Asked  7 Months ago    Answers:  5   Viewed   17 times

Playing around with Swift, coming from a Java background, why would you want to choose a Struct instead of a Class? Seems like they are the same thing, with a Struct offering less functionality. Why choose it then?

 Answers

67

According to the very popular WWDC 2015 talk Protocol Oriented Programming in Swift (video, transcript), Swift provides a number of features that make structs better than classes in many circumstances.

Structs are preferable if they are relatively small and copiable because copying is way safer than having multiple references to the same instance as happens with classes. This is especially important when passing around a variable to many classes and/or in a multithreaded environment. If you can always send a copy of your variable to other places, you never have to worry about that other place changing the value of your variable underneath you.

With Structs, there is much less need to worry about memory leaks or multiple threads racing to access/modify a single instance of a variable. (For the more technically minded, the exception to that is when capturing a struct inside a closure because then it is actually capturing a reference to the instance unless you explicitly mark it to be copied).

Classes can also become bloated because a class can only inherit from a single superclass. That encourages us to create huge superclasses that encompass many different abilities that are only loosely related. Using protocols, especially with protocol extensions where you can provide implementations to protocols, allows you to eliminate the need for classes to achieve this sort of behavior.

The talk lays out these scenarios where classes are preferred:

  • Copying or comparing instances doesn't make sense (e.g., Window)
  • Instance lifetime is tied to external effects (e.g., TemporaryFile)
  • Instances are just "sinks"--write-only conduits to external state (e.g.CGContext)

It implies that structs should be the default and classes should be a fallback.

On the other hand, The Swift Programming Language documentation is somewhat contradictory:

Structure instances are always passed by value, and class instances are always passed by reference. This means that they are suited to different kinds of tasks. As you consider the data constructs and functionality that you need for a project, decide whether each data construct should be defined as a class or as a structure.

As a general guideline, consider creating a structure when one or more of these conditions apply:

  • The structure’s primary purpose is to encapsulate a few relatively simple data values.
  • It is reasonable to expect that the encapsulated values will be copied rather than referenced when you assign or pass around an instance of that structure.
  • Any properties stored by the structure are themselves value types, which would also be expected to be copied rather than referenced.
  • The structure does not need to inherit properties or behavior from another existing type.

Examples of good candidates for structures include:

  • The size of a geometric shape, perhaps encapsulating a width property and a height property, both of type Double.
  • A way to refer to ranges within a series, perhaps encapsulating a start property and a length property, both of type Int.
  • A point in a 3D coordinate system, perhaps encapsulating x, y and z properties, each of type Double.

In all other cases, define a class, and create instances of that class to be managed and passed by reference. In practice, this means that most custom data constructs should be classes, not structures.

Here it is claiming that we should default to using classes and use structures only in specific circumstances. Ultimately, you need to understand the real world implication of value types vs. reference types and then you can make an informed decision about when to use structs or classes. Also, keep in mind that these concepts are always evolving and The Swift Programming Language documentation was written before the Protocol Oriented Programming talk was given.

Tuesday, June 1, 2021
 
mdevils
answered 7 Months ago
19

Judging by the wording of your question (you used the word "hide"), you already know what is going on here. The phenomenon is called "name hiding". For some reason, every time someone asks a question about why name hiding happens, people who respond either say that this called "name hiding" and explain how it works (which you probably already know), or explain how to override it (which you never asked about), but nobody seems to care to address the actual "why" question.

The decision, the rationale behind the name hiding, i.e. why it actually was designed into C++, is to avoid certain counter-intuitive, unforeseen and potentially dangerous behavior that might take place if the inherited set of overloaded functions were allowed to mix with the current set of overloads in the given class. You probably know that in C++ overload resolution works by choosing the best function from the set of candidates. This is done by matching the types of arguments to the types of parameters. The matching rules could be complicated at times, and often lead to results that might be perceived as illogical by an unprepared user. Adding new functions to a set of previously existing ones might result in a rather drastic shift in overload resolution results.

