Official docs says that class can be built dynamically like so:
constant A := Metamodel::ClassHOW.new_type( name => 'A' );
A.^add_method('x', my method x(A:D:) { say 42 });
A.^compose;
A.new.x(); # x will be only called on instances
But what if I am building a class and don't assign it to a constant but rather store it in a var (for instance when I need to create a bunch of classes in loop) like so:
my $x = Metamodel::ClassHOW.new_type( name => 'some custom string' );
$x.^add_method('x', my method ($y:) { say $y });
$x.^compose;
But in this case I can call method x both on class ($x.x) and on instance ($x.new.x) though I want it to only be called on instances.
I tried to define method like so:
$x.^add_method('x', my method ($y:D:) { say $y });
but that produces an error:
Invalid typename 'D' in parameter declaration.
Of course I can check defindness of the value inside the method but I want some compile-time guarantees (I want to believe that type checking is done in compile time).
I tried to play with signatures and parameters but couldn't find a way to create an invocant parameter but what is more important I am not sure how to assign signature which I have in a variable to some method.
Change:
my $x = ...
to:
my constant x = my $ = ...
In full:
my constant x = my $ = Metamodel::ClassHOW.new_type( name => 'some custom string' );
x.^add_method('x', my method (x:D $y:) { say $y });
x.^compose;
x = Metamodel::ClassHOW.new_type( name => 'another custom string' );
...
I want some compile-time guarantees (I want to believe that type checking is done in compile time).
By making the constant's RHS be a variable declaration, you blend static compile-time aspects with dynamic run-time ones.
(BTW, the my before constant is just me being pedantic. A plain constant like you've used is equivalent to our constant which is less strict than my constant.)
I note the error message for a non-instance is different:
Type check failed in binding to parameter '$y';
expected type some custom string cannot be itself
At a guess that's because the usual message comes from Mu or Any and your class isn't inheriting from either of them.
I am not sure how to assign signature which I have in a variable to some method.
I'll leave that part unanswered.
The best way I can think of to produce a method with types substituted into a signature with Raku today is to use a parametric role to help out, like this:
my role Helper[::T] {
method foo(T $inv:) {}
}
my &meth = Helper.^parameterize(Int).^pun.^lookup("foo");
say &meth.signature;
Which outputs (Int $inv: *%_). Substitute Int with the type you are building.
As I know Java is pass-by-value from this post. I am from Java background I wonder what Kotlin is using for passing values in between. Like in Extensions or Methods etc.
Every time I hear about the "pass-by-value" vs "pass-by-reference" Java debate I always think the same. The answer I give: "Java passes a copy (pass-by-value) of the reference (pass-by-reference)". So everyone is happy. I would say Kotlin does the same as it is JVM based language.
UPDATE
OK, so it's been a while since this answer and I think some clarification should be included. As #robert-liberatore is mentioning in the comments, the behaviour I'm describing is true for objects. Whenever your methods expect any object, you can assume that the JVM internally will make a copy of the reference to the object and pass it to your method. That's why having code like
void doSomething(List<Integer> x) {
x = new ArrayList<Integer>()
}
List<Integer> x = Arrays.asList(1, 2, 3);
doSomething(x);
x.length() == 3
behaves like it does. You're copying the reference to the list, so "reassigning it" will take no effect in the real object. But since you're referring to the same object, modifying its inner content will affect the outer object.
This is something you may miss when defining your attributes as final in order to achieve immutability. You won't be able to reassign them, but there's nothing preventing you from changing its content
Of course, this is true for objects where you have a reference. In case of primitives, which are not a reference to an object containing something but "something" themselves, the thing is different. Java will still make a copy of the whole value (as it does with the whole reference) and pass it to the method. But primitives are just values, you can't "modify its inner values". So any change inside a method will not have effect in the outer values
Now, talking about Kotlin
In Kotlin you "don't have" primitive values. But you "do have" primitive classes. Internally, the compiler will try to use JVM primitive values where needed but you can assume that you always work with the boxed version of the JVM primitives. Because of that, when possible the compiler will just make a copy of the primitive value and, in other scenarios, it will copy the reference to the object. Or with code
fun aJvmPrimitiveWillBeUsedHere(x: Int): Int = x * 2
fun aJvmObjectWillBeUsedHere(x: Int?): Int = if (x != null) x * 2 else 1
I'd say that Kotlin scenario is a bit safer than Java because it forces its arguments to be final. So you can modify its inner content but not reassign it
fun doSomething(x: MutableList<Int>) {
x.add(2) // this works, you can modify the inner state
x = mutableListOf(1, 2) // this doesn't work, you can't reassign an argument
}
It uses the same principles like Java. It is always pass-by-value, you can imagine that a copy is passed. For primitive types, e.g. Int this is obvious, the value of such an argument will be passed into a function and the outer variable will not be modified. Please note that parameters in Kotlin cannot be reassigned since they act like vals:
fun takeInt(a: Int) {
a = 5
}
This code will not compile because a cannot be reassigned.
