Anorm is just locking on executeUpdate - scala

I have a very simple Play 2.1 Scala project. As in, this is the only code in it so far. I have a task which I am running with an Akka.system.scheduler. I have some code to select from the database (currently the standard test H2 instance) and I'm following the documentation example almost exactly.
DB.withConnection { implicit c =>
Logger.info("2")
var x = SQL("insert into x (a, b, c) values ({a, b, c})").on(
'a -> a,
'b -> b,
'c -> c
)
Logger.info("2.5")
x.executeUpdate()
Logger.info("3")
It never gets past 2.5. I haven't got any other database operations happening (except for evolutions).
Help?

Based on your link, shouldn't your SQL statement look like:
var x = SQL("insert into x (a, b, c) values ({a}, {b}, {c})").on(
"a" -> a,
"b" -> b,
"c" -> c
)
In the question the values don't have individual braces: {a, b, c}.

Related

Is there a function that transforms/maps both Either's Left and Right cases taking two transformation functions respectively?

I have not found a function in Scala or Haskell that can transform/map both Either's Left and Right cases taking two transformation functions at the same time, namely a function that is of the type
(A => C, B => D) => Either[C, D]
for Either[A, B] in Scala, or the type
(a -> c, b -> d) -> Either a b -> Either c d
in Haskell. In Scala, it would be equivalent to calling fold like this:
def mapLeftOrRight[A, B, C, D](e: Either[A, B], fa: A => C, fb: B => D): Either[C, D] =
e.fold(a => Left(fa(a)), b => Right(fb(b)))
Or in Haskell, it would be equivalent to calling either like this:
mapLeftOrRight :: (a -> c) -> (b -> d) -> Either a b -> Either c d
mapLeftOrRight fa fb = either (Left . fa) (Right . fb)
Does a function like this exist in the library? If not, I think something like this is quite practical, why do the language designers choose not to put it there?
Don't know about Scala, but Haskell has a search engine for type signatures. It doesn't give results for the one you wrote, but that's just because you take a tuple argument while Haskell functions are by convention curried†. https://hoogle.haskell.org/?hoogle=(a -> c) -> (b -> d) -> Either a b -> Either c d does give matches, the most obvious being:
mapBoth :: (a -> c) -> (b -> d) -> Either a b -> Either c d
...actually, even Google finds that, because the type variables happen to be exactly as you thought. (Hoogle also finds it if you write it (x -> y) -> (p -> q) -> Either x p -> Either y q.)
But actually, as Martijn said, this behaviour for Either is only a special case of a bifunctor, and indeed Hoogle also gives you the more general form, which is defined in the base library:
bimap :: Bifunctor p => (a -> b) -> (c -> d) -> p a c -> p b d
†TBH I'm a bit disappointed that Hoogle doesn't by itself figure out to curry the signature or to swap arguments. Pretty sure it actually used to do that automatically, but at some point they simplified the algorithm because with the huge number of libraries the time it took and number of results got out of hand.
Cats provides Bifunctor, for example
import cats.implicits._
val e: Either[String, Int] = Right(41)
e.bimap(e => s"boom: $e", v => 1 + v)
// res0: Either[String,Int] = Right(42)
The behaviour you are talking about is a bifunctor behaviour, and would commonly be called bimap. In Haskell, a bifunctor for either is available: https://hackage.haskell.org/package/bifunctors-5/docs/Data-Bifunctor.html
Apart from the fold you show, another implementation in scala would be either.map(fb).left.map(fa)
There isn't such a method in the scala stdlib, probably because it wasn't found useful or fundamental enough. I can somewhat relate to that: mapping both sides in one operation instead of mapping each side individually doesn't come across as fundamental or useful enough to warrant inclusion in the scala stdlib to me either. The bifunctor is available in Cats though.
In Haskell, the method exists on Either as mapBoth and BiFunctor is in base.
In Haskell, you can use Control.Arrow.(+++), which works on any ArrowChoice:
(+++) :: (ArrowChoice arr) => arr a b -> arr c d -> arr (Either a c) (Either b d)
infixr 2 +++
Specialised to the function arrow arr ~ (->), that is:
(+++) :: (a -> b) -> (c -> d) -> Either a c -> Either b d
Hoogle won’t find +++ if you search for the type specialised to functions, but you can find generalised operators like this by replacing -> in the signature you want with a type variable: x a c -> x b d -> x (Either a b) (Either c d).
An example of usage:
renderResults
:: FilePath
-> Int
-> Int
-> [Either String Int]
-> [Either String String]
renderResults file line column
= fmap ((prefix ++) +++ show)
where
prefix = concat [file, ":", show line, ":", show column, ": error: "]
renderResults "test" 12 34 [Right 1, Left "beans", Right 2, Left "bears"]
==
[ Right "1"
, Left "test:12:34: error: beans"
, Right "2"
, Left "test:12:34: error: bears"
]
There is also the related operator Control.Arrow.(|||) which does not tag the result with Either:
(|||) :: arr a c -> a b c -> arr (Either a b) c
infixr 2 |||
Specialised to (->):
(|||) :: (a -> c) -> (b -> c) -> Either a b -> c
Example:
assertRights :: [Either String a] -> [a]
assertRights = fmap (error ||| id)
sum $ assertRights [Right 1, Right 2]
==
3
sum $ assertRights [Right 1, Left "oh no"]
==
error "oh no"
(|||) is a generalisation of the either function in the Haskell Prelude for matching on Eithers. It’s used in the desugaring of if and case in arrow proc notation.

