What is the equivalent Scala code for this Imperative coding style? - scala

Find below the imperative style code:
private boolean placePieceOnBoard(Results results,
ChessLayout chessLayout,
List<ChessPiece> piecesToPlace,
int pieceIndex,
int startOffset) {
if(pieceIndex == piecesToPlace.size()) {
results.addLayout(chessLayout);
return true;
}
else {
ChessPiece chessPiece = piecesToPlace.get(pieceIndex);
int offset = startOffset;
while(offset < chessLayout.getBoardLength()) {
int placedOffset = chessLayout.placePieceInNextAvailablePosition(chessPiece, offset);
if( placedOffset == ChessLayout.NULL_OFFSET )
break;
else {
logger.debug("piece {} ({}) moved", pieceIndex, chessPiece);
placePieceOnBoard(results, chessLayout.clone(), piecesToPlace, pieceIndex + 1, placedOffset + 1);
chessLayout.removeChessPiece(chessPiece);
offset = placedOffset + 1;
}
}
return false;
}
}
Here I am interested in knowing how to convert changes made to offset within the loop in a functional way, how to do early break from the loop etc.

This is a nice demonstration of one of Scala's great strengths - gradual, easy migration from imperative Java code to functional code.
To convert to Scala, you don't need to go 100% functional straight away. You can easily convert what you have to Scala as follows:
def placePieceOnBoard(results: Results,
chessLayout: ChessLayout,
piecesToPlace: List[ChessPiece],
pieceIndex: Int,
startOffset: Int) {
if (pieceIndex == piecesToPlace.size()) {
results.addLayout(chessLayout)
true
}
else {
val chessPiece = piecesToPlace.get(pieceIndex)
var offset = startOffset
while(offset < chessLayout.getBoardLength) {
val placedOffset = chessLayout.placePieceInNextAvailablePosition(chessPiece, offset)
if( placedOffset == ChessLayout.NULL_OFFSET )
break
else {
logger.debug("piece {} ({}) moved", pieceIndex, chessPiece)
placePieceOnBoard(results, chessLayout.clone(), piecesToPlace, pieceIndex + 1, placedOffset + 1)
chessLayout.removeChessPiece(chessPiece)
offset = placedOffset + 1
}
}
false
}
}
Note we already have some small improvements in verbosity: no more ;s, no redundant type signatures, mutable variables converted to constants where possible, eliminated return statements and function brackets. Note I changed the argument type of piecesToPlace to a scala list, you'll need to use the scala-java converters at the call site to make this compile.
Next, we can start making this more functional... however, this function probably isn't the place you should start, as you are calling state-mutating functions in several places. For example, you could start with ChessLayout:
chessLayout.removeChessPiece(chessPiece)
val placedOffset = chessLayout.placePieceInNextAvailablePosition(chessPiece, offset)
These methods mutate ChessLayout, which is not functional style. You could modify removeChessPiece to return a new ChessLayout with the piece removed, and placePieceInNextAvailablePosition could return a tuple of (ChessLayout, Int). Once you have stopped any functions called from placePieceOnBoard from mutating state, you can then convert it to functional style.

