I have a method which swaps the value of two variables.
void switchValue(int* a , int* b){
//logic here...
}
C++ works on a lower level than Kotlin but can I do this on Kotlin?
It's impossible to do this with a function in Kotlin or Java, because references can only be passed by value. (Unless you're satisfied with using a wrapper class and you swap what two wrapper instances are referencing, but this would be clumsy.)
This is probably the easiest way to swap two variables' values:
var x = 0
var y = 1
//swap:
x = y.also { y = x }
Edit: A bit of correction. You can sort of do it with properties using reflection. This wouldn't work on local variables, though.
fun <T> swap(first: KMutableProperty0<T>, second: KMutableProperty0<T>) {
first.set(second.get().also { second.set(first.get()) })
}
swap(::x, ::y)
In my opinion, this kind of practice (remotely changing variable values), although common in C/C++, should be avoided.
Related
What is the difference between a var and val definition in Scala and why does the language need both? Why would you choose a val over a var and vice versa?
As so many others have said, the object assigned to a val cannot be replaced, and the object assigned to a var can. However, said object can have its internal state modified. For example:
class A(n: Int) {
var value = n
}
class B(n: Int) {
val value = new A(n)
}
object Test {
def main(args: Array[String]) {
val x = new B(5)
x = new B(6) // Doesn't work, because I can't replace the object created on the line above with this new one.
x.value = new A(6) // Doesn't work, because I can't replace the object assigned to B.value for a new one.
x.value.value = 6 // Works, because A.value can receive a new object.
}
}
So, even though we can't change the object assigned to x, we could change the state of that object. At the root of it, however, there was a var.
Now, immutability is a good thing for many reasons. First, if an object doesn't change internal state, you don't have to worry if some other part of your code is changing it. For example:
x = new B(0)
f(x)
if (x.value.value == 0)
println("f didn't do anything to x")
else
println("f did something to x")
This becomes particularly important with multithreaded systems. In a multithreaded system, the following can happen:
x = new B(1)
f(x)
if (x.value.value == 1) {
print(x.value.value) // Can be different than 1!
}
If you use val exclusively, and only use immutable data structures (that is, avoid arrays, everything in scala.collection.mutable, etc.), you can rest assured this won't happen. That is, unless there's some code, perhaps even a framework, doing reflection tricks -- reflection can change "immutable" values, unfortunately.
That's one reason, but there is another reason for it. When you use var, you can be tempted into reusing the same var for multiple purposes. This has some problems:
It will be more difficult for people reading the code to know what is the value of a variable in a certain part of the code.
You may forget to re-initialize the variable in some code path, and end up passing wrong values downstream in the code.
Simply put, using val is safer and leads to more readable code.
We can, then, go the other direction. If val is that better, why have var at all? Well, some languages did take that route, but there are situations in which mutability improves performance, a lot.
For example, take an immutable Queue. When you either enqueue or dequeue things in it, you get a new Queue object. How then, would you go about processing all items in it?
I'll go through that with an example. Let's say you have a queue of digits, and you want to compose a number out of them. For example, if I have a queue with 2, 1, 3, in that order, I want to get back the number 213. Let's first solve it with a mutable.Queue:
def toNum(q: scala.collection.mutable.Queue[Int]) = {
var num = 0
while (!q.isEmpty) {
num *= 10
num += q.dequeue
}
num
}
This code is fast and easy to understand. Its main drawback is that the queue that is passed is modified by toNum, so you have to make a copy of it beforehand. That's the kind of object management that immutability makes you free from.
Now, let's covert it to an immutable.Queue:
def toNum(q: scala.collection.immutable.Queue[Int]) = {
def recurse(qr: scala.collection.immutable.Queue[Int], num: Int): Int = {
if (qr.isEmpty)
num
else {
val (digit, newQ) = qr.dequeue
recurse(newQ, num * 10 + digit)
}
}
recurse(q, 0)
}
Because I can't reuse some variable to keep track of my num, like in the previous example, I need to resort to recursion. In this case, it is a tail-recursion, which has pretty good performance. But that is not always the case: sometimes there is just no good (readable, simple) tail recursion solution.
Note, however, that I can rewrite that code to use an immutable.Queue and a var at the same time! For example:
def toNum(q: scala.collection.immutable.Queue[Int]) = {
var qr = q
var num = 0
while (!qr.isEmpty) {
val (digit, newQ) = qr.dequeue
num *= 10
num += digit
qr = newQ
}
num
}
This code is still efficient, does not require recursion, and you don't need to worry whether you have to make a copy of your queue or not before calling toNum. Naturally, I avoided reusing variables for other purposes, and no code outside this function sees them, so I don't need to worry about their values changing from one line to the next -- except when I explicitly do so.
