Say I have an associative array of pointers to structs to strings, where the struct is called Foo; the type will be Foo*[string]. Suppose I also have a function with the following signature: void bar (Foo*[string] baz). Will baz be passed to bar by value or by reference? I've not found any documentation about this, and I'm curious.
By reference - any modifications to an existing AA will be observed outside of the function.
However, there is a corner case for when the associative array is null. In that case, initializing the AA by adding the first element will not be observed outside of the function. If the AA to be modified might be null, you should pass it by-ref.
It's like in Java. It's pass-by-value, but the value is a reference.
So you can't change the reference itself, but you can modify the AA it refers to.
AAs in D are implemented as a struct https://github.com/D-Programming-Language/druntime/blob/master/src/rt/aaA.d#L82-85, which is pass-by-value. But this struct contains pointer to other struct. And if this pointer to struct is not null, than all modification inside function will be visible in caller side.
Arrays in D will be passed by reference but goes in as a "fat pointer". You should watch out for this situation particularly.
In this case the call-site is not affected by re initializing the array ex.:
arrX = [ new, values, here ];
While this case does affect the call-site: arrX[index] = newVal;
If what you want is to disallow either case to occur, you can prefix them with the in or const modifiers.
Source: The D Programming Language - Andrei Alexandrescu - Addison Wesley.
Related
I'm really new to Swift and I just read that classes are passed by reference and arrays/strings etc. are copied.
Is the pass by reference the same way as in Objective-C or Java wherein you actually pass "a" reference or is it proper pass by reference?
Types of Things in Swift
The rule is:
Class instances are reference types (i.e. your reference to a class instance is effectively a pointer)
Functions are reference types
Everything else is a value type; "everything else" simply means instances of structs and instances of enums, because that's all there is in Swift. Arrays and strings are struct instances, for example. You can pass a reference to one of those things (as a function argument) by using inout and taking the address, as newacct has pointed out. But the type is itself a value type.
What Reference Types Mean For You
A reference type object is special in practice because:
Mere assignment or passing to function can yield multiple references to the same object
The object itself is mutable even if the reference to it is a constant (let, either explicit or implied).
A mutation to the object affects that object as seen by all references to it.
Those can be dangers, so keep an eye out. On the other hand, passing a reference type is clearly efficient because only a pointer is copied and passed, which is trivial.
What Value Types Mean For You
Clearly, passing a value type is "safer", and let means what it says: you can't mutate a struct instance or enum instance through a let reference. On the other hand, that safety is achieved by making a separate copy of the value, isn't it? Doesn't that make passing a value type potentially expensive?
Well, yes and no. It isn't as bad as you might think. As Nate Cook has said, passing a value type does not necessarily imply copying, because let (explicit or implied) guarantees immutability so there's no need to copy anything. And even passing into a var reference doesn't mean that things will be copied, only that they can be if necessary (because there's a mutation). The docs specifically advise you not to get your knickers in a twist.
Everything in Swift is passed by "copy" by default, so when you pass a value-type you get a copy of the value, and when you pass a reference type you get a copy of the reference, with all that that implies. (That is, the copy of the reference still points to the same instance as the original reference.)
I use scare quotes around the "copy" above because Swift does a lot of optimization; wherever possible, it doesn't copy until there's a mutation or the possibility of mutation. Since parameters are immutable by default, this means that most of the time no copy actually happens.
It is always pass-by-value when the parameter is not inout.
It is always pass-by-reference if the parameter is inout. However, this is somewhat complicated by the fact you need to explicitly use the & operator on the argument when passing to an inout parameter, so it may not fit the traditional definition of pass-by-reference, where you pass the variable directly.
Here is a small code sample for passing by reference.