Related

How to return a variable in a function in kotlin

I created a function that recieves input and compare it to a list, when find a match it return the match, in this case this match is the attribute of a class that i created.
I understand that the problem is with the return statement, so in the beginning of the function I declare the return as "Any", further more than that I'm kinda lost.
The error is this: A 'return' expression required in a function with a block body ('{...}')
class Class1(var self: String)
var test_class = Class1("")
fun giver(){
test_class.self = "Anything"
}
class Funciones(){
fun match_finder(texto: String): Any{
var lista = listOf<String>(test_class.self)
var lista_de_listas = listOf<String>("test_class.self")
var count = -1
for (i in lista_de_listas){
count = count + 1
if (texto == i){
lista_de_listas = lista
var variable = lista_de_listas[count]
return variable
}
}
}
}
fun main(){
giver()
var x = "test_class.self"
var funcion = Funciones()
var y = funcion.match_finder(x)
println(y)
}
To explain you what the problem is, let's consider the following code:
class MyClass {
fun doSomething(): String {
val numbers = listOf(1, 2, 3)
for (number in numbers) {
if (number % 2 == 0) {
return "There is at least one even number in the list"
}
}
}
}
If you try compiling it you'll get the same error message as in your question: A 'return' expression required in a function with a block body ('{...}'). Why is that?
Well, we defined a function doSomething returning a String (it could be any other type) but we're returning a result only if the list of numbers contains at least one even number. What should it return if there's no even number? The compiler doesn't know that (how could it know?), so it prompts us that message. We can fix the code by returning a value or by throwing an exception:
class MyClass {
fun doSomething(): String {
val numbers = listOf(1, 2, 3)
for (number in numbers) {
if (number % 2 == 0) {
return "There is at least one even number in the list"
}
}
// return something if the list doesn't contain any even number
return "There is no even number in the list"
}
}
The same logic applies to your original code: what should the function return if there is no i such that texto == i?
Please also note that the solution you proposed may be syntactically correct - meaning it compiles correctly - but will probably do something unexpected. The for loop is useless since the if/else statement will always cause the function to return during the first iteration, so the value "There is no match" could be returned even if a match actually exists later in the list.
I searched online, if someone has the same problem, the correct code is as follows:
class Funciones(){
fun match_finder(texto: String): Any{
var lista = listOf<String>(test_class.self)
var lista_de_listas = listOf<String>("test_class.self")
var count = -1
var variable = " "
for (i in lista_de_listas){
count = count + 1
if (texto == i){
lista_de_listas = lista
var variable = lista_de_listas[count]
return variable
} else {
return "There is no match"
}
}
return variable
}
}

LinkedList in Scala

For exercise I'm trying to implement a LinkedList in Scala.
Main problem is about Null reference.
But first some code:
class Node(xkey: String, xnext: Option[Node], xinfo: Int) {
val key: String = xkey;
var next = xnext.getOrElse(None);
var info: Int = xinfo;
def this(xkey: String, xinfo: Int) {
this(xkey, None, xinfo);
}
def this(xkey: String) {
this(xkey, None, -1);
}
#Override
override def toString: String = key + ":" + info
}
At this point, I'm already concerned about things.
I declare xnext in construct as a Option[Node], because the tail in this linkedList does not have a next.
In my first try, it was just a Node, but had problem with null object because compilator just told me that "null can't cast to Node" (or something like that, I do not remember now) - And so I switch to this Option.
But, is it ok? Because, you know, for me next should be a Node, not a Option, otherwise, I don't know, in the linkedList how to reference to next Node.
Whatever, second class (i.e. my Linked List)
class LinkedNode {
private var first: Option[Node] = None;
private var last: Option[Node] = None;
def addNode(newNode: Node) = {
if (first == null) {
first = Some(newNode);
last = Some(newNode);
first.next = last;
}
else {
last.next = newNode;
newNode.next = null;
last = newNode
}
}
def size(): Long = {
var currentNode : = first;
var size = 0L;
while (currentNode != null) {
size+=1;
currentNode = currentNode.next;
}
size
}
def findNodeByKey(key: String) : Node = {
var currentNode = first;
while(currentNode != null) {
if (currentNode.key.equals(key))
currentNode
else {
currentNode = currentNode.next;
}
}
currentNode;
}
def delNodeByKey(key : String) : Boolean = {
var currentNode = first;
var previousNode = first;
while(currentNode != null) {
if (currentNode.key.equals(key)) {
previousNode = currentNode.next;
return true;
}
previousNode = currentNode;
currentNode = currentNode.next;
}
return false;
}
}
And nothing. I'm already block to my constructor because first and last.
How should I declare them? Node? Or Option[Node]?
Problems are also in Add method.
When I add a node, I want to add a Node object, not an Option[Node].
And I don't get how to achieve things I want with all Option, Some and None classes.
I know I should not be so vague with my request, but any help?
P.S. I've already read this Q/A and it didn't help me
At this point, I'm already concerned about things. I declare xnext in construct as a Option[Node], because the tail in this linkedList does not have a next.
[...]
But, is ok? because, you know, for me next should be a Node, not a Option, otherwise, I don't know, in the linkedList how to reference to next Node.
This is a good solution to replacing null, which you definitely want to do to prevent null-pointer exceptions and the like. An Option[Node] is simply a Node wrapped in a container (or None). You can check whether or not it has a value with isEmpty or get its value with get (which will throw an exception if the Option is empty).
The only difference to null, as you'd use it in Java, is that you need to check if it isEmpty instead of checking for null, and that you need to unwrap (option.get) it explicitly when you're sure that it is not None.
A more paradigmatic (scala-typical) way of retrieving the value from an option is pattern matching:
option match {
case Some(x) => println(x)
case None => println("Whoops, no value :(")
}
Regarding your other questions, they are indeed a little vague.
How should I declere them? Node? or Option[Node]?
Use Option[Node] if the possibility exists that there's no value for the variable (i.e., if you'd set it to null sometimes in Java).
When I add a node, I want to add a Node object, not a Option[Node].
No, you want to add an Option[Node] internally, because you will need to check later on if a node is set or not. In Scala it is preferrable to do this via Option[Node].isEmpty compared to setting things to null. You're already doing this in some parts of your code (e.g., addNode), where you do Some(newNode) (I'd call this "wrapping the node in an Option", but I'm not entirely sure if that's the correct terminology).
And I don't get how to achieve things I want with all Option, Some and None class.
What you're doing in your addNode does seem correct to a degree, but you somehow try to use null again in the else branch. What about:
// We don't need Option[Node] for the parameter, because there
// _must_ be a Node value to be added
def addNode(newNode: Node) = {
if (first.isEmpty) {
first = Some(newNode)
last = Some(newNode)
first.next = last
} else {
newNode.next = None
last.next = Some(newNode)
last = Some(newNode)
}
}
(I didn't run that code, nor did I do an thorough check of your logic)

