subscript-operator

I have overloaded [] operator in my class Interval to return minutes or seconds . But I am not sure how to assign values to minutes or second using [] operator. For example : I can use this statement cout << a[1] << "min and " << a[0] << "sec" << endl; but I want to overload [] operator, so that I can even assign values to minutes or seconds using a[1] = 5; a[0] = 10; My code : #include <iostream> using namespace std; class Interval { public: long minutes; long seconds; Interval(long m, long s) { minutes = m + s / 60; seconds = s % 60; } void Print() const { cout << minutes << ':' << seconds <

I am beginner with the Swift having no advance knowledge with operators. I have the following class class Container { var list: [Any] = []; } I want to implement the operator subscript [] in order to access the data from list . I need something like this: var data: Container = Container() var value = data[5] // also data[5] = 5 Also I want to be able to write something like this: data[1][2] Is it possible considering that element 1 from Container is an array ? Thanx for help. It looks like there are 2 questions here. 1. How can I enable subscripting on my own custom class? To enable

This question asks whether one can use subscripting with CKRecord in Swift. While I already knew how to do what the questioner wanted, every permutation of it gives me a stack overflow: subscript(key: String) -> CKRecordValue? { get { return objectForKey(key) as CKRecordValue? } set { setObject(newValue, forKey: key) } } The stack overflow occurs in the getter. (I've never tried the setter, so it may occur there, too.) I've tried implementing with objectForKey: , objectForKeyedSubscript: , and valueForKey: . All produce the same result: a stack overflow. This is very strange, since CKRecord is

This question already has an answer here: With arrays, why is it the case that a[5] == 5[a]? 18 answers Why does x[y] == y[x] in c++? [duplicate] 3 answers As I have learned, one can write the following code: char *a = new char[50]; for (int i = 0; i < 50; ++i) { i[a] = '5'; } It compiles. It works. It does exactly the same as char *a = new char[50]; for (int i = 0; i < 50; ++i) { a[i] = '5'; } Is it just because: a[b] is implemented as a macro *(a + b) by default and the fact that both code samples are valid is just an accident/compiler specific it's standardized somewhere and the outcome of