Related
I try to get names of all trait a class extends using getInterfaces which returns an array of trait's names. When I manually access each member of the array, the method getName returns simple names like this
trait A
trait B
class C() extends A, B
val c = C()
val arr = c.getClass.getInterfaces
arr(0).getName // : String = A
arr(1).getName // : String = B
However, when I use map function on arr. The resulting array contains a cryptic version of trait's names
arr.map(t => t.getName) // : Array[String] = Array(repl$.rs$line$1$A, repl$.rs$line$2$B)
The goal of this question is not about how to get the resulting array that contains simple names (for that purpose, I can just use arr.map(t => t.getSimpleName).) What I'm curious about is that why accessing array manually and using a map do not yield a compatible result. Am I wrong to think that both ways are equivalent?
I believe you run things in Scala REPL or Ammonite.
When you define:
trait A
trait B
class C() extends A, B
classes A, B and C aren't defined in top level of root package. REPL creates some isolated environment, compiles the code and loads the results into some inner "anonymous" namespace.
Except this is not true. Where this bytecode was created is reflected in class name. So apparently there was something similar (not necessarily identical) to
// repl$ suggest object
object repl {
// .rs sound like nested object(?)
object rs {
// $line sounds like nested class
class line { /* ... */ }
// $line$1 sounds like the first anonymous instance of line
new line { trait A }
// import from `above
// $line$2 sounds like the second anonymous instance of line
new line { trait B }
// import from above
//...
}
}
which was made because of how scoping works in REPL: new line creates a new scope with previous definitions seen and new added (possibly overshadowing some old definition). This could be achieved by creating a new piece of code as code of new anonymous class, compiling it, reading into classpath, instantiating and importing its content. Byt putting each new line into separate class REPL is able to compile and run things in steps, without waiting for you to tell it that the script is completed and closed.
When you are accessing class names with runtime reflection you are seeing the artifacts of how things are being evaluated. One path might go trough REPLs prettifiers which hide such things, while the other bypass them so you see the raw value as JVM sees it.
The problem is not with map rather with Array, especially its toString method (which is one among the many reasons for not using Array).
Actually, in this case it is even worse since the REPL does some weird things to try to pretty-print Arrays which in this case didn't work well (and, IMHO, just add to the confusion)
You can fix this problem calling mkString directly like:
val arr = c.getClass.getInterfaces
val result = arr.map(t => t.getName)
val text = result.mkString("[", ", ", "]")
println(text)
However, I would rather suggest just not using Array at all, instead convert it to a proper collection (e.g. List) as soon as possible like:
val interfaces = c.getClass.getInterfaces.toList
interfaces .map(t => t.getName)
Note: About the other reasons for not using Arrays
They are mutable.
Thet are invariant.
They are not part of the collections hierarchy thus you can't use them on generic methods (well, you actually can but that requires more tricks).
Their equals is by reference instead of by value.
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
With a background from object-oriented programming I am not able to understand how to make immutable lists in Scala.
Example; I want to make a list of 10 random people:
object MyApplication extends App {
val numberOfPersons = 10 : Int
val listOfPersons = makeListOfPersons(numberOfPersons) : List[Person]
def makeListOfPersons( numberOfPersons : Int ) : List[Person] = {
// TODO: return a immutable list of 10 persons
}
}
class Person {
/**
Generic content,
like age and name.
* */
}
What is the "correct" way of making an immutable list in Scala?
If you know what collection type you want, you may be able to use the tabulate method on that type:
List.tabulate(10)(makePerson)
In this case makePerson is a function that takes an Int and returns the Person object for that Int.
If you don't care about the collection type, you can call map on the range 1 to 10 like this:
(1 to 10).map(makePerson)
If you don't need to use the Int parameter, you can do this:
List.tabulate(10)(_ => makeRandomPerson())
In this particular case,
List.fill(numberOfPersons){ codeThatCreatesASinglePerson }
seems most appropriate.
In most other cases: Nil creates an empty list, x :: y prepends an element x to list y.
If you want to append to list, instead of prepending to it, then you can take a collection.mutable.ListBuffer, append to it all the elements that you want to have in the list, and then call toList when you're done... or just use the built-in factory methods that do exactly that.
As the default List in Scala is immutable, the right way to add an element is to return a new list with the new element plus the older elements.
