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CS253 Struct And Class

struct

struct is the old C way of doing things.
In C, there were no methods—only data members.
A struct is still useful for holding unadorned data.

struct Point {
    int x[100], y;
};
Point p;
p.x[90] = 12345;
p.y = 67890;
cout << p.x[90] << ',' << p.y << '\n';

12345,67890

class

A class has methods and data. Data is usually private.
A class should have public methods to be useful, (e.g., a ctor) but private methods are also common.
public: and private: define sections of the class.
- It’s not a label on each item, as in Java.
A class is the same as a struct, except methods, data, and inheritance are, by default, private for a class, and public for a struct.

class Point {
  public:
    Point(int a, int b) { x = a; y = b; }   // constructor (alias ctor)
    int get_x() const { return x; }         // accessor (getter)
    int get_y() const { return y; }         // accessor (getter)
    void go_right() { x++; }                // mutator (setter)
  private:
    int x, y;                               // Hands off!
};

this

The special variable this is a pointer (not a reference, as in Java) to the current object. It’s not often needed. Here is an unnecessary use of this:

class Point {
  public:
    Point(int a, int b) { x = a; y = b; }
    double radius() const { return sqrt(x*x + y*this->y); }
  private:
    int x, y;
};
Point p(300,400);
cout << p.radius() << '\n';

Inside Point::radius(), this is of type const Point *, because the method is const.

Output

To make a class work for output, define operator<< appropriately:

class Point {
  public:
    Point(int a, int b) { x = a; y = b; }   // ctor
    int get_x() const { return x; }         // accessor (getter)
    int get_y() const { return y; }         // accessor (getter)
    void go_right() { x++; }                // mutator (setter)
  private:                                  // Mine!
    int x, y;
};

ostream &operator<<(ostream &out, const Point &p) {
    return out << p.get_x() << ',' << p.get_y();
}

int main() {
    Point p(12, 34);
    cout << p << '\n';      // invoke operator<<
}

12,34

Example

class Quoted {
    string s;
  public:
    Quoted(const string &word) : s(word) { }
    string get() const { return "“" + s + "”"; }
};

ostream &operator<<(ostream &os,
                    const Quoted &rhs) {
    return os << rhs.get();
}

int main() {
    Quoted name("Bob");
    cout << "Hi, " << name << "!\n";
}

Hi, “Bob”!

s is private
const reference argument
member initialization
const accessor (getter)
operator<< not a method
half-indent for public:
return ostream reference

mutable

mutable is a hack. It indicates that, even if a method is const, and hence should not modify data members, it’s ok to modify this data member. mutable is used when you want to change a data member without really changing the state of the object: caching a computed value, counting instances, etc.

mutable #1

class Process {
  public:
    pid_t get_pid() const {
        return getpid(); // assume this is slow
    }
};

Process p;
cout << p.get_pid() << ' ' << p.get_pid() << '\n';

1748869 1748869

This is the straightforward solution. However, if getpid() is slow, then we’re calling the slow function twice. The process id (pid) doesn’t change! We should only call getpid() once.

mutable #2

class Process {
    const pid_t pid = getpid();
  public:
    pid_t get_pid() const {
        return pid;
    }
};

Process p;
cout << p.get_pid() << ' ' << p.get_pid() << '\n';

1748870 1748870

That works, but always calls getpid(). What if we never need the process id? We called getpid() for no reason!

mutable #3

class Process {
    mutable pid_t pid = -1;
  public:
    pid_t get_pid() const {
        if (pid == -1)
            pid = getpid(); // assume this is slow
        return pid;
    }
};

Process p;
cout << p.get_pid() << ' ' << p.get_pid() << '\n';

1748871 1748871

This is the solution. getpid() is called once, at most, and not called at all if the process ID is never required. (Arguably, a local static variable inside the method would be simpler.)

methods: in-line and not

Methods are declared in the class, but can be defined inside or outside of the class.

class Quoted {
    string s;
  public:
    Quoted(const string &word) : s(word) { }
    string get() const; // declaration only
};
string Quoted::get() const {
    return "“" + s + "”";
}
ostream &operator<<(ostream &os, const Quoted &rhs) {
    return os << rhs.get();
}

int main() {
    Quoted name("Slartibartfast");
    cout << "I am " << name << ".\n";
}

I am “Slartibartfast”.

