Swift 2.0 has new feature called indirectly recursive enum. Can someone explain what it is?
From Swift Docs
A recursive enumeration is an enumeration that has another instance of the enumeration as the associated value for one or more of the enumeration cases.
The example given highlights a simplified use case:
indirect enum ArithmeticExpression {
case Number(Int)
case Addition(ArithmeticExpression, ArithmeticExpression)
case Multiplication(ArithmeticExpression, ArithmeticExpression)
}
func evaluate(expression: ArithmeticExpression) -> Int {
switch expression {
case .Number(let value):
return value
case .Addition(let left, let right):
return evaluate(left) + evaluate(right)
case .Multiplication(let left, let right):
return evaluate(left) * evaluate(right)
}
}
// evaluate (5 + 4) * 2
let five = ArithmeticExpression.Number(5)
let four = ArithmeticExpression.Number(4)
let sum = ArithmeticExpression.Addition(five, four)
let product = ArithmeticExpression.Multiplication(sum, ArithmeticExpression.Number(2))
print(evaluate(product))
// prints "18"
Related
So I got a question that was delivered as a 2D List
val SPE = listOf(
listOf('w', 'x'),
listOf('x', 'y'),
listOf('z', 'y'),
listOf('z', 'v'),
listOf('w', 'v')
)
It asks to find the shortest path between w and z. So obviously, BFS would be the best course of action here to find that path the fastest. Here's my code for it
fun shortestPath(edges: List<List<Char>>, root: Char, destination: Char): Int {
val graph = buildGraph3(edges)
val visited = hashSetOf(root)
val queue = mutableListOf(mutableListOf(root, 0))
while (queue.size > 0){
val node = queue[0].removeFirst()
val distance = queue[0].removeAt(1)
if (node == destination) return distance as Int
graph[node]!!.forEach{
if (!visited.contains(it)){
visited.add(it)
queue.add(mutableListOf(it, distance + 1))
}
}
}
queue.sortedByDescending { it.size }
return queue[0][1]
}
fun buildGraph3(edges: List<List<Char>>): HashMap<Char, MutableList<Char>> {
val graph = HashMap<Char, MutableList<Char>>()
for (i in edges.indices){
for (n in 0 until edges[i].size){
var a = edges[i][0]
var b = edges[i][1]
if (!graph.containsKey(a)) { graph[a] = mutableListOf() }
if (!graph.containsKey(b)) { graph[b] = mutableListOf() }
graph[a]?.add(b)
graph[b]?.add(b)
}
}
return graph
}
I am stuck on the return part. I wanted to use a list to keep track of the incrementation of the char, but it wont let me return the number. I could have done this wrong, so any help is appreciated. Thanks.
If I paste your code into an editor I get this warning on your return queue[0][1] statement:
Type mismatch: inferred type is {Comparable<*> & java.io.Serializable} but Int was expected
The problem here is queue contains lists that hold Chars and Int distances, mixed together. You haven't specified the type that list holds, so Kotlin has to infer it from the types of the things you've put in the list. The most general type that covers both is Any?, but the compiler tries to be as specific as it can, inferring the most specific type that covers both Char and Int.
In this case, that's Comparable<*> & java.io.Serializable. So when you pull an item out with queue[0][1], the value you get is a Comparable<*> & java.io.Serializable, not an Int, which is what your function is supposed to be returning.
You can "fix" this by casting - since you know how your list is meant to be organised, two elements with a Char then an Int, you can provide that information to the compiler, since it has no idea what you're doing beyond what it can infer:
val node = queue[0].removeFirst() as Char
val distance = queue[0].removeAt(1) as Int
...
return queue[0][1] as Int
But ideally you'd be using the type system to create some structure around your data, so the compiler knows exactly what everything is. The most simple, generic one of these is a Pair (or a Triple if you need 3 elements):
val queue = mutableListOf(Pair<Char, Int>(root, 0))
// or if you don't want to explicitly specify the type
val queue = mutableListOf(root to 0)
Now the type system knows that the items in your queue are Pairs where the first element is a Char, and the second is an Int. No need to cast anything, and it will be able to help you as you try to work with that data, and tell you if you're doing the wrong thing.