Is destructuring assignment guaranteed to be parallel assignment in Swift?

I'm looking at the description of the assignment operator in the Swift language reference. Is it guaranteed that destructuring assignment is made in parallel? As opposed to serial assignment. I don't see that point addressed in the description of the assignment operator.
Just to be clear, is it guaranteed that (a, b) = (b, a) (where a and b are var) is equivalent to foo = a, a = b, b = foo, where foo is a variable which is guaranteed not to exist yet, and not a = b, b = a (which would yield a and b both with the same value). I tried (a, b) = (b, a) and it appears to work in parallel, but that's not the same as a documented description of the behavior.
Tuples are value types, and once constructed they are independent of the values that are part of them.
var a = 2
var b = 3
var t1 = (a, b)
print(t1) // (2, 3)
a = 4
print(t1) // (2, 3)
Thus the tuple (a, b) carries the actual values of the two variables, which is why the (b, a) = (a, b) assignment works without problems. Behind the scenes, the actual assigment is (b, a) = (2, 3) (assuming a and b have the values from the above code snippet).

map3 in scala in Parallelism

def map2[A,B,C] (a: Par[A], b: Par[B]) (f: (A,B) => C) : Par[C] =
(es: ExecutorService) => {
val af = a (es)
val bf = b (es)
UnitFuture (f(af.get, bf.get))
}
def map3[A,B,C,D] (pa :Par[A], pb: Par[B], pc: Par[C]) (f: (A,B,C) => D) :Par[D] =
map2(map2(pa,pb)((a,b)=>(c:C)=>f(a,b,c)),pc)(_(_))
I have map2 and need to produce map3 in terms of map2. I found the solution in GitHub but it is hard to understand. Can anyone put a sight on it and explain map3 and also what this does (())?
On a purely abstract level, map2 means you can run two tasks in parallel, and that is a new task in itself. The implementation provided for map3 is: run in parallel (the task that consist in running in parallel the two first ones) and (the third task).
Now down to the code: first, let's give name to all the objects created (I also extended _ notations for clarity):
def map3[A,B,C,D] (pa :Par[A], pb: Par[B], pc: Par[C]) (f: (A,B,C) => D) :Par[D] = {
def partialCurry(a: A, b: B)(c: C): D = f(a, b, c)
val pc2d: Par[C => D] = map2(pa, pb)((a, b) => partialCurry(a, b))
def applyFunc(func: C => D, c: C): D = func(c)
map2(pc2d, pc)((c2d, c) => applyFunc(c2d, c)
}
Now remember that map2 takes two Par[_], and a function to combine the eventual values, to get a Par[_] of the result.
The first time you use map2 (the inside one), you parallelize the first two tasks, and combine them into a function. Indeed, using f, if you have a value of type A and a value of type B, you just need a value of type C to build one of type D, so this exactly means that partialCurry(a, b) is a function of type C => D (partialCurry itself is of type (A, B) => C => D).
Now you have again two values of type Par[_], so you can again map2 on them, and there is only one natural way to combine them to get the final value.
The previous answer is correct but I found it easier to think about like this:
def map3[A, B, C, D](a: Par[A], b: Par[B], c: Par[C])(f: (A, B, C) => D): Par[D] = {
val f1 = (a: A, b: B) => (c: C) => f(a, b, c)
val f2: Par[C => D] = map2(a, b)(f1)
map2(f2, c)((f3: C => D, c: C) => f3(c))
}
Create a function f1 that is a version of f with the first 2 arguments partially applied, then we can map2 that with a and b to give us a function of type C => D in the Par context (f1).
Finally we can use f2 and c as arguments to map2 then apply f3(C => D) to c to give us a D in the Par context.
Hope this helps someone!