First steps in Swift, performance issue when allocating small objects in a BST

In trying to learn Swift 2.2, I am facing a serious performance drop when trying to allocate many small objects (fundamentally, a BST of 262144 elements). My current benchmark is a Java 1.8.0_74 compile of an old snip that I wrote some years ago, which, on my 2012 Retina Macbook Pro executes in 59 seconds (59036178 microsecond). The Issue I can observe via Instruments is that I get literally dozens of swift_retain_ and swift_release per iteration. Not sure how to avoid them:
import Foundation
import Darwin;
import Foundation
public class BinarySearchTree<T : Comparable> {
private var _value : T?;
private var _leftTree : BinarySearchTree<T>?;
private var _rightTree : BinarySearchTree<T>?;
public init(value : T) {
_value = value;
}
var value : T? {
get {
return self._value;
}
set {
self._value = newValue;
}
}
var leftTree : BinarySearchTree<T>? {
get {
return self._leftTree;
}
set {
self._leftTree = newValue;
}
}
var rightTree : BinarySearchTree<T>? {
get {
return self._rightTree;
}
set {
self._rightTree = newValue;
}
}
public func add(newValue : T) -> BinarySearchTree<T> {
var navigator : BinarySearchTree<T>?;
var subtree : BinarySearchTree<T>?;
var done : Bool?;
done = false;
navigator = self;
while (!done!) {
if (newValue < navigator?.value) {
subtree = navigator?.leftTree;
if (subtree != nil) {
navigator = subtree;
} else {
let newNode = BinarySearchTree<T>(value: newValue);
navigator!.leftTree = newNode;
done = true;
}
} else if (newValue > navigator?.value) {
subtree = navigator?.rightTree;
if (subtree != nil) {
navigator = subtree;
} else {
let newNode = BinarySearchTree<T>(value: newValue);
navigator?.rightTree = newNode;
done = true;
}
} else {
done = true;
}
}
return self;
}
} /* cut remove/search methods */
And this is the test code I wrote for the test run
let count : Int32 = 262144;
let base : Int32 = 65536;
let target : Int32 = count + 1;
var info = mach_timebase_info(numer:0, denom:0);
var timebase = mach_timebase_info(&info);
let numer = UInt64(info.numer);
let denom = UInt64(info.denom);
let norm = UInt64(numer/denom);
let check1 = (mach_absolute_time() * norm);
var root = BinarySearchTree<Int32>(value:base);
for var loop in 0 ... count-1 {
if (loop % 1000 == 0) {
print(loop);
}
root = root.add(loop);
}
let check2 = (mach_absolute_time() * norm);
print("Creation phase microseconds: [" + String((check2 - check1) / 1000) + "]");
I tried searching for the specific swift release/retain issue with no luck, I am not sure how to proceed. Thanks everyone
The issue, as you note is the retain/release (though it isn't really, retain/release is insignificant next to the power of ummm... we'll get there at the end). This isn't really related to allocations. You're not allocating extra objects, you're just retaining them briefly and then releasing them. I'll start with Kenneth's code, which optimizes out a lot of performance problems in the original, but still has this issue. (I'm not considering the recursive code because it crashes in your current use case. It does dodge some redundant retains, though.)
It is worth saying that Kenneth's code is good and is generally the way you should do things (as you'll see even more as we go along).
First note: when you mentioned -Ofast, that's for ObjC, not Swift. The flag for Swift is just -O. You also want -whole-module-optimization, but that doesn't really help anything here.
One more little thing, then we're get to it. Mark classes final any time you can. This makes sure there's no dynamic dispatch. This doesn't matter very much here compared the retain/release, but hey, take the easy stuff.
Does 30% off sound good?
OK, now a big one, and it's a trick. I find that I can knock out about 30% of the time (from ~6min to ~4min for the full import) by rewriting this:
guard let subtree = navigator.leftTree else {
navigator.leftTree = BinarySearchTree<T>(value: newValue)
break
}
navigator = subtree
continue
As this:
let subtree = navigator.leftTree
if subtree == nil {
navigator.leftTree = BinarySearchTree(value: newValue)
break
}
navigator = subtree!
continue
This is a something to be very careful with. It turns out to be faster in this case, but that may not be as fast across other inputs. That may not be as fast across changes to the optimizer (the SIL generation is a bit weird, and I suspect may actually be a mistake because it seems to double-retain navigator in the second case, but only after the if has succeeded). But it does seem to currently be faster. (EDIT: The Swift team was surprised by this finding, and there is now a bug opened against it. Do not expect this to work in the future.)
How about 85% How's that sound?
But like you said, couldn't we avoid all this with structs? But it'd be insanely expensive to copy the whole tree every time we touch it. Of course we could dramatically improve that with copy-on-write like Array uses. But COW is pretty complicated. If only there were a way to reuse the existing stuff. What if we use Array?