It might be better to make actual classes that reflect your data, e.g.
data class Step(node: Char, distance: Int)
because a Pair is pretty general, but it's up to you. You can pull the data out of it like this:
val node = queue[0].first
val distance = queue[0].second
// or use destructuring to assign the components to multiple variables at once
val (node, distance) = queue[0]
If you make those changes, you'll have to rework some of your algorithm - but you'll have to do that anyway, it's broken in a few ways. I'll just give you some pointers:
your return queue[0][1] line can only be reached when queue is empty
queue[0].removeAt(1) is happening on a list that now has 1 element (i.e. at index 0)
don't you need to remove items from your queue instead?
when building your graph, you call add(b) twice
try printing your graph, the queue at each stage in the loop etc to see what's happening! Make sure it's doing what you expect. Comment out any code that doesn't work so you can make sure the stuff that runs before that is working.
Good luck with it! Hopefully once you get your types sorted out things will start to fall into place more easily
I am having trouble trying to understand pattern matching rules in Rust. I originally thought that the idea behind patterns are to match the left-hand side and right-hand side like so:
struct S {
x: i32,
y: (i32, i32)
}
let S { x: a, y: (b, c) } = S { x: 1, y: (2, 3) };
// `a` matches `1`, `(b, c)` matches `(2, 3)`
However, when we want to bind a reference to a value on the right-hand side, we need to use the ref keyword.
let &(ref a, ref b) = &(3, 4);
This feels rather inconsistent.
Why can't we use the dereferencing operator * to match the left-hand side and right-hand side like this?
let &(*a, *b) = &(3, 4);
// `*a` matches `3`, `*b` matches `4`
Why isn't this the way patterns work in Rust? Is there a reason why this isn't the case, or have I totally misunderstood something?
Using the dereferencing operator would be very confusing in this case. ref effectively takes a reference to the value. These are more-or-less equivalent:
let bar1 = &42;
let ref bar2 = 42;
Note that in let &(ref a, ref b) = &(3, 4), a and b both have the type &i32 — they are references. Also note that since match ergonomics, let (a, b) = &(3, 4) is the same and shorter.
Furthermore, the ampersand (&) and asterisk (*) symbols are used for types. As you mention, pattern matching wants to "line up" the value with the pattern. The ampersand is already used to match and remove one layer of references in patterns:
let foo: &i32 = &42;
match foo {
&v => println!("{}", v),
}
By analogy, it's possible that some variant of this syntax might be supported in the future for raw pointers:
let foo: *const i32 = std::ptr::null();
match foo {
*v => println!("{}", v),
}
Since both ampersand and asterisk could be used to remove one layer of reference/pointer, they cannot be used to add one layer. Thus some new keyword was needed and ref was chosen.
See also:
Meaning of '&variable' in arguments/patterns
What is the syntax to match on a reference to an enum?
How can the ref keyword be avoided when pattern matching in a function taking &self or &mut self?
How does Rust pattern matching determine if the bound variable will be a reference or a value?
Why does pattern matching on &Option<T> yield something of type Some(&T)?
In this specific case, you can achieve the same with neither ref nor asterisk:
fn main() {
let (a, b) = &(3, 4);
show_type_name(a);
show_type_name(b);
}
fn show_type_name<T>(_: T) {
println!("{}", std::any::type_name::<T>()); // rust 1.38.0 and above
}
It shows both a and b to be of type &i32. This ergonomics feature is called binding modes.
But it still doesn't answer the question of why ref pattern in the first place. I don't think there is a definite answer to that. The syntax simply settled on what it is now regarding identifier patterns.
This question already has an answer here:
Compare enums only by variant, not value
(1 answer)
Closed 5 years ago.
I have an enum with some nested values. I want to check that this enum is of given variant but without specifying what's inside. Check the following program:
enum Test {
Zero,
One(u8),
Two(u16),
Four(u32),
}
fn check(x: Test, y: Test) -> bool {
x == y;
}
fn main() {
let x = Test::Two(10);
let b1 = check(x, Test::One);
let b2 = check(x, Test::Two);
let b3 = match x {
Test::Four(_) => true,
_ => false,
}
}
b3 checks that x is Test::Four with an arbitrary value inside. I want that check to be done in the function check. Current code does not compile and I can't figure out how can I extract only enum variant without corresponding inside values.
I guess that could done with macro transforming to match expression, but is it possible to do that without macro?