Avoid doing this, unless you have a strong reason to.
func ComputeSomeValues(_ value1: inout String, _ value2: inout Int){
value1 = "my great computation 1";
value2 = 123456;
}
Call it like this
var val1: String = "";
var val2: Int = -1;
ComputeSomeValues(&val1, &val2);
The Apple Swift Developer blog has a post called Value and Reference Types that provides a clear and detailed discussion on this very topic.
To quote:
Types in Swift fall into one of two categories: first, “value types”,
where each instance keeps a unique copy of its data, usually defined
as a struct, enum, or tuple. The second, “reference types”, where
instances share a single copy of the data, and the type is usually
defined as a class.
The Swift blog post continues to explain the differences with examples and suggests when you would use one over the other.
When you use inout with an infix operator such as += then the &address symbol can be ignored. I guess the compiler assumes pass by reference?
extension Dictionary {
static func += (left: inout Dictionary, right: Dictionary) {
for (key, value) in right {
left[key] = value
}
}
}
origDictionary += newDictionaryToAdd
And nicely this dictionary 'add' only does one write to the original reference too, so great for locking!
Classes and structures
One of the most important differences between structures and classes is that structures are always copied when they are passed around in your code, but classes are passed by reference.
Closures
If you assign a closure to a property of a class instance, and the closure captures that instance by referring to the instance or its members, you will create a strong reference cycle between the closure and the instance. Swift uses capture lists to break these strong reference cycles
ARC(Automatic Reference Counting)
Reference counting applies only to instances of classes. Structures and enumerations are value types, not reference types, and are not stored and passed by reference.
Classes are passed by references and others are passed by value in default.
You can pass by reference by using the inout keyword.
Swift assign, pass and return a value by reference for reference type and by copy for Value Type
[Value vs Reference type]
If compare with Java you can find matches:
Java Reference type(all objects)
Java primitive type(int, bool...) - Swift extends it using struct
struct is a value type so it's always passed as a value. let create struct
//STEP 1 CREATE PROPERTIES
struct Person{
var raw : String
var name: String
var age: Int
var profession: String
// STEP 2 CREATE FUNCTION
func personInformation(){
print("\(raw)")
print("name : \(name)")
print("age : \(age)")
print("profession : \(profession)")
}
}
//allow equal values
B = A then call the function
A.personInformation()
B.personInformation()
print(B.name)
it have the same result when we change the value of 'B' Only Changes Occured in B Because A Value of A is Copied, like
B.name = "Zainab"
a change occurs in B's name. it is Pass By Value
Pass By Reference
Classes Always Use Pass by reference in which only address of occupied memory is copied, when we change similarly as in struct change the value of B , Both A & B is changed because of reference is copied,.
The description comes from the swift office document
Closures can capture and store references to any constants and variables from the context in which they are defined. This is known as closing over those constants and variables.
I don't fully understand the store references part. Does it means It creates some sort of "Pointer" to a variable? If the value changed the "de-referenced pointer" also changed to the new value. I think Swift has no pointer concept. I borrowed it to express my understandings.
Would be very nice that someone can provide a simple code lines to explain how the closure stores reference to the constants/variables.
Does it means It creates some sort of "Pointer" to a variable? If the
value changed the "de-referenced pointer" also changed to the new
value.
Yes.
I think Swift has no pointer concept.
It most certainly does, in the (implicit) form of reference types (classes), and the (explicit) form of UnsafePointer
Here's an example of a variable being captured in a closure. This all happens to be single threaded, but you could have this closure be dispatched by grand central dispatch. x is captured so that it exists for the entire life of the closure, even if the closure exists after x's declaring scope (f()) exits.
func f() {
var x = 0 //captured variable
_ = {
x = 5 //captures x from parent scope, and modifies it
}()
print(x) //prints 5, as x was modified by the closure
}
f()
I wrote another answer that explains the relationships between functions, closures, and other terms. It's worth reading, IMO.