private struct Node<Element: Comparable> {
let value: Element
var leftIndex = -1 // Ugly, but ~25% faster than using Int? in my tests
var rightIndex = -1
init(_ value: Element) { self.value = value }
}
// This works exactly the same if you make it a `final class`. Your choice.
public struct BinarySearchTree<Element: Comparable> {
private var storage: [Node<Element>] = []
init(value: Element) { storage.append(Node(value)) }
public mutating func add(newValue: Element) {
if storage.isEmpty {
storage.append(Node(newValue))
}
var index = 0
while (true) {
let node = storage[index]
if (newValue < node.value) {
if node.leftIndex < 0 {
storage.append(Node(newValue))
storage[index].leftIndex = storage.count - 1 // Don't use node here; remember value types!
break
}
index = node.leftIndex
continue
} else if (newValue > node.value) {
if node.rightIndex < 0 {
storage.append(Node(newValue))
storage[index].rightIndex = storage.count - 1
break
}
index = node.rightIndex
continue
} else {
break
}
}
}
}
This takes ~45s to run on my system. Of course this makes delete a bit more complicated. You'd either have to accept "leaked" memory (possibly with periodic repacking), or you'd need to maintain a freelist. But a freelist wouldn't be too hard to add.
Let's try 99.97% improvement with no changes to add().
And of course it's important to remember that this is a nearly pathological case for BST. Even if you were often handed data in order, you'd be better off applying a shuffle prior to inserting it, even including the cost of the shuffle. For example, using shuffleInPlace (and counting its time), inserting exactly the same values:
var values = Array(0 ... count - 1)
values.shuffleInPlace()
for (loop, value) in values.enumerate() {
if (loop % 1000 == 0) {
print(loop)
}
root.add(value)
}
This takes us from 45s to about 0.1s. (Kenneth's version and my "!" version are about about 0.2s under this metric; I'd probably use Kenneth's solution, with final added. Even your original code, which has a lot of inefficiencies that Kenneth fixed, only takes 0.5s with this change. Remember, the Kenneth-optimized version with in-order add was 6 minutes on my system.)
It's worth it to shuffle before inserting. If you get things over time, it could be worth it to batch them up and shuffle them before inserting. If the tree changes over time, it'd be worth checking if it's gotten too deep and rebuilding it periodically. Keeping the tree depth reasonable overwhelms every other optimization. Clever ways of working around Swift memory management couldn't touch this one change.
Fix the algorithm. Everything else is peanuts in comparison.
I simplified your code, removing some Optionals and your getter/setters because they were unnecessary and could contribute to slow code.
I profiled both your code and mine and got this result on the same data set of random elements:
1000 elements:
Yours: Creation phase microseconds: [28680771]
Mine : Creation phase microseconds: [8564279]
10000 elements:
Yours: Creation phase microseconds: [426233689]
Mine : Creation phase microseconds: [126725800]
Here is my code:
public class BinarySearchTree2<T : Comparable> {
public init(value : T) {
self.value = value
}
var value : T
var leftTree : BinarySearchTree2<T>?
var rightTree : BinarySearchTree2<T>?
public func add(newValue : T) -> BinarySearchTree2<T> {
var navigator = self
while (true) {
if (newValue < navigator.value) {
guard let subtree = navigator.leftTree else {
navigator.leftTree = BinarySearchTree2<T>(value: newValue)
break
}
navigator = subtree
continue
}
if (newValue > navigator.value) {
guard let subtree = navigator.rightTree else {
navigator.rightTree = BinarySearchTree2<T>(value: newValue)
break
}
navigator = subtree
continue
}
break
}
return self
}
} /* cut remove/search methods */
Edit:
I also did a more optimum balanced tree test where I created a data set of 1001 sequential elements, removed the middle element, used a Fisher-Yates shuffle to randomize the order, initialized the root with the middle element, and ran both sets. Here are my results:
Yours: Creation phase microseconds: [27648219]
Mine: Creation phase microseconds: [8332361]
Edit 2:
I switched the add() method to use recursion with significant gains in speed:
Before (my original code): Creation phase microseconds: [8088804]
After : Creation phase microseconds: [1179398]
Here is the new code:
public class BinarySearchTree3<T : Comparable> {
public init(value : T) {
self.value = value
}
let value : T
var leftTree : BinarySearchTree3<T>?
var rightTree : BinarySearchTree3<T>?
public func add(newValue : T) {
if (newValue < self.value) {
if self.leftTree?.add(newValue) == nil {
self.leftTree = BinarySearchTree3<T>(value: newValue)
}
return
}
if (newValue > self.value) {
if self.rightTree?.add(newValue) == nil {
self.rightTree = BinarySearchTree3<T>(value: newValue)
}
return
}
}
} /* cut remove/search methods */