I can see that Test::One is fn(u16) -> Test {Two}. Can I use that fact? To test that x was created using that function.
This is not supported (yet). There is the active RFC 639 which suggests implementing a function that returns an integer which corresponds to the enum discriminant. With that hypothetical function you could expect the following to work:
assert_eq!(Test::Two(10).discriminant(), Test::Two(42).discriminant());
In the swift book by Apple, there is an enum example. It lets you convert a raw Int to a enum rank. However when I try to remove the if statement, the code gives me
Playground execution failed: error: :30:13: error: 'Rank?' does not have a member named 'simpleDesciption'
enum Rank: Int {
case Ace = 1
case Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten
case Jack, Queen, King
func simpleDesciption() -> String {
switch self {
case .Ace:
return "ace"
case .Jack:
return "jack"
case .Queen:
return "queen"
case .King:
return "king"
default:
return String(self.toRaw())
}
}
}
if let convertedRank = Rank.fromRaw(1){
let threeDescription = convertedRank.simpleDesciption()
}
// Why does it need to be wrapped in a if statement?
let convertedRank = Rank.fromRaw(1)
let threeDescription = convertedRank.simpleDesciption()
The if let construct allows to unwrap an optional type. The Rank.fromRaw returns the Rank? optional type, which means it could either be nil or an actual value of the type.
There are two ways to unwrap the value of an optional type in Swift, one of which is the if let construct. The body of the if let is executed if the optional type is not nil, and the variable that follows if let is bound to its value (and thus has type Rank, not Rank?).
Note that if you simply write let convertedRank = Rank.fromRaw(1) without the if, convertedRank has the type Rank? and, again, cannot be directly used as a Rank without being unwrapped.
If you are sure that the value will never be nil, you can force-unwrap it by suffixing the variable name with !:
let convertedRank = Rank.fromRaw(1)
convertedRank!.simpleDescription()
But this is not recommended style, as it will fail at runtime if the value is not defined.
In my imperative-style Scala code, I have an algorithm:
def myProcessor(val items: List) {
var numProcessed = 0
while(numProcessed < items.size) {
val processedSoFar = items.size - numProcessed
numProcessed += processNextBlockOfItems(items, processedSoFar)
}
}
I would like to keep the "block processing" functionality, and not just do a "takeWhile" on the items list. How can I rewrite this in functional style?
You need to change it to a recursive style wherein you "pass" in the "state" of each loop
#tailrec
def myProcessor(items: List[A], count: Int = 0): Int = items match{
case Nil => count
case x :: xs =>
processNextBlockOfItems(items, count)
myProcessor(xs, count + 1)
}
assuming that "processedSoFar" is not an index. If you can work with the current "head" of the list:
#tailrec
def myProcessor(items: List[A], count: Int = 0): Int = items match{
case Nil => count
case x :: xs =>
process(x)
myProcessor(xs, count + 1)
}
where process would only process the current "head" of the List.
So, this depends on what you consider to be more functional, but here's a version without the 'var'
def myProcessorFunctional(items: List[Int]) {
def myProcessorHelper(items: List[Int], numProcessed: Int) {
if (numProcessed < items.size) {
val processedSoFar = items.size - numProcessed
myProcessorHelper(items,
numProcessed + processNextBlockOfItems(items, processedSoFar))
}
}
myProcessorHelper(items, 0)
}
(making it a list of Ints just for simplicity, it would be easy to make it work with a generic List)
I have to say it's one of those cases where I don't mind the mutable variable - it's clear, no reference to it escapes the method.
But as I said in a comment above, processNextBlockOfItems is inherently non-functional anyway, since it's called for its side effects. A more functional way would be for it to return the state of its processing so far, and this state would be updated (and returned) on a subsequent call. Right now, if you in the middle of processing two different items lists, you'd have the issue of maintaining two different partially-processed states within processNextBlockOfItems...
Later:
Still ignoring the state issue, one convenient change would be if processNextBlockOfItems always processed the first block of the items list passed to it, returned the remaining items it had not processed (this is convenient and efficient if using List, so much so I'm wondering why you're using indicies).
This would yield something like:
def myProcessorMoreFunctional(items: List[Int]) {
if (!items.isEmpty) {
myProcessorMoreFunctional(processNextBlockOfItems(items))
}
}