Why does my CoffeeScript program issue 'number is not a function' error?

In the program below the first two logs work fine. I'm not doing anything new in the third and final log but somehow it crashes at runtime. Where is the error in my script? I have looked it over a large number of times and it seems like a fairly trivial modification of the proven working code above it.
sumSq = (n) -> ([0..n].map (i) -> i * i).reduce (a, b) -> a + b
sq = (n) -> n * n
sqSum = ((n) -> ([0..n].reduce (a, b) -> a + b))
console.log(sqSum 5)
console.log(sq(sqSum 5))
newSqSum = sq ((n) -> ([0..n].reduce (a, b) -> a + b))
console.log(newSqSum(5))
This is a function, not a number:
(n) -> ([0..n].reduce (a, b) -> a + b)
so when you say this:
newSqSum = sq ((n) -> ([0..n].reduce (a, b) -> a + b))
you're calling sq with a function as its argument. Then sq will try to multiply that function with itself and the result will be NaN because there is no sensible numeric representation of a function. And finally, your third console.log tries to call that NaN value as function and there's your error message.
Something of the form fn1 fn2, for functions f1 and f2, is not a function composition, it is in fact the same as writing fn1(fn2) and that won't produce a new function unless fn1 is explicitly constructed to return a function. If you want to compose the functions then I think you need to do it by hand:
newSqSum = (n) -> sq ((n) -> ([0..n].reduce (a, b) -> a + b)) n
# Or with less hate for the people maintaining your code:
newSqSum = (n) -> sq sqSum n

