How to overload the ++ operator for a enum in C++

Question

This is what I tried, but I see that overloading only increments the variable if I assign it to another variable. I.e, The value of the variable on which I do the increment does not increase. So, in the example below variable newDay is THU but currentDay remains WED.

How do I define the overload operator to increment variable I am doing the operation on?

typedef enum days {MON, TUE, WED, THU, FRI, SAT, SUN} days;

inline days operator++ (days& d) {
    return static_cast<days>((static_cast<int>(d) + 1) % 7);
}

int main(int argc, const char * argv[]) {
   days currentDay = WED;
   days newDay = ++currentDay;
   cout << "Current day: " << currentDay << ", Stored day: " << calendar[0] << ", New day: " << newDay << endl;
}

Why do you expect `static_cast(d) + 1) % 7` to increment something, and what it that something? — n. m. could be an AI, Jul 23 '17 at 06:19
`typedef enum days {...} days;` is an exercise in redundancy. Just define `enum days {...};`. C++ doesn't require the same acrobatics as C to introduce the enum as a type. — StoryTeller - Unslander Monica, Jul 23 '17 at 06:21
Consider using [scoped enum](http://en.cppreference.com/w/cpp/language/enum#Scoped_enumerations). — suhdonghwi, Jul 23 '17 at 06:25

score 3 · Accepted Answer · edited Oct 19 '21 at 13:42

If I modify your overloaded operator to this:

inline days operator++ (days const& d) {
    return static_cast<days>((static_cast<int>(d) + 1) % 7);
}

It still compiles, despite the fact I added a const specifier there. That's because you are not modifying d like the semantics of prefix ++ demand.

So make sure you do modify it, if you want the operator to have the desired effect:

inline days operator++ (days& d) {
    d = static_cast<days>((static_cast<int>(d) + 1) % 7);
    return d;
}

Without commenting on the validity of your own design, note that it is a widely held opinion that prefix operator++ should return a modifiable lvalue, like the built-ins do. Bear in mind if you find yourself writing code like ++x = y, you need to return a reference, i.e. date& operator++(date&).

Just be aware that the modulo operator is very slow. Increasing the value and comparing to the maximum is much faster. — The Science Boy, Feb 03 '19 at 08:51

Johnathan Gross · Answer 2 · 2017-07-23T07:00:12.057

You defined the postfix operator. The normal behavior of the postfix operator is to increment the value of its argument but return the original value. It should behave like this:

days operator++(days& d,int){
    days temp=d;
    d=static_cast<days>((static_cast<int>(d) + 1) % 7);
    return temp;
}

What you want is the prefix operator. This increments the value of the argument and returns a reference to its arguments. It should look like this:

days& operator++(days& d){
    d=static_cast<days>((static_cast<int>(d) + 1) % 7);
    return d;
}

How to overload the ++ operator for a enum in C++

2 Answers2