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
In C (or C++), is it possible to create an array a (or something that "looks like" an array), such that a[0], a[1], etc., all point to the same memory location? So if you do
a[0] = 0.0f;
a[1] += 1.0f;
then a[0] will be equal to 1.0f, because it's the same memory location as a[1].
I do have a reason for wanting to do this. It probably isn't a good reason. Therefore, please treat this question as if it were asked purely out of curiosity.
I should have said: I want to do this without overloading the [] operator. The reason for this has to do with avoiding a dynamic dispatch. (I already told you my reason for wanting to do this is probably not a good one. There's no need to tell me I shouldn't want to do it. I already know this.)
I suppose a class like this is what you need
template <typename T>
struct strange_array
{
T & operator [] (int) { return value; }
private:
T value;
};
You can always define an array of pointers which points towards the same variable :
typedef int* special;
int i = 0;
unsigned int var = 0xdeadbeef;
special arr[5];
for (i=0; i<5; i++)
arr[i] = &var;
*(arr[0]) = 0;
*(arr[3]) += 3;
printf("%d\n", *(arr[2]));
// -> 3
In C, I don't think so.
The expression a[i] simply means *(a + i), so it's hard to avoid the addition due to the indexing.
You might be able to glue something together by making a (the array name) a macro, but I'm not sure how: you wouldn't have access to the index in order to compensate for it.
Without overloading operator[]?
No, it's not possible.
Fortunately.
From all that conversation here, I now understand the problem as follows:
You want to have the syntax of an array, e.g.
a[n] // only lookup
a[n]++ // lookup and write
but you want to have the semantics changed to all of those map to the same element, like
a[0]
a[0]++
The C++ way to achieve this is IMHO to overload the index access operator [].
But, you don't want it for performance reasons.
I join the opinon of user Lightness Races in Orbit that you can not do this within C++.
As you don't provide more information about the use case it is hard to come up with a solution.
Best I can imagine is that you have lots of written code which uses array semantics which you can not change.
What is left (wanting to keep performance) are code transformation techniques (CPP, sed, ..), generating a source code from the given source code with the desired behaviour, e.g. by forcing all index values to 0.
I'm trying to use a wrapper for a library that wants an Array as an input parameter.
I tried casting the Array, but I get an error: Cannot convert 'any[]' to 'Array'
Is there a way to make this work?
var rows = new Array(10);
var rows2 = <Array>rows; //<--- Cannot convert 'any[]' to 'Array'
There are 4 possible conversion methods in TypeScript for arrays:
let x = []; //any[]
let y1 = x as number[];
let z1 = x as Array<number>;
let y2 = <number[]>x;
let z2 = <Array<number>>x;
The as operator's mostly designed for *.tsx files to avoid the syntax ambiguity.
I think the right syntax is:
var rows2 = <Array<any>>rows;
That's how you cast to interface Array<T>
I think this is just a bug - can you log an issue on the CodePlex site?
As a workaround, you can write <Array><any>rows;
A simple solution for all types
const myArray = <MyType[]>value;
I'm working a small C++ JSON library to help sharpen my rusty C++ skills, and I'm having trouble understanding some behavior with initialization lists.
The core of the library is a variant class (named "var") that stores any of the various JSON datatypes (null, boolean, number, string, object, array).
The goal is for var to work as closely as possible to a JavaScript variable, so there's lots of operator overloading going on. The primitive datatypes are easy to take care of...
var fee = "something";
var fie = 123.45;
var foe = false;
The problem is with objects (maps) and arrays (vectors).
To get something close to a JavaScript object and array literal syntax, I'm using initialization lists. It looks like this:
// in my headers
typedef var object[][2];
typedef var array[];
// in user code
var foo = (array){ 1, "b", true };
var bar = (object){ { "name", "Bob" }, { "age", 42 } };
This works out pretty nicely. The problem comes in with nested lists.
var foo = (array){ 1, "b", (array){ 3.1, 3.2 } };
For some reason my variant class interprets the nested "array" as a boolean, giving:
[1, "b", true]
Instead of:
[1, "b", [3.1, 3.2]]
If I explicitly cast the inner list to a var, it works:
var foo = (array){ 1, "b", (var)(array){ 3.1, 3.2 } };
Why do I have to explicitly cast the inner list to a var after I cast it to an array, and how can I get around this extra cast? As far as I can tell it should be implicitly converting the array to my var class anyway, since it's using the constructor for an array of vars:
template <size_t length>
var(const var(&start)[length]) {
// internal stuff
init();
setFromArray(vector<var>(start, start + length));
}
It seems that without the explicit cast to var, the initialization list somehow gets cast to something else on its way from being cast from an array to a var. I'm trying to understand why this happens, and how to avoid it.
Here's a gist with the complete source. Let me know if I should add anything relevant to the question.
Update
Apparently (foo){1, "two"} does not actually cast an initialiation list; it's a complete expression called a compound literal. It seems that it's only available in C, although g++ doesn't complain unless you give it -pedantic.
It looks like my options are:
Find another concise initialization syntax that is officially supported.
Use compound literals and hope they work in other compilers.
Drop support for C++ < 11 and use initializer_list.
Don't offer a concise initialization syntax.
Any help with the first option would be the sort of answer I'm looking for at this point.
Macros are another sort of last-ditch option, and I've written some that do the job, but I'd like to not have to use them.
You need to use the facilities already provided to you by Boost.
typedef boost::optional<boost::make_recursive_variant<
float, int, bool, //.. etc
std::unordered_map<std::string, boost::optional<boost::recursive_variant_>>,
std::vector<boost::recursive_variant_>
> JSONType;
They can easily define recursive variant types.
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.