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CS253 Rule Of Three

The Rule of Three

The Rule of Three states that if a class defines any of these methods, called the “Big Three”, then it should probably define all of them:

• copy constructor
• copy assignment operator
• destructor

There are certainly exceptions to this rule. Fortunately, the rule says “probably”.

When none of the Big Three are needed

• Often, none of the Big Three are required. Consider:

class Hero {
    string name;
  public:
    Hero(string n) : name(n) { }
    auto who() const { return name; }
};
Hero h1("Superman"), h2(h1);
cout << h1.who() << ' ' << h2.who();

Superman Superman

• For class Hero, the default big three are all fine:
  • copy constructor: copy-constructs the two strings
  • copy assignment operator: copies the two strings
  • destructor: does nothing
• Great!

No problem

• class Hero worked fine, because the string objects didn’t require any maintenance.
• Sure, the string objects, internally, do all sorts of dynamic memory manipulation, with calls to new[] and delete[], but that’s string’s business. Hero’s dtor does not have to delete anything.
• What if I (foolishly) decide that string objects have too much overhead, and so I rewrite Hero to use const char * instead?

Problem

class Hero {
    char *name;
  public:
    Hero(const char *n) {
        name = new char[strlen(n)+1];
        copy(n, n+strlen(n)+1, name);
    }
    ~Hero() { delete[] name; }
    auto who() const { return name; }
};
Hero h1("Superman");
cout << h1.who();

Superman

It works well so far. Let’s try using the compiler-generated copy ctor.

Problem

class Hero {
    char *name;
  public:
    Hero(const char *n) {
        name = new char[strlen(n)+1];
        copy(n, n+strlen(n)+1, name);
    }
    ~Hero() { delete[] name; }
    auto who() const { return name; }
};
Hero h1("Superman"), h2(h1);
cout << h1.who() << ' ' << h2.who();

free(): double free detected in tcache 2
SIGABRT: Aborted

Problem with copy ctor

• What did the compiler-generated copy ctor do? This:

  Hero(const Hero &rhs) : name(rhs.name) { }
  

• It simply copied the member variable.
• Trouble is, copying the pointers doesn’t copy the data. It just copies the pointers. After calling the copy ctor, we have two pointers that both point to the same memory.
• When h2 falls out of scope, its dtor is called, which does delete[] name.
• Then, h1 falls out of scope. Its dtor is called, which does delete[] name. But … that memory has already been freed!
• >boom<

Problem with assignment operator

• Of course, the default assignment operator would be similar:

  Hero &operator=(const Hero &rhs) { name = rhs.name; }
  

• Like the default copy ctor, it just copies the pointer, so we have two pointers pointing to the same location.
• One gets deleted, the other is dangling.
• >boom<

Solution #1

class Hero {
    char *name;
  public:
    Hero(const char *n) {
        name = new char[strlen(n)+1];
        copy(n, n+strlen(n)+1, name);
    }
    Hero(const Hero &rhs) {
        name = new char[strlen(rhs.name)+1];
        copy(rhs.name, rhs.name+strlen(rhs.name)+1, name);
    }
    Hero &operator=(const Hero &rhs) {
        delete[] name;
        name = new char[strlen(rhs.name)+1];
        copy(rhs.name, rhs.name+strlen(rhs.name)+1, name);
        return *this;
    }
    ~Hero() { delete[] name; }
    auto who() const { return name; }
};
Hero h1("Superman"), h2(h1);
cout << h1.who() << ' ' << h2.who();

Superman Superman

Solution #2

class Hero {
    char *name = nullptr;
    void set(const char *p) {
        delete[] name;
        name = new char[strlen(p)+1];
        copy(p, p+strlen(p)+1, name);
    }
  public:
    Hero(const char *n) { set(n); }
    Hero(const Hero &rhs) { set(rhs.name); }
    Hero &operator=(const Hero &rhs) {
        set(rhs.name);
        return *this;
    }
    ~Hero() { delete[] name; }
    auto who() const { return name; }
};
Hero h1("Superman"), h2(h1);
cout << h1.who() << ' ' << h2.who();

Superman Superman

It gets worse

• There are two more methods of interest:
  • the move constructor: Hero(Hero &&rhs)
  • the move assignment operator: operator=(Hero &&rhs)
• These methods don’t copy from the right-hand-side object, they move from it. They steal the value.
• Why on earth would you want to do that?

Move methods

• Consider this code:

string where="base", what="ball";
string sport = where+what;
cout << sport;

baseball

• This built a temporary string containing "base" + "ball", and copied that temporary string to sport.
• Now we have two copies of the string "baseball": one in the unnamed temporary variable, and one in sport.
• That’s a shame, given that the string inside of the temporary variable is no longer needed.
• Wouldn’t it be nice if we could just move the string from the temporary variable to sport?

Move methods

• That’s what the move ctor and move assignment operator do.
• They only get called for variables that the compiler has designated as temporaries—no longer needed, doomed to die soon.
• They steal the value from the temporary variable.
• They must leave the temporary variable in a consistent state, such that when its dtor is called, we won’t have a problem.
• For a string, steal the char *, set the source char * to nullptr.
• Neither move method is necessary. They’re of no use for plain scalar data.
• Big Three + two move methods = Big Five.