Variadic opIndex Override

Following my previous question, I'm now having trouble overriding opIndex with variadic parameters. I've tried multiple methods (even hack-ish ones) but to no avail.
The code I'm using to generate the identifier string
static string array_decl(D...)(string identifier, D dimensions)
{
static if(dimensions.length == 0)
{
return identifier;
}
else
{
return array_decl(identifier ~ "[" ~ to!(string)(dimensions[0]) ~ "]", dimensions[1..$]);
}
}
What my opIndex override looks like:
T opIndex(D...)(D indices)
{
mixin("return " ~ array_decl("Data", indices) ~ ";");
}
Fails with:
./inheritance.d(81): Error: tuple D is used as a type
./inheritance.d(89): Error: template instance inheritance.array_ident!(int, int, int).array_ident.array_ident!(_param_2, _param_3) error instantiating
./inheritance.d(112): instantiated from here: array_ident!(int, int, int)
./inheritance.d(174): instantiated from here: opIndex!(int, int, int)
./inheritance.d(112): Error: CTFE failed because of previous errors in array_ident
./inheritance.d(112): Error: argument to mixin must be a string, not ("return " ~ array_ident("Data", _param_0, _param_1, _param_2) ~ ";") of type string
The question is how (or is it possible) to implement the opIndex operator for this situation.
I think mixins are the way to go since I only have to generate a string with the format of:
type[index0][index1]...[indexN] Data
for the opIndex overload.
Apparently this is not possible as the tuple passed to opIndex is inaccessible at compile-time. A few solutions I came up with (by Adam D. Ruppe's suggestions):
1. Hard-coding Array Access
Using compile-time conditions to index the array, a bit ugly and the amount of dimensions that can be accessed depends on the amount of conditions implemented.
T opIndex(D...)(D indices)
{
static if(indices.length == 1)
{
return Data[indices[0]];
}
static if(indices.length == 2)
{
return Data[indices[0]][indices[1]];
}
static if(indices.length == 3)
{
return Data[indices[0]][indices[1]][indices[2]];
}
static if(indices.length == 4)
{
return Data[indices[0]][indices[1]][indices[2]][indices[3]];
}
}
2. Pointer(s)
The only other method was to cast the array to a pointer then use offsets. The offset is computed and is then used to index the pointer.
To be able to access the template parameters at run-time:
struct Vector_MultiDim(T, D...)
{
enum dimensions = [D];
static const size_t DimCount = D.length;
... Other members here
}
Function to compute offset (the size of each dimension must be known at run-time):
size_t GetIndex(size_t[] indices)
{
size_t index;
for(size_t i = 0; i < DimCount; i++)
{
size_t factor = 1;
for(size_t j = i + 1; j < DimCount; j++)
{
factor *= dimensions[j];
}
index += indices[i] * factor;
}
return index;
}
opIndex Override:
T opIndex(D...)(D indices)
{
T* arr = cast(T*)Data;
return arr[GetIndex([indices])];
}