Protection

Class members (data & methods) can be public, private, or protected.
The default is private for a class, public for a struct.
These are sections, not labels on individual members.
public: anyone can access them
private: nobody except other methods of this class can access them
protected: can be accessed by this class, and by immediately derived class

Protection

class Foo {     // A class, with a variable of each type
  public:
    int pub;    // I’m public!
  private:
    int priv;   // I’m private!
  protected:
    int prot;   // I’m a little teapot, short & stout.
};

int main() {
    Foo f;

    f.pub = 1;          // great
    // f.priv = 2;      // nope
    // f.prot = 2;      // nope

    return f.pub;
}

Friends

Friend Declarations

One class can declare another class/method/function to be its friend:

class Foo {
    friend class Bar;
    friend double Zip::zulu(int) const;
    friend int alpha();
    friend std::ostream &operator<<(std::ostream &os, const Foo &);
  private:
    int x,y;
};

All the methods of class Bar can access our data.
So can the method zulu(int) const of the class Zip.
So can the function (not method) alpha().
So can the function operator<<(ostream &, const Foo &).

Restrictions

Friendship is offered, not demanded
- class A, via friend class B, offers friendship to class B.
Friendship is not symmetric
- If C is D’s friend, then D is not necessarily C’s friend.
Friendship is not transitive
- If E & F are friends, and F & G are friends, then E & G are strangers.
Friendship is not inherited
- If H is I’s friend, and I is the parent of J, then H & J are strangers.
Inheritance does not imply friendship
- If K is L’s parent, then K & L are strangers.

Don’t go nuts

C++ programmers tend to fall in love with friend declarations, and overuse them.
friend should be your last tool, not your first one.
- Like casting.
Minimize access: prefer method friendship to class friendship.
Create accessors (getters) and mutators (setters) instead.
If you’re using friend declarations to avoid the overhead of method calls, then you have no faith in current compilers. They’re quite good at inlining methods.

Counting

Imagine that, for some reason, we wanted to keep track of how many instances of a class are extant.
We’d need a counter.
Every time a ctor is called, increment the counter.
Every time a dtor is called, decrement the counter.

Try #1

class Foo {
  public:
    Foo() { counter++; }
    ~Foo() { counter--; }
    int get_count() const { return counter; }
  private:
    int counter=0;
};

int main() {
    Foo a, b, c, d, e;
    cout << e.get_count() << '\n';
}

Why doesn’t this work?

Each instance of Foo has its own counter, and each one gets initialized to zero, then incremented to one.

Try #2

int counter=0;

class Foo {
  public:
    Foo() { counter++; }
    ~Foo() { counter--; }
    int get_count() const { return counter; }
};

int main() {
    Foo a, b, c, d, e;
    cout << e.get_count() << '\n';
}

That works, but it has an evil global variable. 👹

Try #3

class Foo {
  public:
    Foo() { counter++; }
    ~Foo() { counter--; }
    int get_count() const { return counter; }
  private:
    static int counter=0;
};

int main() {
    Foo a, b, c, d, e;
    cout << e.get_count() << '\n';
}

c.cc:7: error: ISO C++ forbids in-class initialization of non-const static 
   member 'Foo::counter'

That failed.

Try #4

class Foo {
  public:
    Foo() { counter++; }
    ~Foo() { counter--; }
    int get_count() const { return counter; }
  private:
    static int counter;
};
int Foo::counter = 0;

int main() {
    Foo a, b, c, d, e;
    cout << e.get_count() << '\n';
}

There we go! Only a single, shared, counter, with class scope.

static class variables

A static member variable:

has class scope
can be public, protected, or private
is shared between all instances of this class
must be externally defined, typically in the implementation file.

static methods

A static method:

has class scope
can be public, protected, or private
can be called without an instance of the class
can’t access non-static data members, because those go with instances
can’t call non-static data methods, because those go with instances.

const methods

class Ratio {
    int top, bottom;
  public:
    Ratio(int a, int b) : top(a), bottom(b) { }
    double get_real() {
        return top / double(bottom);
    }
};

int main() {
    const Ratio f(355,113);
    cout << f.get_real() << '\n';
}

c.cc:12: error: passing 'const Ratio' as 'this' argument discards qualifiers

Oops—can’t call a non-const method on a const object.

const methods

class Ratio {
    int top, bottom;
  public:
    Ratio(int a, int b) : top(a), bottom(b) { }
    double get_real() const {
        return top / double(bottom);
    }
};

int main() {
    const Ratio f(355,113);
    cout << f.get_real() << '\n';
}

3.14159

Making Ratio::get_real() const fixed it. It also properly indicates that get_real() doesn’t change the object.

