Asked  7 Months ago    Answers:  5   Viewed   31 times

I heard a few people recommending to use enum classes in C++ because of their type safety.

But what does that really mean?

 Answers

14

C++ has two kinds of enum:

  1. enum classes
  2. Plain enums

Here are a couple of examples on how to declare them:

 enum class Color { red, green, blue }; // enum class
 enum Animal { dog, cat, bird, human }; // plain enum 

What is the difference between the two?

  • enum classes - enumerator names are local to the enum and their values do not implicitly convert to other types (like another enum or int)

  • Plain enums - where enumerator names are in the same scope as the enum and their values implicitly convert to integers and other types

Example:

enum Color { red, green, blue };                    // plain enum 
enum Card { red_card, green_card, yellow_card };    // another plain enum 
enum class Animal { dog, deer, cat, bird, human };  // enum class
enum class Mammal { kangaroo, deer, human };        // another enum class

void fun() {

    // examples of bad use of plain enums:
    Color color = Color::red;
    Card card = Card::green_card;

    int num = color;    // no problem

    if (color == Card::red_card) // no problem (bad)
        cout << "bad" << endl;

    if (card == Color::green)   // no problem (bad)
        cout << "bad" << endl;

    // examples of good use of enum classes (safe)
    Animal a = Animal::deer;
    Mammal m = Mammal::deer;

    int num2 = a;   // error
    if (m == a)         // error (good)
        cout << "bad" << endl;

    if (a == Mammal::deer) // error (good)
        cout << "bad" << endl;

}

Conclusion:

enum classes should be preferred because they cause fewer surprises that could potentially lead to bugs.

Tuesday, June 1, 2021
 
Daveel
answered 7 Months ago
53

It's a "pointer to member" - the following code illustrates its use:

#include <iostream>
using namespace std;

class Car
{
    public:
    int speed;
};

int main()
{
    int Car::*pSpeed = &Car::speed;

    Car c1;
    c1.speed = 1;       // direct access
    cout << "speed is " << c1.speed << endl;
    c1.*pSpeed = 2;     // access via pointer to member
    cout << "speed is " << c1.speed << endl;
    return 0;
}

As to why you would want to do that, well it gives you another level of indirection that can solve some tricky problems. But to be honest, I've never had to use them in my own code.

Edit: I can't think off-hand of a convincing use for pointers to member data. Pointer to member functions can be used in pluggable architectures, but once again producing an example in a small space defeats me. The following is my best (untested) try - an Apply function that would do some pre &post processing before applying a user-selected member function to an object:

void Apply( SomeClass * c, void (SomeClass::*func)() ) {
    // do hefty pre-call processing
    (c->*func)();  // call user specified function
    // do hefty post-call processing
}

The parentheses around c->*func are necessary because the ->* operator has lower precedence than the function call operator.

Tuesday, June 1, 2021
 
lewiguez
answered 7 Months ago
89

Because dynamic_cast can only downcast polymorphic types, so sayeth the Standard.

You can make your class polymoprphic by adding a virtual destructor to the base class. In fact, you probably should anyway (See Footnote). Else if you try to delete a B object through an A pointer, you'll evoke Undefined Behavior.

class A
{
public:
  virtual ~A() {};
};

et voila!

Footnote

There are exceptions to the "rule" about needing a virtual destructor in polymorphic types.
One such exception is when using boost::shared_ptr as pointed out by Steve Jessop in the comments below. For more information about when you need a virtual destructor, read this Herb Sutter article.

Thursday, June 10, 2021
 
ojrac
answered 6 Months ago
20

Varargs methods create an array.

public static void foo(Object... args) {
  System.out.println(args.length);
}

This works, because of the implicit array creation. EnumSet is a class designed to be very, very fast, so by creating all the extra overloads they can skip the array creation step in the first few cases. This is especially true since in many cases Enum don't have that many elements, and if they do, the EnumSet might not contain all of them.

Javadoc for EnumSet<E> of(E e1, E e2, E e3, E e4, E e5):

Creates an enum set initially containing the specified elements. Overloadings of this method exist to initialize an enum set with one through five elements. A sixth overloading is provided that uses the varargs feature. This overloading may be used to create an enum set initially containing an arbitrary number of elements, but is likely to run slower than the overloadings that do not use varargs.

Tuesday, August 3, 2021
 
Mirko
answered 4 Months ago
40

As far as I know non-template function is always preferred to template function during overload resolution.

This is true, only when the specialization and the non template are exactly the same. This is not the case here though. When you call uct u3(u1) The overload sets gets

uct(const uct &)
uct(uct &) // from the template

Now, since u1 is not const it would have to apply a const transformation to call the copy constructor. To call the template specialization it needs to do nothing since it is an exact match. That means the template wins as it is the better match.

To stop this one thing you can do is use SFINAE to limit the template function to only be called when T is not a uct. That would look like

template <typename T, std::enable_if_t<!std::is_same_v<uct, std::decay_t<T>>, bool> = true>
uct(T &&) { std::cerr << "template" << std::endl; }
Saturday, September 11, 2021
 
ramdemon
answered 3 Months ago
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