Parsing text and representing it with tokens using Scala

I'm getting frustrated trying to convert a small part of the Golang templating language to Scala.
Below are the key parts of the lex.go source code: https://github.com/golang/go/blob/master/src/text/template/parse/lex.go
The tests are here: https://github.com/golang/go/blob/master/src/text/template/parse/lex_test.go
Basically this "class" takes a string and returns an Array of "itemType". In the template string, the start and end of special tokens is using curly braces {{ and }}.
For for example:
"{{for}}"
returns an array of 4 items:
item{itemLeftDelim, 0, "{{" } // scala case class would be Item(ItemLeftDelim, 0, "")
item{itemIdentifier, 0, "for"}
item{itemRightDelim, 0, "}}"}
item{itemEOF, 0, ""}
The actual call would look like:
l := lex("for", `{{for}}`, "{{", "}}") // you pass in the start and end delimeters {{ and }}
for {
item := l.nextItem()
items = append(items, item)
if item.typ == itemEOF || item.typ == itemError {
break
}
}
return
The key parts of the source code are below:
// itemType identifies the type of lex items.
type itemType int
const (
itemError itemType = iota // error occurred; value is text of error
itemEOF
itemLeftDelim // left action delimiter
// .............. skipped
)
const (
leftDelim = "{{"
rightDelim = "}}"
leftComment = "/*"
rightComment = "*/"
)
// item represents a token or text string returned from the scanner.
type item struct {
typ itemType // The type of this item.
pos Pos // The starting position, in bytes, of this item in the input string.
val string // The value of this item.
}
// stateFn represents the state of the scanner as a function that returns the next state.
type stateFn func(*lexer) stateFn
// lexer holds the state of the scanner.
type lexer struct {
name string // the name of the input; used only for error reports
input string // the string being scanned
leftDelim string // start of action
rightDelim string // end of action
state stateFn // the next lexing function to enter
pos Pos // current position in the input
start Pos // start position of this item
width Pos // width of last rune read from input
lastPos Pos // position of most recent item returned by nextItem
items chan item // channel of scanned items
parenDepth int // nesting depth of ( ) exprs
}
// lex creates a new scanner for the input string.
func lex(name, input, left, right string) *lexer {
if left == "" {
left = leftDelim
}
if right == "" {
right = rightDelim
}
l := &lexer{
name: name,
input: input,
leftDelim: left,
rightDelim: right,
items: make(chan item),
}
go l.run()
return l
}
// run runs the state machine for the lexer.
func (l *lexer) run() {
for l.state = lexText; l.state != nil; {
l.state = l.state(l)
}
}
// nextItem returns the next item from the input.
func (l *lexer) nextItem() item {
item := <-l.items
l.lastPos = item.pos
return item
}
// emit passes an item back to the client.
func (l *lexer) emit(t itemType) {
l.items <- item{t, l.start, l.input[l.start:l.pos]}
l.start = l.pos
}
// lexText scans until an opening action delimiter, "{{".
func lexText(l *lexer) stateFn {
for {
if strings.HasPrefix(l.input[l.pos:], l.leftDelim) {
if l.pos > l.start {
l.emit(itemText)
}
return lexLeftDelim
}
if l.next() == eof {
break
}
}
// Correctly reached EOF.
if l.pos > l.start {
l.emit(itemText)
}
l.emit(itemEOF)
return nil
}
// next returns the next rune in the input.
func (l *lexer) next() rune {
if int(l.pos) >= len(l.input) {
l.width = 0
return eof
}
r, w := utf8.DecodeRuneInString(l.input[l.pos:])
l.width = Pos(w)
l.pos += l.width
return r
}
// lexLeftDelim scans the left delimiter, which is known to be present.
func lexLeftDelim(l *lexer) stateFn {
l.pos += Pos(len(l.leftDelim))
if strings.HasPrefix(l.input[l.pos:], leftComment) {
return lexComment
}
l.emit(itemLeftDelim)
l.parenDepth = 0
return lexInsideAction
}
// lexRightDelim scans the right delimiter, which is known to be present.
func lexRightDelim(l *lexer) stateFn {
