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How base-class is defined in Swift?
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Answer / iosraj
In Swift the classes are not inherited from the base class and the classes that you define without specifying its superclass, automatically becomes the base-class.
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What do you do when you realize that your app is prone to crashing?
Swift defines the AnyObject type alias to represent instances of any reference type, and it’s internally defined as a protocol. Consider the following code: var array = [AnyObject]() struct Test {} array.append(Test()) This code generates a compilation error, with the following error message: Type 'Test' does not conform to protocol 'AnyObject' The failure is obvious because a struct is a value and not a reference type, and as such it doesn’t implement and cannot be cast to the AnyObject protocol. Now consider the following code: var array = [AnyObject]() array.append(1) array.append(2.0) array.append("3") array.append([4, 5, 6]) array.append([7: "7", 8: "8"]) struct Test {} array.append(Test()) The array array is filled in with values of type respectively int, double, string, array and dictionary. All of them are value types and not reference types, and in all cases no error is reported by the compiler. Why?
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