For example, let's say the base class B has a member function foo that takes a parameter of type void *, and all calls to foo(NULL) are resolved to B::foo(void *). Let's say there's no name hiding and this B::foo(void *) is visible in many different classes descending from B. However, let's say in some [indirect, remote] descendant D of class B a function foo(int) is defined. Now, without name hiding D has both foo(void *) and foo(int) visible and participating in overload resolution. Which function will the calls to foo(NULL) resolve to, if made through an object of type D? They will resolve to D::foo(int), since int is a better match for integral zero (i.e. NULL) than any pointer type. So, throughout the hierarchy calls to foo(NULL) resolve to one function, while in D (and under) they suddenly resolve to another.

Another example is given in The Design and Evolution of C++, page 77:

class Base {
    int x;
public:
    virtual void copy(Base* p) { x = p-> x; }
};

class Derived : public Base{
    int xx;
public:
    virtual void copy(Derived* p) { xx = p->xx; Base::copy(p); }
};

void f(Base a, Derived b)
{
    a.copy(&b); // ok: copy Base part of b
    b.copy(&a); // error: copy(Base*) is hidden by copy(Derived*)
}

Without this rule, b's state would be partially updated, leading to slicing.

This behavior was deemed undesirable when the language was designed. As a better approach, it was decided to follow the "name hiding" specification, meaning that each class starts with a "clean sheet" with respect to each method name it declares. In order to override this behavior, an explicit action is required from the user: originally a redeclaration of inherited method(s) (currently deprecated), now an explicit use of using-declaration.

As you correctly observed in your original post (I'm referring to the "Not polymorphic" remark), this behavior might be seen as a violation of IS-A relationship between the classes. This is true, but apparently back then it was decided that in the end name hiding would prove to be a lesser evil.

Tuesday, June 1, 2021
 
Nate
answered 7 Months ago
42

Is it faster to create these objects as class or as struct?

You are the only person who can determine the answer to that question. Try it both ways, measure a meaningful, user-focused, relevant performance metric, and then you'll know whether the change has a meaningful effect on real users in relevant scenarios.

Structs consume less heap memory (because they are smaller and more easily compacted, not because they are "on the stack"). But they take longer to copy than a reference copy. I don't know what your performance metrics are for memory usage or speed; there's a tradeoff here and you're the person who knows what it is.

Is it better to create these objects as class or as struct?

Maybe class, maybe struct. As a rule of thumb: If the object is :
1. Small
2. Logically an immutable value
3. There's a lot of them
Then I'd consider making it a struct. Otherwise I'd stick with a reference type.

If you need to mutate some field of a struct it is usually better to build a constructor that returns an entire new struct with the field set correctly. That's perhaps slightly slower (measure it!) but logically much easier to reason about.

Are objects on the heap and the stack processed equally by the garbage collector?

No, they are not the same because objects on the stack are the roots of the collection. The garbage collector does not need to ever ask "is this thing on the stack alive?" because the answer to that question is always "Yes, it's on the stack". (Now, you can't rely on that to keep an object alive because the stack is an implementation detail. The jitter is allowed to introduce optimizations that, say, enregister what would normally be a stack value, and then it's never on the stack so the GC doesn't know that it is still alive. An enregistered object can have its descendents collected aggressively, as soon as the register holding onto it is not going to be read again.)

But the garbage collector does have to treat objects on the stack as alive, the same way that it treats any object known to be alive as alive. The object on the stack can refer to heap-allocated objects that need to be kept alive, so the GC has to treat stack objects like living heap-allocated objects for the purposes of determining the live set. But obviously they are not treated as "live objects" for the purposes of compacting the heap, because they're not on the heap in the first place.

Is that clear?

Saturday, June 5, 2021
 
exxed
answered 6 Months ago
22

I ran into the same issue when trying to decode/encode the "edited" field on a Reddit Listing JSON response. I created a struct that represents the dynamic type that could exist for the given key. The key can have either a boolean or an integer.