Rationale

Yeah, but …, that’s stupid.
Nobody would ever declare a const Ratio like that.
Fair enough.
Let’s look at a realistic scenario.

const methods

class Ratio {
    int top, bottom;
  public:
    Ratio(int a, int b) : top(a), bottom(b) { }
    double get_real() const {
        return top / double(bottom);
    }
};

void show_ratio(Ratio r) {
    cout << r.get_real() << '\n';
}

int main() {
    Ratio f(355,113);
    show_ratio(f);
}

3.14159

That call to show_ratio copies an object by value. Expensive! 😱 Well, OK, it’s only two ints—big deal. If the object had a lot of data, or used dynamic memory, then copying would be expensive.

const methods

class Ratio {
    int top, bottom;
  public:
    Ratio(int a, int b) : top(a), bottom(b) { }
    double get_real() const {
        return top / double(bottom);
    }
};

void show_ratio(const Ratio &r) {
    cout << r.get_real() << '\n';
}

int main() {
    Ratio f(355,113);
    show_ratio(f);
}

3.14159

Look! A const Ratio! r is now const, so get_real() must be const.

Implementation

The compiler’s implementation of a const method is simple; it regards the hidden implicit pointer argument this as pointing to a const object, which explains the message:

class Foo {
  public:
    void zip() {}
};

int main() {
    const Foo f;
    f.zip();
}

c.cc:8: error: passing 'const Foo' as 'this' argument discards qualifiers

In Foo::zip(), this is type Foo *, which can’t point to const Foo f.

Poor Strategy

Some students think, “figuring out which methods should be const is too much work; I’ll add const later”.
- This is a disaster in the making.
- Half an hour before the homework is due, the student decides that it’s const time.
- The student adds const to a single method.
- This causes problems with all the non-const methods called by the newly-const method.
- Making those methods const causes problems with the non-const methods that they call.
- The student gives up, and turns in uncompilable code.
- The student remembers only that it’s all const’s fault, and vows never to use const again.

Good Strategy

If, instead, the student had declared methods const as they were written, then there wouldn’t be an “OMG, I’m doomed” moment.

constexpr variable

We’ve already seen a constexpr variable, whose value is fixed at compile-time:

constexpr double pi=3.1415926535897932;
pi = 22.0/7.0;

c.cc:2: error: assignment of read-only variable 'pi'

It’s a crime against language to call it a “variable”, because it doesn’t vary.
const means you don’t change it (but others can).
constexpr means that it simply does not change. The compiler can use that information to produce smaller/faster code, by substituting the actual value for the constexpr variable.
There are also constexpr methods.

constexpr methods

class Foo {
  public:
    constexpr int bar() { return 42; }
};

Foo f;
cout << f.bar();

I would’ve thought that the constexpr would go after the (). Too bad for me!
Think of the function returning a constexpr int, that is, an int that can be computed at compile-time.
This means that the compiler must calculate this value at compile-time.

constexpr methods

class Foo {
  public:
    constexpr int bar() { return getpid(); }
};

Foo f;
cout << f.bar();

c.cc:3: error: call to non-'constexpr' function '__pid_t getpid()'

constexpr means “Listen, compiler: you must execute this function at compile-time (given constexpr arguments). If you can’t, complain and die.”
Hence, the compiler verifies your claim that the function is doable at compile-time. Think of it as an assertion.

Temporary files

Often, in programming, you need a temporary file, with a unique name.
In Linux, these typically reside in the directory /tmp.
Give it a meaningful name, in case of a misbehaving program.
Two people might run your program simultaneously, but they can’t use the same temporary file.
Functions such as mkstemp help create a unique file.
You should remove the file, even if your code takes an early return or an exception is caught.
A class can help with that—dtors love doing cleanup.

Conversion methods

class Tempfile {
  public:
    Tempfile() { close(mkstemp(name)); }
    ~Tempfile() { remove(name); }
    string tempname() const { return name; }
  private:
    char name[18] = "/tmp/cs253-XXXXXX";
};

Tempfile t;
const string fname = t.tempname();
cout << fname << '\n';

/tmp/cs253-otSuy9

Six alphanumerics means more than 56 billion possible filenames.
Ctor creates a temporary file; dtor destroys it.
The method .tempname() gets the name of the temporary file.
- Really, that’s the only thing that you do with it.
Can we make this even easier?

Conversion methods

Let’s add an operator string method:

class Tempfile {
  public:
    Tempfile() { close(mkstemp(name)); }
    ~Tempfile() { remove(name); }
    operator string() const { return name; }
  private:
    char name[18] = "/tmp/cs253-XXXXXX";
};

Tempfile t;
const string fname = t;
cout << fname << '\n';

/tmp/cs253-qsNoyb

Just treat it like a std::string, and it works!