I'm doing something like this:
someFunction(&myClass)
where someFunction sorts an array on myClass.
someFunction(inout someclass:ClassA) {
someClass.sort({$0.price > $1.price})
}
If I print myClass after the function call, I notice the array is still unsorted. From what I know, Swift passes values by copy. But when I use inout, shouldn't it change to pass by reference?
This is because class instances and functions are reference types. Ints, structs, and everything else are value types. When you pass a reference type into a function as a parameter, you are already going to be referencing that instance. When you pass a value type as a parameter, the function gets a copy of that variable (by default), so inout is usually (see edit) only needed if you want to alter a value type from inside of a function.
Altering a class instance without & or inout:
More details
When you create a reference type var t = myClass(), you're really creating a variable t that is a pointer to a myClass instance in memory. By using an ampersand &t in front of a reference type, you are really saying "give me the pointer to the pointer of a myClass instance"
More info on reference vs value types: https://stackoverflow.com/a/27366050/580487
EDIT
As was pointed out in the comments, you can still use inout with reference types if you want to alter a pointer, etc, but I was trying to shed light on the general use case.
Below is an example of sorting an array inside of a function:
If you post your code here, it would be more meaningful. BTW, look at below links that might helpful for you,
https://developer.apple.com/library/prerelease/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Functions.html#//apple_ref/doc/uid/TP40014097-CH10-ID173
https://developer.apple.com/library/prerelease/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Declarations.html#//apple_ref/doc/uid/TP40014097-CH34-ID545
I'm really new to Swift and I just read that classes are passed by reference and arrays/strings etc. are copied.
Is the pass by reference the same way as in Objective-C or Java wherein you actually pass "a" reference or is it proper pass by reference?
Types of Things in Swift
The rule is:
Class instances are reference types (i.e. your reference to a class instance is effectively a pointer)
Functions are reference types
Everything else is a value type; "everything else" simply means instances of structs and instances of enums, because that's all there is in Swift. Arrays and strings are struct instances, for example. You can pass a reference to one of those things (as a function argument) by using inout and taking the address, as newacct has pointed out. But the type is itself a value type.
What Reference Types Mean For You
A reference type object is special in practice because:
Mere assignment or passing to function can yield multiple references to the same object
The object itself is mutable even if the reference to it is a constant (let, either explicit or implied).
A mutation to the object affects that object as seen by all references to it.
Those can be dangers, so keep an eye out. On the other hand, passing a reference type is clearly efficient because only a pointer is copied and passed, which is trivial.
What Value Types Mean For You
Clearly, passing a value type is "safer", and let means what it says: you can't mutate a struct instance or enum instance through a let reference. On the other hand, that safety is achieved by making a separate copy of the value, isn't it? Doesn't that make passing a value type potentially expensive?
Well, yes and no. It isn't as bad as you might think. As Nate Cook has said, passing a value type does not necessarily imply copying, because let (explicit or implied) guarantees immutability so there's no need to copy anything. And even passing into a var reference doesn't mean that things will be copied, only that they can be if necessary (because there's a mutation). The docs specifically advise you not to get your knickers in a twist.
Everything in Swift is passed by "copy" by default, so when you pass a value-type you get a copy of the value, and when you pass a reference type you get a copy of the reference, with all that that implies. (That is, the copy of the reference still points to the same instance as the original reference.)
I use scare quotes around the "copy" above because Swift does a lot of optimization; wherever possible, it doesn't copy until there's a mutation or the possibility of mutation. Since parameters are immutable by default, this means that most of the time no copy actually happens.
It is always pass-by-value when the parameter is not inout.
It is always pass-by-reference if the parameter is inout. However, this is somewhat complicated by the fact you need to explicitly use the & operator on the argument when passing to an inout parameter, so it may not fit the traditional definition of pass-by-reference, where you pass the variable directly.
Here is a small code sample for passing by reference.