{ "edited": false }
{ "edited": 123456 }

If you only need to be able to decode, just implement init(from:). If you need to go both ways, you will need to implement encode(to:) function.

struct Edited: Codable {
    let isEdited: Bool
    let editedTime: Int

    // Where we determine what type the value is
    init(from decoder: Decoder) throws {
        let container =  try decoder.singleValueContainer()

        // Check for a boolean
        do {
            isEdited = try container.decode(Bool.self)
            editedTime = 0
        } catch {
            // Check for an integer
            editedTime = try container.decode(Int.self)
            isEdited = true
        }
    }

    // We need to go back to a dynamic type, so based on the data we have stored, encode to the proper type
    func encode(to encoder: Encoder) throws {
        var container = encoder.singleValueContainer()
        try isEdited ? container.encode(editedTime) : container.encode(false)
    }
}

Inside my Codable class, I then use my struct.

struct Listing: Codable {
    let edited: Edited
}

Edit: A more specific solution for your scenario

I recommend using the CodingKey protocol and an enum to store all the properties when decoding. When you create something that conforms to Codable the compiler will create a private enum CodingKeys for you. This lets you decide on what to do based on the JSON Object property key.

Just for example, this is the JSON I am decoding:

{"type": "1.234"}
{"type": 1.234}

If you want to cast from a String to a Double because you only want the double value, just decode the string and then create a double from it. (This is what Itai Ferber is doing, you would then have to decode all properties as well using try decoder.decode(type:forKey:))

struct JSONObjectCasted: Codable {
    let type: Double?

    init(from decoder: Decoder) throws {
        // Decode all fields and store them
        let container = try decoder.container(keyedBy: CodingKeys.self) // The compiler creates coding keys for each property, so as long as the keys are the same as the property names, we don't need to define our own enum.

        // First check for a Double
        do {
            type = try container.decode(Double.self, forKey: .type)

        } catch {
            // The check for a String and then cast it, this will throw if decoding fails
            if let typeValue = Double(try container.decode(String.self, forKey: .type)) {
                type = typeValue
            } else {
                // You may want to throw here if you don't want to default the value(in the case that it you can't have an optional).
                type = nil
            }
        }

        // Perform other decoding for other properties.
    }
}

If you need to store the type along with the value, you can use an enum that conforms to Codable instead of the struct. You could then just use a switch statement with the "type" property of JSONObjectCustomEnum and perform actions based upon the case.

struct JSONObjectCustomEnum: Codable {
    let type: DynamicJSONProperty
}

// Where I can represent all the types that the JSON property can be. 
enum DynamicJSONProperty: Codable {
    case double(Double)
    case string(String)

    init(from decoder: Decoder) throws {
        let container =  try decoder.singleValueContainer()

        // Decode the double
        do {
            let doubleVal = try container.decode(Double.self)
            self = .double(doubleVal)
        } catch DecodingError.typeMismatch {
            // Decode the string
            let stringVal = try container.decode(String.self)
            self = .string(stringVal)
        }
    }

    func encode(to encoder: Encoder) throws {
        var container = encoder.singleValueContainer()
        switch self {
        case .double(let value):
            try container.encode(value)
        case .string(let value):
            try container.encode(value)
        }
    }
}
Monday, August 2, 2021
 
Gil
answered 4 Months ago
Gil
60

Principles Of Object Oriented Class Design (the "SOLID" principles)

  • SRP: The Single Responsibility Principle A class should have one, and only one, reason to change.
  • OCP: The Open Closed Principle You should be able to extend a classes behavior, without modifying it.
  • LSP: The Liskov Substitution Principle Derived classes must be substitutable for their base classes.
  • ISP: The Interface Segregation Principle Make fine grained interfaces that are client specific.
  • DIP: The Dependency Inversion Principle Depend on abstractions, not on concretions.

Source: http://butunclebob.com/ArticleS.UncleBob.PrinciplesOfOod

Sunday, August 8, 2021
 
user435216
answered 4 Months ago
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