For objects it's a bit more difficult but it's also call-by-value. If you call a function with an object, a copy of its reference is passed into that function:
data class SomeObj(var x: Int = 0)
fun takeObject(o: SomeObj) {
o.x = 1
}
fun main(args: Array<String>) {
val obj = SomeObj()
takeObject(obj)
println("obj after call: $obj") // SomeObj(x=1)
}
You can use a reference passed into a function to change the actual object.
The semantics is identical to Java.
In Java, when you have an instance of an object, and you pass it to a method, that method can change the state of that object, and when the method is done, the changes would have been applied to the object at the call site.
The same applies in Kotlin.
For primitives value is passed, and for non-primitives a reference to the object is passed. I'll explain with an example:
The code:
fun main() {
var a = 5
var b = a
a = 6
println("b = $b")
}
prints: b = 5
Kotlin passes the value of a to b, because a is a primitive. So changing a afterwards won't impact b.
The code:
fun main() {
var a = Dog(5)
var b = a
a.value = 6
println("b = ${b.value}")
}
class Dog (var value: Int)
prints b = 6, because this time a is not a primitive and so the reference to the object (Dog) was passed to b and not its value. Therefore changing a would affect all objects that point to it.
In Java primitive types like int, float, double, boolean are passed to a method by value, if you modify them inside the receiver method they doesn't change into the calling method. But if the property/variable type isn't a primitive, like arrays of primitives or other classes when they are changed inside the method that receive them as parameter they also change in the caller method.
But with Kotlin nothing seems to be primitive, so I think all is passed by reference.
This might be a little bit confusing.
The correct answer, IMHO, is that everything passes by reference, but no assignment is possible so it will be similar to passing by value in C++.
Note that function parameters are constant, i.e., they cannot be assigned.
Remember that in Kotlin there are no primitive types. Everything is an object.
When you write:
var x: Int = 3
x += 10
You actually create an object of type Int, assign it the value 3, and get a reference, or pointer, named x.
When you write
x += 10
You reassign a new Int object, with the value 13, to x. The older object becomes a garbage (and garbage-collected).
Of course, the compiler optimizes it, and creates no objects in the heap in this particular case, but conceptually it is as explained.
So what is the meaning of passing by reference function parameters?
Since no assignment is possible for function parameters, the main advantage of passing by reference in C++ does not exist in Kotlin.
If the object (passed to the function) has a method which changes its internal state, it will affect the original object.
No such method exists for Int, String, etc. They are immutable objects.
No copy is ever generated when passing objects to functions.
Bear in mind, am quite new to Kotlin. In my opinion, primitives are passed-by-value, but objects are passed-by-reference.
A primitive passed to a class works by default, but if you pass an object from a list, for example, and that object changes, the class object changes too. Because, in fact, it is the same object.
Additionally, if objects gets removed from the list, the class object IS STILL A REFERENCE. So it can still change due to references in other places.
Example below explaines. You can run it here.
fun main() {
val listObjects = mutableListOf(ClassB(), ClassB(), ClassB())
val listPrimitives = mutableListOf(111, 222, 333)
val test = ClassA()
test.ownedObject = listObjects[0]
test.ownedPrimitive = listPrimitives[0]
println("ownedObject: " + test.ownedObject.isEnabled +", ownedPrimitive: " +
test.ownedPrimitive)
listObjects[0].isEnabled = true
println("ownedObject: " + test.ownedObject.isEnabled +", ownedPrimitive: " +
test.ownedPrimitive)
listPrimitives[0] = 999
println("ownedObject: " + test.ownedObject.isEnabled +", ownedPrimitive: " +
test.ownedPrimitive)
}
class ClassA {
var ownedObject: ClassB = ClassB()
var ownedPrimitive: Int = 0
}
class ClassB {
var isEnabled = false
}
Since Kotlin is a new language for JVM, like Java it is pass-by-value. The confusing part is with object, at first it looks like that it is passed-by-reference but the actuality is that the reference/pointer itself is pass-by-value (a copy of a reference is passed to a method) hence when a method receives a reference to an object, the method can manipulate the original object.
I'm making a program for the game 15 puzzle.
My function headers look like this:
void leftSlide(vector< vector<int> >& puzzle);
void rightSlide(vector< vector<int> >& puzzle);
void upSlide(vector< vector<int> >& puzzle);
void downSlide(vector< vector<int> >& puzzle);
my main function also has a vector< vector<int> > puzzle. Am I allowed to do this, or will this cause problems?