Functional equivalent of if (p(f(a), f(b)) a else b

I'm guessing that there must be a better functional way of expressing the following:
def foo(i: Any) : Int
if (foo(a) < foo(b)) a else b
So in this example f == foo and p == _ < _. There's bound to be some masterful cleverness in scalaz for this! I can see that using BooleanW I can write:
p(f(a), f(b)).option(a).getOrElse(b)
But I was sure that I would be able to write some code which only referred to a and b once. If this exists it must be on some combination of Function1W and something else but scalaz is a bit of a mystery to me!
EDIT: I guess what I'm asking here is not "how do I write this?" but "What is the correct name and signature for such a function and does it have anything to do with FP stuff I do not yet understand like Kleisli, Comonad etc?"
Just in case it's not in Scalaz:
def x[T,R](f : T => R)(p : (R,R) => Boolean)(x : T*) =
x reduceLeft ((l, r) => if(p(f(l),f(r))) r else l)
scala> x(Math.pow(_ : Int,2))(_ < _)(-2, 0, 1)
res0: Int = -2
Alternative with some overhead but nicer syntax.
class MappedExpression[T,R](i : (T,T), m : (R,R)) {
def select(p : (R,R) => Boolean ) = if(p(m._1, m._2)) i._1 else i._2
}
class Expression[T](i : (T,T)){
def map[R](f: T => R) = new MappedExpression(i, (f(i._1), f(i._2)))
}
implicit def tupleTo[T](i : (T,T)) = new Expression(i)
scala> ("a", "bc") map (_.length) select (_ < _)
res0: java.lang.String = a
I don't think that Arrows or any other special type of computation can be useful here. Afterall, you're calculating with normal values and you can usually lift a pure computation that into the special type of computation (using arr for arrows or return for monads).
However, one very simple arrow is arr a b is simply a function a -> b. You could then use arrows to split your code into more primitive operations. However, there is probably no reason for doing that and it only makes your code more complicated.
You could for example lift the call to foo so that it is done separately from the comparison. Here is a simiple definition of arrows in F# - it declares *** and >>> arrow combinators and also arr for turning pure functions into arrows:
type Arr<'a, 'b> = Arr of ('a -> 'b)
let arr f = Arr f
let ( *** ) (Arr fa) (Arr fb) = Arr (fun (a, b) -> (fa a, fb b))
let ( >>> ) (Arr fa) (Arr fb) = Arr (fa >> fb)
Now you can write your code like this:
let calcFoo = arr <| fun a -> (a, foo a)
let compareVals = arr <| fun ((a, fa), (b, fb)) -> if fa < fb then a else b
(calcFoo *** calcFoo) >>> compareVals
The *** combinator takes two inputs and runs the first and second specified function on the first, respectively second argument. >>> then composes this arrow with the one that does comparison.
But as I said - there is probably no reason at all for writing this.
Here's the Arrow based solution, implemented with Scalaz. This requires trunk.
You don't get a huge win from using the arrow abstraction with plain old functions, but it is a good way to learn them before moving to Kleisli or Cokleisli arrows.
import scalaz._
import Scalaz._
def mod(n: Int)(x: Int) = x % n
def mod10 = mod(10) _
def first[A, B](pair: (A, B)): A = pair._1
def selectBy[A](p: (A, A))(f: (A, A) => Boolean): A = if (f.tupled(p)) p._1 else p._2
def selectByFirst[A, B](f: (A, A) => Boolean)(p: ((A, B), (A, B))): (A, B) =
selectBy(p)(f comap first) // comap adapts the input to f with function first.
val pair = (7, 16)
// Using the Function1 arrow to apply two functions to a single value, resulting in a Tuple2
((mod10 &&& identity) apply 16) assert_≟ (6, 16)
// Using the Function1 arrow to perform mod10 and identity respectively on the first and second element of a `Tuple2`.
val pairs = ((mod10 &&& identity) product) apply pair
pairs assert_≟ ((7, 7), (6, 16))
// Select the tuple with the smaller value in the first element.
selectByFirst[Int, Int](_ < _)(pairs)._2 assert_≟ 16
// Using the Function1 Arrow Category to compose the calculation of mod10 with the
// selection of desired element.
val calc = ((mod10 &&& identity) product) ⋙ selectByFirst[Int, Int](_ < _)
calc(pair)._2 assert_≟ 16
Well, I looked up Hoogle for a type signature like the one in Thomas Jung's answer, and there is on. This is what I searched for:
(a -> b) -> (b -> b -> Bool) -> a -> a -> a
Where (a -> b) is the equivalent of foo, (b -> b -> Bool) is the equivalent of <. Unfortunately, the signature for on returns something else:
(b -> b -> c) -> (a -> b) -> a -> a -> c
This is almost the same, if you replace c with Bool and a in the two places it appears, respectively.