Scala opted to let the programmer do that, if the programmer deemed it to be the best solution. Other languages have chosen to make such code difficult. The price Scala (and any language with widespread mutability) pays is that the compiler doesn't have as much leeway in optimizing the code as it could otherwise. Java's answer to that is optimizing the code based on the run-time profile. We could go on and on about pros and cons to each side.
Personally, I think Scala strikes the right balance, for now. It is not perfect, by far. I think both Clojure and Haskell have very interesting notions not adopted by Scala, but Scala has its own strengths as well. We'll see what comes up on the future.
val is final, that is, cannot be set. Think final in java.
In simple terms:
var = variable
val = variable + final
val means immutable and var means mutable.
Full discussion.
The difference is that a var can be re-assigned to whereas a val cannot. The mutability, or otherwise of whatever is actually assigned, is a side issue:
import collection.immutable
import collection.mutable
var m = immutable.Set("London", "Paris")
m = immutable.Set("New York") //Reassignment - I have change the "value" at m.
Whereas:
val n = immutable.Set("London", "Paris")
n = immutable.Set("New York") //Will not compile as n is a val.
And hence:
val n = mutable.Set("London", "Paris")
n = mutable.Set("New York") //Will not compile, even though the type of n is mutable.
If you are building a data structure and all of its fields are vals, then that data structure is therefore immutable, as its state cannot change.
Thinking in terms of C++,
val x: T
is analogous to constant pointer to non-constant data
T* const x;
while
var x: T
is analogous to non-constant pointer to non-constant data
T* x;
Favoring val over var increases immutability of the codebase which can facilitate its correctness, concurrency and understandability.
To understand the meaning of having a constant pointer to non-constant data consider the following Scala snippet:
val m = scala.collection.mutable.Map(1 -> "picard")
m // res0: scala.collection.mutable.Map[Int,String] = HashMap(1 -> picard)
Here the "pointer" val m is constant so we cannot re-assign it to point to something else like so
m = n // error: reassignment to val
however we can indeed change the non-constant data itself that m points to like so
m.put(2, "worf")
m // res1: scala.collection.mutable.Map[Int,String] = HashMap(1 -> picard, 2 -> worf)
"val means immutable and var means mutable."
To paraphrase, "val means value and var means variable".
A distinction that happens to be extremely important in computing (because those two concepts define the very essence of what programming is all about), and that OO has managed to blur almost completely, because in OO, the only axiom is that "everything is an object". And that as a consequence, lots of programmers these days tend not to understand/appreciate/recognize, because they have been brainwashed into "thinking the OO way" exclusively. Often leading to variable/mutable objects being used like everywhere, when value/immutable objects might/would often have been better.
val means immutable and var means mutable
you can think val as java programming language final key world or c++ language const key world。
Val means its final, cannot be reassigned
Whereas, Var can be reassigned later.
It's as simple as it name.
var means it can vary
val means invariable
Val - values are typed storage constants. Once created its value cant be re-assigned. a new value can be defined with keyword val.
eg. val x: Int = 5
Here type is optional as scala can infer it from the assigned value.
Var - variables are typed storage units which can be assigned values again as long as memory space is reserved.
eg. var x: Int = 5
Data stored in both the storage units are automatically de-allocated by JVM once these are no longer needed.
In scala values are preferred over variables due to stability these brings to the code particularly in concurrent and multithreaded code.
Though many have already answered the difference between Val and var.
But one point to notice is that val is not exactly like final keyword.
We can change the value of val using recursion but we can never change value of final. Final is more constant than Val.
def factorial(num: Int): Int = {
if(num == 0) 1
else factorial(num - 1) * num
}
Method parameters are by default val and at every call value is being changed.
In terms of javascript , it same as
val -> const
var -> var
I'm a newbie in C++, I was learning to code in python.
I believe the solution is simple but I have no idea how to do it.
Here is what I was trying to do in C++ (not working):
int createBoard(int x, int y) {
int l[x];
int board[y, l[x]];
return board;
}
int main() {
int x = 5;
int y = 6;
board = createBoard(x,y);
return 0;
}
Here is what I wanted to replicate (working, but in python):
def createBoard(x,y):
length = [i for i in range(0,10)]
area = [y,length]
return area
area = createBoard(5,6)
Basically I want to create a function that returns an array with the y value and an array counting until x.