Avoid doing this, unless you have a strong reason to.
func ComputeSomeValues(_ value1: inout String, _ value2: inout Int){
value1 = "my great computation 1";
value2 = 123456;
}
Call it like this
var val1: String = "";
var val2: Int = -1;
ComputeSomeValues(&val1, &val2);
The Apple Swift Developer blog has a post called Value and Reference Types that provides a clear and detailed discussion on this very topic.
To quote:
Types in Swift fall into one of two categories: first, “value types”,
where each instance keeps a unique copy of its data, usually defined
as a struct, enum, or tuple. The second, “reference types”, where
instances share a single copy of the data, and the type is usually
defined as a class.
The Swift blog post continues to explain the differences with examples and suggests when you would use one over the other.
When you use inout with an infix operator such as += then the &address symbol can be ignored. I guess the compiler assumes pass by reference?
extension Dictionary {
static func += (left: inout Dictionary, right: Dictionary) {
for (key, value) in right {
left[key] = value
}
}
}
origDictionary += newDictionaryToAdd
And nicely this dictionary 'add' only does one write to the original reference too, so great for locking!
Classes and structures
One of the most important differences between structures and classes is that structures are always copied when they are passed around in your code, but classes are passed by reference.
Closures
If you assign a closure to a property of a class instance, and the closure captures that instance by referring to the instance or its members, you will create a strong reference cycle between the closure and the instance. Swift uses capture lists to break these strong reference cycles
ARC(Automatic Reference Counting)
Reference counting applies only to instances of classes. Structures and enumerations are value types, not reference types, and are not stored and passed by reference.
Classes are passed by references and others are passed by value in default.
You can pass by reference by using the inout keyword.
Swift assign, pass and return a value by reference for reference type and by copy for Value Type
[Value vs Reference type]
If compare with Java you can find matches:
Java Reference type(all objects)
Java primitive type(int, bool...) - Swift extends it using struct
struct is a value type so it's always passed as a value. let create struct
//STEP 1 CREATE PROPERTIES
struct Person{
var raw : String
var name: String
var age: Int
var profession: String
// STEP 2 CREATE FUNCTION
func personInformation(){
print("\(raw)")
print("name : \(name)")
print("age : \(age)")
print("profession : \(profession)")
}
}
//allow equal values
B = A then call the function
A.personInformation()
B.personInformation()
print(B.name)
it have the same result when we change the value of 'B' Only Changes Occured in B Because A Value of A is Copied, like
B.name = "Zainab"
a change occurs in B's name. it is Pass By Value
Pass By Reference
Classes Always Use Pass by reference in which only address of occupied memory is copied, when we change similarly as in struct change the value of B , Both A & B is changed because of reference is copied,.
So I just finished reading this post and realized that Java is actually entirely pass-by-value. In one of the answers it noted some languages are pass by reference, but C and Java are not. Since I am mostly familiar with C and Java and not familiar with the languages listed as languages which support pass by reference, I suppose it's possible that I've never seen pass by reference. My question is whether passing a pointer to a pointer to an object as a parameter in C and then modifying what the pointer to the object points to is essentially achieving pass-by-reference.
Example:
If I have a pointer bar and wanted to "pass it by reference" so I could change the object it points to, I could pass a pointer to bar which we can call foo. foo would be passed to the function by value, but we could change the value bar points to within that function. Is that essentially pass by reference from bar's perspective? Or does pass by reference work differently?
Passing by reference is not only different from passing by value. It's also different from "passing by pointer", i. e. passing a pointer to an object by value.
However, you are correctly assuming that one can emulate passing by reference using pointers. Namely, if you have an object of type T, and you pass a pointer of type T * as a function argument that points to that object, then the function can modify the object. Something like this:
void change(int *n)
{
*n = 1337;
}
int i = 42;
printf("Before: %d\n", i); /* prints 42 */
change(&i);
printf("After: %d\n", i); /* prints 1337 */
That's why reference types are often implemented using pointers in some languages, but conceptually (and often syntactically and in some semantical details as well) they are different.