So, right now, I suspect it doesn't exist. It occured to me that there's a more general type signature, so I tried it as well:
(a -> b) -> ([b] -> b) -> [a] -> a
This one yielded nothing.
EDIT:
Now I don't think I was that far at all. Consider, for instance, this:
Data.List.maximumBy (on compare length) ["abcd", "ab", "abc"]
The function maximumBy signature is (a -> a -> Ordering) -> [a] -> a, which, combined with on, is pretty close to what you originally specified, given that Ordering is has three values -- almost a boolean! :-)
So, say you wrote on in Scala:
def on[A, B, C](f: ((B, B) => C), g: A => B): (A, A) => C = (a: A, b: A) => f(g(a), g(b))
The you could write select like this:
def select[A](p: (A, A) => Boolean)(a: A, b: A) = if (p(a, b)) a else b
And use it like this:
select(on((_: Int) < (_: Int), (_: String).length))("a", "ab")
Which really works better with currying and dot-free notation. :-) But let's try it with implicits:
implicit def toFor[A, B](g: A => B) = new {
def For[C](f: (B, B) => C) = (a1: A, a2: A) => f(g(a1), g(a2))
}
implicit def toSelect[A](t: (A, A)) = new {
def select(p: (A, A) => Boolean) = t match {
case (a, b) => if (p(a, b)) a else b
}
}
Then you can write
("a", "ab") select (((_: String).length) For (_ < _))
Very close. I haven't figured any way to remove the type qualifier from there, though I suspect it is possible. I mean, without going the way of Thomas answer. But maybe that is the way. In fact, I think on (_.length) select (_ < _) reads better than map (_.length) select (_ < _).
This expression can be written very elegantly in Factor programming language - a language where function composition is the way of doing things, and most code is written in point-free manner. The stack semantics and row polymorphism facilitates this style of programming. This is what the solution to your problem will look like in Factor:
# We find the longer of two lists here. The expression returns { 4 5 6 7 8 }
{ 1 2 3 } { 4 5 6 7 8 } [ [ length ] bi# > ] 2keep ?
# We find the shroter of two lists here. The expression returns { 1 2 3 }.
{ 1 2 3 } { 4 5 6 7 8 } [ [ length ] bi# < ] 2keep ?
Of our interest here is the combinator 2keep. It is a "preserving dataflow-combinator", which means that it retains its inputs after the given function is performed on them.
Let's try to translate (sort of) this solution to Scala.
First of all, we define an arity-2 preserving combinator.
scala> def keep2[A, B, C](f: (A, B) => C)(a: A, b: B) = (f(a, b), a, b)
keep2: [A, B, C](f: (A, B) => C)(a: A, b: B)(C, A, B)
And an eagerIf combinator. if being a control structure cannot be used in function composition; hence this construct.
scala> def eagerIf[A](cond: Boolean, x: A, y: A) = if(cond) x else y
eagerIf: [A](cond: Boolean, x: A, y: A)A
Also, the on combinator. Since it clashes with a method with the same name from Scalaz, I'll name it upon instead.
scala> class RichFunction2[A, B, C](f: (A, B) => C) {
| def upon[D](g: D => A)(implicit eq: A =:= B) = (x: D, y: D) => f(g(x), g(y))
| }
defined class RichFunction2
scala> implicit def enrichFunction2[A, B, C](f: (A, B) => C) = new RichFunction2(f)
enrichFunction2: [A, B, C](f: (A, B) => C)RichFunction2[A,B,C]
And now put this machinery to use!
scala> def length: List[Int] => Int = _.length
length: List[Int] => Int
scala> def smaller: (Int, Int) => Boolean = _ < _
smaller: (Int, Int) => Boolean
scala> keep2(smaller upon length)(List(1, 2), List(3, 4, 5)) |> Function.tupled(eagerIf)
res139: List[Int] = List(1, 2)
scala> def greater: (Int, Int) => Boolean = _ > _
greater: (Int, Int) => Boolean
scala> keep2(greater upon length)(List(1, 2), List(3, 4, 5)) |> Function.tupled(eagerIf)
res140: List[Int] = List(3, 4, 5)
This approach does not look particularly elegant in Scala, but at least it shows you one more way of doing things.
There's a nice-ish way of doing this with on and Monad, but Scala is unfortunately very bad at point-free programming. Your question is basically: "can I reduce the number of points in this program?"
Imagine if on and if were differently curried and tupled:
def on2[A,B,C](f: A => B)(g: (B, B) => C): ((A, A)) => C = {
case (a, b) => f.on(g, a, b)
}
def if2[A](b: Boolean): ((A, A)) => A = {
case (p, q) => if (b) p else q
}
Then you could use the reader monad:
on2(f)(_ < _) >>= if2
The Haskell equivalent would be:
on' (<) f >>= if'
where on' f g = uncurry $ on f g
if' x (y,z) = if x then y else z
Or...
flip =<< flip =<< (if' .) . on (<) f
where if' x y z = if x then y else z