l.pos += Pos(len(l.rightDelim))
l.emit(itemRightDelim)
return lexText
}
// there are more stateFn
So I was able to write the item and itemType:
case class Item(typ: ItemType, pos: Int, v: String)
sealed trait ItemType
case object ItemError extends ItemType
case object ItemEOF extends ItemType
case object ItemLeftDelim extends ItemType
...
..
.
The stateFn and Lex definitions:
trait StateFn extends (Lexer => StateFn) {
}
I'm basically really stuck on the main parts here. So things seem to be kicked of like this:
A Lex is created, then "go l.run()" is called.
Run is a loop, which keeps looping until EOF or an error is found.
The loop initializes with lexText, which scans until it finds an {{, and then it sends a message to a channel with all the preceding text of type 'itemText', passing it an 'item'. It then returns the function lexLeftDelim. lexLeftDelim does the same sort of thing, it sends a message 'item' of type itemLeftDelim.
It keeps parsing the string until it reaches EOF basically.
I can't think in scala that well, but I know I can use an Actor here to pass it a message 'Item'.
The part of returning a function, I asked I got some good ideas here: How to model recursive function types?
Even after this, I am really frustrated and I can seem to glue these concepts together.
I'm not looking for someone to implement the entire thing for me, but if someone could write just enough code to parse a simple string like "{{}}" that would be awesome. And if they could explain why they did a certain design that would be great.
I created a case class for Lex:
case class Lex(
name: String,
input: String,
leftDelim: String,
rightDelim: String,
state: StateFn,
var pos: Int = 0,
var start: Int = 0,
var width: Int = 0,
var lastPos: Int = 0,
var parenDepth: Int = 0
) {
def next(): Option[String] = {
if (this.pos >= this.input.length) {
this.width = 0
return None
}
this.width = 1
val nextChar = this.input.drop(this.pos).take(1)
this.pos += 1
Some(nextChar)
}
}
The first stateFn is LexText and so far I have:
object LexText extends StateFn {
def apply(l: Lexer) = {
while {
if (l.input.startsWith(l.leftDelim)) {
if (l.pos > l.start) {
// ????????? emit itemText using an actor?
}
return LexLeftDelim
}
if (l.next() == None) {
break
}
}
if(l.pos > l.start) {
// emit itemText
}
// emit EOF
return None // ?? nil? how can I support an Option[StateFn]
}
}
I need guidance on getting the Actor's setup, along with the main run loop:
func (l *lexer) run() {
for l.state = lexText; l.state != nil; {
l.state = l.state(l)
}
}
This is an interesting problem domain that I tried to tackle using Scala, and so far I am a bit confused hoping some else finds it interesting and can work with what little I have so far and provide some code and critique if I am doing it correctly or not.
I know deep down I shouldn't be mutating, but I'm still on the first few pages of the functional book :)
If you translate the go code literally into Scala, you'll get very unidiomatic piece of code. You'll probably get much more maintainable (and shorter!) Scala version by using parser combinators. There are plenty of resources about them on the internet.
import scala.util.parsing.combinator._
sealed trait ItemType
case object LeftDelim extends ItemType
case object RightDelim extends ItemType
case object Identifier extends ItemType
case class Item(ty: ItemType, token: String)
object ItemParser extends RegexParsers {
def left: Parser[Item] = """\{\{""".r ^^ { _ => Item(LeftDelim, "{{") }
def right: Parser[Item] = """\}\}""".r ^^ { _ => Item(RightDelim, "}}") }
def ident: Parser[Item] = """[a-z]+""".r ^^ { x => Item(Identifier, x) }
def item: Parser[Item] = left | right | ident
def items: Parser[List[Item]] = rep(item)
}
// ItemParser.parse(ItemParser.items, "{{foo}}")
// res5: ItemParser.ParseResult[List[Item]] =
// [1.8] parsed: List(Item(LeftDelim,{{), Item(Identifier,foo), Item(RightDelim,}}))
Adding whitespace skipping, or configurable left and right delimiters is trivial.