As far as I understood from your Python code, you want to create a 2D array. For a complete beginner in C++ that might be a challenging task. Many recommend to use std::vector and they are right but 2D "array" using such container could be very slow. So this example will work but undesirable in the future case when you gain more experience in C++:
#include <vector>
std::vector< std::vector<int> > createBoard(size_t x, size_t y)
{
return std::vector< std::vector<int> >(x, std::vector<int>(y));
}
So if you want to use a more efficient way of creating 2D arrays, see this example:
LINK
Translating code line by line is almost guaranteed to fail. You better do it in two steps: 1) fully understand what code in language A does. 1a) forget about the code in language A. 2) Write the same in language B.
I am not very proficient with python so I start from this:
Basically I want to create a function that returns an array with the y
value and an array counting until x.
You declared the function to return a single int. A single int is not two arrays.
Next, this
int l[x];
Is not standard c++ because x is not a compile time constant. Some compilers offer it as an extension, but there is no reason to use it because c++ has std::vector.
Then, this
int board[y, l[x]];
is problematic in multiple ways. l[x] is accessing an element in the array l that is out of bounds. Valid indexes are 0 till x-1 because l has x elements. Accessing the array out of bounds is undefined behaviour. We could stop at this point, because in the presece of undefined behaviour anything can happen. However, y, l[x] invokes the comma operator. It evaluates both sides and results in the right operand. Then you again have the same problem, l[x] is no compile time constant.
In this place I had code in c++, but it turned out that I completely misunderstood what your code is supposed to do. I'll leave the answer and refer you to others for the solution.
There are several problems with your code. The main one is that the Python array area contains objects of two different types: The first is the integer y, the second is the array length. All elements of a C++ array must have the same type.
Depending on what you want to use it for, you can replace the board array with a std::pair. This is an object containing two elements of different types.
Also, in C++ arrays with non-constant lengths must be dynamically created. Either using the new operator or (better) using std::unique_ptr. (Or you might want to use std::vector instead.)
Here's a small C++ program that does something like what you want to do:
#include <utility>
#include <memory>
auto createBoard(int x, int y) {
return std::make_pair(y, std::make_unique<int[]>(x));
}
int main() {
auto board = createBoard(5,6);
return 0;
}
(This will only work if your compiler supports C++14 or newer.)
But this is actually rather much above "newbie" level, and I doubt that you will find it very useful.
It would be better to start with a specification of what your program should do, rather than try to translate code from Python.
EDIT
Same code with std::vector instead of a dynamic array:
#include <utility>
#include <vector>
auto createBoard(int x, int y) {
return std::make_pair(y, std::vector<int>(x));
}
int main() {
auto board = createBoard(5,6);
return 0;
}
Wanted to toy with adding some sugar in Swift3. Basically, I wanted to be able to do something like:
let randomAdjust = (-10...10).random
To do that, I decided I would need to extend ClosedRange. But then found it would probably be even better for my case, I really just plan on doing Int's for now, to use CountableClosedRange. My latest of multiple attempts looked like:
extension CountableClosedRange where Bound == Int {
var random:Int {
return Int(arc4random_uniform(UInt32(self.count) + 1)) + self.lowerBound
}
}
But the playground complains:
error: same-type requirement makes generic parameter 'Bound' non-generic
extension CountableClosedRange where Bound == Int {
I don't even know what it's telling me there.
The way this roadblock is commonly encountered is when attempting to extend Array. This is legal:
extension Array where Element : Comparable {
}
But this is illegal:
extension Array where Element == Int {
}
The compiler complains:
Same-type requirement makes generic parameter 'Element' non-generic
The problem is the use of == here in combination with Array's parameterized type Element, because Array is a generic struct.
One workaround with Array is to rise up the hierarchy of Array's inheritance to reach something that is not a generic struct:
extension Sequence where Iterator.Element == Int {
}
That's legal because Sequence and Iterator are generic protocols.
Another solution, though, is to rise up the hierarchy from the target type, namely Int. If we can find a protocol to which Int conforms, then we can use the : operator instead of ==. Well, there is one:
extension CountableClosedRange where Bound : Integer {
}
That's the real difference between our two attempts to implement random on a range. The reason your attempt hits a roadblock and mine doesn't is that you are using == whereas I am using :. I can do that because there's a protocol (FloatingPoint) to which Double conforms.
But, as you've been told, with luck all this trickery will soon be a thing of the past.
