
whenever a nested set of bindings is
required. As a simple example,
consider:
let y = a*b
f x = (x+y)/y
in f c + f d
The set of bindings created by a let
expression is mutually recursive, and
pattern bindings are treated as lazy
patterns (i.e. they carry an implicit
~). The only kind of declarations
permitted are type signatures,
function bindings, and pattern
bindings.
Where Clauses.
Sometimes it is convenient to scope
bindings over several guarded
equations, which requires a where
clause:
f x y | y>z = ...
| y==z = ...
| y<z = ...
where z = x*x
Note that this cannot be done with a let expression, which only scopes over the expression which it encloses. A where clause is only allowed at the top level of a set of equations or case expression. The same properties and constraints on bindings in let expressions apply to those in where clauses. These two forms of nested scope seem very similar, but remember that a let expression is an expression, whereas a where clause is not -- it is part of the syntax of function declarations and case expressions.
1: The problem in the example
f :: State s a
f = State $ \x -> y
where y = ... x ...
is the parameter x. Things in the where clause can refer only to the parameters of the function f (there are none) and things in outer scopes.
2: To use a where in the first example, you can introduce a second named function
that takes the x as a parameter, like this:
f = State f'
f' x = y
where y = ... x ...
or like this:
f = State f'
where
f' x = y
where y = ... x ...
3: Here is a complete example without the ...'s:
module StateExample where
data State a s = State (s -> (a, s))
f1 :: State Int (Int, Int)
f1 = State $ \state#(a, b) ->
let
hypot = a^2 + b^2
result = (hypot, state)
in result
f2 :: State Int (Int, Int)
f2 = State f
where
f state#(a, b) = result
where
hypot = a^2 + b^2
result = (hypot, state)
4: When to use let or where is a matter of taste. I use let to emphasize a computation (by moving it to the front) and where to emphasize the program flow (by moving the computation to the back).
While there is the technical difference with respect to guards that ephemient pointed out, there is also a conceptual difference in whether you want to put the main formula upfront with extra variables defined below (where) or whether you want to define everything upfront and put the formula below (let). Each style has a different emphasis and you see both used in math papers, textbooks, etc. Generally, variables that are sufficiently unintuitive that the formula doesn't make sense without them should be defined above; variables that are intuitive due to context or their names should be defined below. For example, in ephemient's hasVowel example, the meaning of vowels is obvious and so it need not be defined above its usage (disregarding the fact that let wouldn't work due to the guard).
Legal:
main = print (1 + (let i = 10 in 2 * i + 1))
Not legal:
main = print (1 + (2 * i + 1 where i = 10))
Legal:
hasVowel [] = False
hasVowel (x:xs)
| x `elem` vowels = True
| otherwise = False
where vowels = "AEIOUaeiou"
Not legal: (unlike ML)
let vowels = "AEIOUaeiou"
in hasVowel = ...
Sadly, most of the answers here are too technical for a beginner.
LHYFGG has a relevant chapter on it -which you should read if you haven't already, but in essence:
where is just a syntactic construct (not a sugar) that are useful only at function definitions.
let ... in is an expression itself, thus you can use them wherever you can put an expression. Being an expression itself, it cannot be used for binding things for guards.
Lastly, you can use let in list comprehensions too:
calcBmis :: (RealFloat a) => [(a, a)] -> [a]
calcBmis xs = [bmi | (w, h) <- xs, let bmi = w / h ^ 2, bmi >= 25.0]
-- w: width
-- h: height
We include a let inside a list comprehension much like we would a predicate, only it doesn't filter the list, it only binds to names. The names defined in a let inside a list comprehension are visible to the output function (the part before the |) and all predicates and sections that come after of the binding. So we could make our function return only the BMIs of people >= 25:
I found this example from LYHFGG helpful:
ghci> 4 * (let a = 9 in a + 1) + 2
42
let is an expression so you can put a let anywhere(!) where expressions can go.
In other words, in the example above it is not possible to use where to simply replace let (without perhaps using some more verbose case expression combined with where).