As a matter of fact, List has two methods, among others:
+:
++
The first one takes an element, add it as the first element and the rest of the list as it's tail and then returns the resulting list.
The other one takes another "collection" as parameter and adds it to the first list at the start.
List has another methods for adding the new element as the last one.
In Scala, these operations are permitted but take into consideration that always a new instance will be retrieved with the requested modifications as all objects are immutable by default.
As for your code goes, you could try with something like this:
object MyApplication extends App {
val numberOfPersons: Int = 10
val listOfPersons: List[Person] = makeListOfPersons(numberOfPersons)
def makeListOfPersons( numberOfPersons : Int ) : List[Person] = {
(1 to numberOfPersons).foldLeft(List.empty[Person]){ (accum, elem) =>
new Person() :: accum
}
}
}
(1 to numberOfPersons) creates a range, which could be seen as a List of ints, which will be traversed by foldLeft. This method will iterate through that list, and receives a seed, in this case an empty list of Person. Then, for every element in the int's list, a new Person is created and add to the list, returned as is the last expression and used the accumulator for the next iteration. Finally, a list of ten instances of Person is retrieved.
There are 5 ways to create List in scala:
Lisp style:
val list = 1::2::3::Nil
this style can also be thought of as a Haskell or functional programming (FP) style.
Java Style:
val list = List(1,2,3)
Scala List with range method
List.range(1, 10)
Create scala List with fill
List.fill(3)(5)
Scala List with tabulate
List.tabulate(5)(n => n * n)
element of the list are created according to the function we supply.
for more info please read this :
Preferred way to create a Scala list
I'm writing a data structure that converts the results of a database query. The raw structure is a java ResultSet and it would be converted to a map or class which permits accessing different fields on that data structure by either a named method call or passing a string into apply(). Clearly different values may have different types. In order to reduce burden on the clients of this data structure, my preference is that one not need to cast the values of the data structure but the value fetched still has the correct type.
For example, suppose I'm doing a query that fetches two column values, one an Int, the other a String. The result then names of the columns are "a" and "b" respectively. Some ideal syntax might be the following:
val javaResultSet = dbQuery("select a, b from table limit 1")
// with ResultSet, particular values can be accessed like this:
val a = javaResultSet.getInt("a")
val b = javaResultSet.getString("b")
// but this syntax is undesirable.
// since I want to convert this to a single data structure,
// the preferred syntax might look something like this:
val newStructure = toDataStructure[Int, String](javaResultSet)("a", "b")
// that is, I'm willing to state the types during the instantiation
// of such a data structure.
// then,
val a: Int = newStructure("a") // OR
val a: Int = newStructure.a
// in both cases, "val a" does not require asInstanceOf[Int].
I've been trying to determine what sort of data structure might allow this and I could not figure out a way around the casting.
The other requirement is obviously that I would like to define a single data structure used for all db queries. I realize I could easily define a case class or similar per call and that solves the typing issue, but such a solution does not scale well when many db queries are being written. I suspect some people are going to propose using some sort of ORM, but let us assume for my case that it is preferred to maintain the query in the form of a string.
Anyone have any suggestions? Thanks!
To do this without casting, one needs more information about the query and one needs that information at compiole time.
I suspect some people are going to propose using some sort of ORM, but let us assume for my case that it is preferred to maintain the query in the form of a string.
Your suspicion is right and you will not get around this. If current ORMs or DSLs like squeryl don't suit your fancy, you can create your own one. But I doubt you will be able to use query strings.
The basic problem is that you don't know how many columns there will be in any given query, and so you don't know how many type parameters the data structure should have and it's not possible to abstract over the number of type parameters.
There is however, a data structure that exists in different variants for different numbers of type parameters: the tuple. (E.g. Tuple2, Tuple3 etc.) You could define parameterized mapping functions for different numbers of parameters that returns tuples like this:
def toDataStructure2[T1, T2](rs: ResultSet)(c1: String, c2: String) =
(rs.getObject(c1).asInstanceOf[T1],
rs.getObject(c2).asInstanceOf[T2])
def toDataStructure3[T1, T2, T3](rs: ResultSet)(c1: String, c2: String, c3: String) =
(rs.getObject(c1).asInstanceOf[T1],
rs.getObject(c2).asInstanceOf[T2],
rs.getObject(c3).asInstanceOf[T3])
You would have to define these for as many columns you expect to have in your tables (max 22).