In Swift 4, what you are attempting is now completely supported. Hooray!
extension Stack where Element: Equatable {
func isTop(_ item: Element) -> Bool {
guard let topItem = items.last else {
return false
}
return topItem == item
}
}
Example from Swift docs: https://docs.swift.org/swift-book/LanguageGuide/Generics.html#ID553
Simple example: I have some functions and I need to call them all, to modify a structure, only in one function. With these simple functions the task can be made in ways that don't use void, but in others tasks you have to use void. So what can you do?
type player = { mutable name : string; mutable points : int } ;;
let putname brad = match brad with
{ name = x; points = y } -> { name = brad; points = y } ;;
let putpoint guy score = match guy with
{ name = x; points = y } -> { name = x; points = score } ;;
let loosers listplayer guy = guy :: listplayer ;;
Here is the problem - How can I do the next function?
let someoneloses guy = void
guy = putpoint guy 0 ;;
listplayer = loosers (listplayer guy) ;;
Given you are using the name "void" I'm assuming you are more familiar with C (or C++). In OCaml the equivalent of "void" (the name of the type for no value) is "unit". There is another difference though: while in C the syntax is complex enough that it have constructs for no values (for instance, you can either "return a_value;" or "return;", two differents yet syntactically valid use cases for the keyword "return"), in OCaml the syntax is simpler and always require a value. So we have a notation for "nothing", which is, astutely but maybe also confusedly, is written "()".
So, the OCaml equivalent of the C:
void do_nothing(void) { return; }
is written:
let do_nothing () = ()
(notice how OCaml syntax is simpler and easier to grok once you got the "()" trick).
Now that this is hopefully clearer, back to your question.
A function that returns nothing is a function that return "()", either explicitly (as "do_nothing" above) or because it ends with an expression that has "()" as its value. For instance, an assignment (something tell me you'll love assignments), such as:
let putpoint guy score = guy.points <- score
Now back to your problem. You seem to be doing some kind of game with players represented as mutable records and some functions modifying those records as the game develop. You need not use pattern matching for that. Actually, functions such as "putpoint" above is probably what you want. But then you need some more state in your program: the list of loosers for instance is probably going to be a reference to a list that you modify etc.
This is the "imperative" side of OCaml but there is another one, which is usually regarded as more elegant although often slower in general (but not in functional languages which are optimised for this technique), consisting of refraining from altering state (changing values of things) but instead using functions merely taking values and returning values. Implemented like this, a player would be represented as an immutable record and each function acting a user would take an "old user" and return a "new user", and the same goes with the list of loosers, and so on. Actually, the whole game state would be represented as a big value that the "main loop" of your program would, given the previous value, and possible also the time and user inputs, would compute the "new state" and return it.
Have fun!
Also, your question has nothing to do with ocaml-batteries.
since you are using mutable data, you just have to assigned the value directly.
let p = {name = "me";points=0};;
let update x = x.name <- "you";
x.points <- 3;;
update p ;;
I am having problems translating C++ data structures to Scala. Scala is really different from C++, but I like a lot of it.
I have the following Code fragment in C++:
struct Output
{
double point;
double solution[6];
};
struct Coeff
{
double rcont1[6];
double rcont2[6];
double rcont3[6];
double rcont4[6];
double rcont5[6];
double rcont6[6];
};
std::list<Output> output;
std::list<Coeff> coeff;
I now fill the list in a while loop with data
while(n<nmax) {
if step successfull
Output out;
out.point = some values;
out.solution[0] = some value;
output.push_back(out);
}
I tried creating a simple class in Scala to hold the data.
class Output
{
var point: Double
var solution: Array[Double] = new Array(6)
}
But this doens't work since point is not initialized. Is there a way around this? I just want to define the variable but not initialize it.
Another quick thing. I am looking for an equivalent to stl::lower_bound.
Is finds the right position to insert an element in an sorted container to maintain the order.
Thanks for helping a Scala beginner
Why don't you want to initialize it? For efficiency? I'm afraid that the JVM doesn't let you get away with having random junk in your variables based on whatever was there originally. So since you have to initialize it anyway, why not specify what your "uninitialized" value is?
class Output {
var point = 0.0
var solution = new Array[Double](6)
}
(You could use Double.NaN and check for point.isNaN if you later need to see whether the value has been initialized or not.)
You could use _ as the default initialization, but unless you use it in generic code:
class Holder[T] {
var held: T = _
}
then you're just obscuring what the value really will be set to. (Or you are announcing "I really don't care what goes here, it could be anything"--which could be useful.)
I just found the answer to the intialistion:
class Output
{
var point: Double = _
var solution: Array[Double] = Array(6)
}
Puh Scala has a lot of syntx to get used to :-)
Anyone have a solution for the lower_bound equivalent ?
It's hard to translate effectively, as you've left a lot of unknowns hidden behind pseudo code, but I'd advocate something along these lines:
// type alias
type Coeff = Seq[Seq[Double]]
// parameters passed to a case class become member fields
case class Output (point: Double, solution: Seq[Double])
val outputs = (0 to nmax) map { n =>
Output(generatePoint(n), generateSolution(n))
}
If you can flesh out your sample code a bit more fully, I'll be able to give a better translation.