Closure Cannot Implicitly Capture a Mutating Self Parameter

Closure cannot implicitly capture a mutating self parameter

The short version

The type owning your call to FirebaseRef.observeSingleEvent(of:with:) is most likely a value type (a struct?), in which case a mutating context may not explicitly capture self in an @escaping closure.

The simple solution is to update your owning type to a reference once (class).

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The longer version

The observeSingleEvent(of:with:) method of Firebase is declared as follows

func observeSingleEvent(of eventType: FIRDataEventType, 
     with block: @escaping (FIRDataSnapshot) -> Void)

The block closure is marked with the @escaping parameter attribute, which means it may escape the body of its function, and even the lifetime of self (in your context). Using this knowledge, we construct a more minimal example which we may analyze:

struct Foo {
    private func bar(with block: @escaping () -> ()) { block() }

    mutating func bax() {
        bar { print(self) } // this closure may outlive 'self'
        /* error: closure cannot implicitly capture a 
                  mutating self parameter              */
    }
}

Now, the error message becomes more telling, and we turn to the following evolution proposal was implemented in Swift 3:

• SE-0035: Limiting inout capture to @noescape contexts

Stating [emphasis mine]:

Capturing an inout parameter, including self in a mutating
method, becomes an error in an escapable closure literal, unless the
capture is made explicit (and thereby immutable).

Now, this is a key point. For a value type (e.g. struct), which I believe is also the case for the type that owns the call to observeSingleEvent(...) in your example, such an explicit capture is not possible, afaik (since we are working with a value type, and not a reference one).

The simplest solution to this issue would be making the type owning the observeSingleEvent(...) a reference type, e.g. a class, rather than a struct:

class Foo {
    init() {}
    private func bar(with block: @escaping () -> ()) { block() }

    func bax() {
        bar { print(self) }
    }
}

Just beware that this will capture self by a strong reference; depending on your context (I haven't used Firebase myself, so I wouldn't know), you might want to explicitly capture self weakly, e.g.

FirebaseRef.observeSingleEvent(of: .value, with: { [weak self] (snapshot) in ...

SwiftUI Escaping closure captures mutating 'self' parameter

An object's initializer cannot do anything asynchronous. Its job is to produce the object immediately, with all its properties initialized.

Escaping closure captures mutating 'self' parameter: struct

As you have found, the quick solution is to use a reference type, a class. But why is this the case?

Swift structs are value types, so they are immutable. You can mark a function as mutating to indicate to the compiler that a function mutates the struct, but what does that actually mean?

Consider a simple struct:

struct Counter {
   var count

   init(_ count: Int = 0)
   {
       self.count = count
   }

   mutating func increment() {
       self.count+=1
   }
}

Now, try and assign an instance of this to a let constant:

let someCounter = Counter()
someCounter.increment()
print(someCounter.count)

You will get an error; you need to use a var.

var someCounter = Counter()
someCounter.increment()
print(someCounter.count)

What actually happens when you call a mutating func is that a new Counter is created, with the new count and it is assigned to someCounter. It is effectively saying someCounter = Counter(someCounter.count+1)

Now, think what would happen if you could mutate self in an escaping closure - That new Counter is going to be created at some unspecified time in the future, but execution has already moved on. It is too late to update someCounter.

The other advantage of using a class, as you have found, is that you can use ObservableObject, which makes updating your SwiftUI views much easier.

SwiftUI: What's 'Escaping closure captures mutating 'self' parameter' and how to fix it

EDIT:
Seems like you cannot mutate structs anymore in escaping closure without removing @escaping which not be possible in your case. You might want to consider changing your implementation to a class.

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Structs are immutable. Which mean they cannot be mutated. In your case you are modifying the value of self.bool1 = true which is changing the value of self.

~~A better way (IMO) would be to create a mutating func to do your firebase call and update the values inside mutating function.~~

Also, you shouldn’t use State property wrappers in models. They should only be used in views.

Closure cannot implicitly capture self parameter. Swift

It's because you're using struct. Since structs are value, they are copied (with COW-CopyOnWrite) inside the closure for your usage. It's obvious now that copied properties are copied by "let" hence you can not change them. If you want to change local variables with callback you have to use class. And beware to capture self weakly ([weak self] in) to avoid retain-cycles.

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