The scope of a variable is within the enclosing curly braces. For example,
void foo()
{
int x; // variable x is not known outside of foo
}
This scoping rule applies even for variables in the argument list. For example,
void boo (int y)
{
// variable y in not known outside of boo
}
Therefore, in your case, the variables will be passed from the main driver to the individual functions by reference. So, yes, you can have variables of the same name in different scopes.
simply yes
The potential scope of a variable introduced by a declaration in a block (compound statement) begins at the point of declaration and ends at the end of the block. Actual scope is the same as potential scope unless there is a nested block with a declaration that introduces identical name (in which case, the entire potential scope of the nested declaration is excluded from the scope of the outer declaration)
This is my friend question from my instructor about how to print
main method value of local variable in method function when it's
variable pushed and pop from stack (because when method function
called it's variable pushed and when it reached to end pop from
stack) then it's local variable storage back to the memory.
Why main method print 100 ?
// Define a global pointer
int *ptr;
int method()
{
// Define a variable local in this method
int local = 100;
// Set address of local variable (name of variable is local)
// in the ptr pointer
ptr = &local;
return -1;
}
int main()
{
// Call method
method();
// Print value of ptr pointer
cout<<*ptr<<"\n";
return -1;
}
If you are asking why the main method is printing 100.
1.The local variable is assigned some memory space. (say at X. Therefore [x]=>100)
2.The pointer which is global is then assigned to the point at X.(say pointers space is Y. [Y]=>X)
3.Therefore the point of the pointer is X.
4.Now you choose to print value of the pointer. i.e. [[y]]=[x] which is 100.
Why is doesn't print a garbage value is because even though the memory space is no longer allocated to local it still contains that value.
If you wrote some more code the point in memory might have been overwritten by another variable
Difference between value parameter and reference parameter ? This question is asked sometime by interviewers during my interviews. Can someone tell me the exact difference that is easy to explain with example? And is reference parameter and pointer parameter are same thing ?
Thanks
Changes to a value parameter are not visible to the caller (also called "pass by value").
Changes to a reference parameter are visible to the caller ("pass by reference").
C++ example:
void by_value(int n) { n = 42; }
void by_ref(int& n) { n = 42; }
void also_value(int const& n); // Even though a reference is used, this is
// semantically a value parameter---though there are implementation
// artifacts, like not being able to write "n = 42" (it's const) and object
// identity (&n here has different ramifications than for by_value above).
One use of pointers is to implement "reference" parameters without using a special reference concept, which some languages, such as C, don't have. (Of course you can also treat pointers as values themselves.)
The main difference is whether the object passed is copied. If it's a value parameter the compiler must generate such code that altering the function parameter inside the function has no effect on the original object passsed, so it will usually copy the object. In case of reference parameters the compiler must generate such code taht all operations are done on the original object being passed.
A pointer is a low-level way of representing a reference, so passing a pointer (by value) is how languages like C typically achieve pass by reference semantics.
The difference is pretty simple: direct parameters are passed by value, and the receiver receives a copy of what is passed; meaning that if the parameter is modified by the receiver, these changes will not be reflected back to the caller. (This is often called, appropriately enough, pass by value, or by copy.
There are basically three kinds of parameters; pointer, reference and direct.
The difference is pretty simple: direct parameters are passed by value, and the receiver receives a copy of what is passed; meaning that if the parameter is modified by the receiver, these changes will not be reflected back to the caller. (This is often called, appropriately enough, pass by value, or bycopy.
Pointers are also passed by value, but rather than sending the actual value, the caller sends the address of the value. This means that by following this pointer, the receiver can modify the argument. Note that changes made to the actual pointer still aren't reflected back to the caller.
The final form, call-by-reference, is sort of a middle ground between these two approaches. Essentially it can be thought of as a pointer that looks like a value.
It is worth mentioning that at the core of it all, parameters are always passed by value, but different languages have different ways of implementing reference semantics (see Kylotans answer).
// Example using C
// bycopy
int multiply(int x, int y) {
return x * y;
}
void multiply_p(int *x, int y) {
*x *= y;
}
int main () {
int i, j, k;
i = 20;
j = 10;
k = multiply(i,j); // k is now 200
multiply_p(&i, k); // i is now 4000 (200 * 20)
return 0;
}
Pseudocode:
Pass by Value:
void setTo4(value) { // value is passed by value
value = 4;
}
int x = 1;
setTo4(x);
// x is still 1
Pass by Reference:
void setTo4(value) { // value is passed by reference
value = 4;
}
int x = 1;
setTo4(x);
// x is 4