This depends of course on that using getObject and casting it to a given type is safe.
In your example you could use the resulting tuple as follows:
val (a, b) = toDataStructure2[Int, String](javaResultSet)("a", "b")
if you decide to go the route of heterogeneous collections, there are some very interesting posts on heterogeneous typed lists:
one for instance is
http://jnordenberg.blogspot.com/2008/08/hlist-in-scala.html
http://jnordenberg.blogspot.com/2008/09/hlist-in-scala-revisited-or-scala.html
with an implementation at
http://www.assembla.com/wiki/show/metascala
a second great series of posts starts with
http://apocalisp.wordpress.com/2010/07/06/type-level-programming-in-scala-part-6a-heterogeneous-list%C2%A0basics/
the series continues with parts "b,c,d" linked from part a
finally, there is a talk by Daniel Spiewak which touches on HOMaps
http://vimeo.com/13518456
so all this to say that perhaps you can build you solution from these ideas. sorry that i don't have a specific example, but i admit i haven't tried these out yet myself!
Joschua Bloch has introduced a heterogeneous collection, which can be written in Java. I once adopted it a little. It now works as a value register. It is basically a wrapper around two maps. Here is the code and this is how you can use it. But this is just FYI, since you are interested in a Scala solution.
In Scala I would start by playing with Tuples. Tuples are kinda heterogeneous collections. The results can be, but not have to be accessed through fields like _1, _2, _3 and so on. But you don't want that, you want names. This is how you can assign names to those:
scala> val tuple = (1, "word")
tuple: ([Int], [String]) = (1, word)
scala> val (a, b) = tuple
a: Int = 1
b: String = word
So as mentioned before I would try to build a ResultSetWrapper around tuples.
If you want "extract the column value by name" on a plain bean instance, you can probably:
use reflects and CASTs, which you(and me) don't like.
use a ResultSetToJavaBeanMapper provided by most ORM libraries, which is a little heavy and coupled.
write a scala compiler plugin, which is too complex to control.
so, I guess a lightweight ORM with following features may satisfy you:
support raw SQL
support a lightweight,declarative and adaptive ResultSetToJavaBeanMapper
nothing else.
I made an experimental project on that idea, but note it's still an ORM, and I just think it may be useful to you, or can bring you some hint.
Usage:
declare the model:
//declare DB schema
trait UserDef extends TableDef {
var name = property[String]("name", title = Some("姓名"))
var age1 = property[Int]("age", primary = true)
}
//declare model, and it mixes in properties as {var name = ""}
#BeanInfo class User extends Model with UserDef
//declare a object.
//it mixes in properties as {var name = Property[String]("name") }
//and, object User is a Mapper[User], thus, it can translate ResultSet to a User instance.
object `package`{
#BeanInfo implicit object User extends Table[User]("users") with UserDef
}
then call raw sql, the implicit Mapper[User] works for you:
val users = SQL("select name, age from users").all[User]
users.foreach{user => println(user.name)}
or even build a type safe query:
val users = User.q.where(User.age > 20).where(User.name like "%liu%").all[User]
for more, see unit test:
https://github.com/liusong1111/soupy-orm/blob/master/src/test/scala/mapper/SoupyMapperSpec.scala
project home:
https://github.com/liusong1111/soupy-orm
It uses "abstract Type" and "implicit" heavily to make the magic happen, and you can check source code of TableDef, Table, Model for detail.
Several million years ago I wrote an example showing how to use Scala's type system to push and pull values from a ResultSet. Check it out; it matches up with what you want to do fairly closely.
implicit val conn = connect("jdbc:h2:f2", "sa", "");
implicit val s: Statement = conn << setup;
val insertPerson = conn prepareStatement "insert into person(type, name) values(?, ?)";
for (val name <- names)
insertPerson<<rnd.nextInt(10)<<name<<!;
for (val person <- query("select * from person", rs => Person(rs,rs,rs)))
println(person.toXML);
for (val person <- "select * from person" <<! (rs => Person(rs,rs,rs)))
println(person.toXML);
Primitives types are used to guide the Scala compiler into selecting the right functions on the ResultSet.
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