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Show Lecture.ExtensibleNotMutable as a slide show.

CS253 Extensible Not Mutable

Extensible, not mutable

• C++ is extensible, not mutable.
• extensible: you can extend the language. You can add functionality.
• mutable: you cannot change the language. 2+2 will always be 4.

Extending

int main() {
    string name = "Jack";
    cout << name + 36782;
}

c.cc:3: error: no match for 'operator+' in 'name + 36782' (operand types are 
   'std::__cxx11::string' {aka 'std::__cxx11::basic_string<char>'} and 'int')

That failed, because you can’t use + on a std::string and an int.

Extending

string operator+(const string &s, int n) {
    return s + to_string(n);
}

int main() {
    string name = "Jack";
    cout << name + 21825;
}

Jack21825

Hooray! C++ is extensible! I extended the language to include string + int.

More Extending

class Foo {
  public:
    Foo(int value) : n(value) { }
    int n;
};

int main() {
    Foo f(42);
    cout << "The foo is " << f << '\n';
}

c.cc:9: error: no match for 'operator<<' in 'std::operator<< 
   <std::char_traits<char> >(std::cout, ((const char*)"The foo is ")) << f' 
   (operand types are 'std::basic_ostream<char>' and 'Foo')

Of course, that failed. << has no idea what to do with an ostream (output stream) and a Foo.

More Extending

class Foo {
  public:
    Foo(int value) : n(value) { }
    int get() const { return n; }
  private:
    int n;
};
ostream & operator<<(ostream &os, const Foo &f) {
    os << f.get();
    return os;
}

int main() {
    Foo f(42);
    cout << "The foo is " << f << '\n';
}

The foo is 42

Hooray! I extended the language to include ostream + Foo. Why the return os in operator<<?

More Extending

class Foo {
  public:
    Foo(int value) : n(value) { }
    int get() const { return n; }
  private:
    int n;
};
ostream & operator<<(ostream &os, const Foo &f) {
    return os << f.get();
}

int main() {
    Foo f(42);
    cout << "The foo is " << f << '\n';
}

The foo is 42

operator<< can simply return the result of <<. Why is operator<< not a method of Foo?

How operators work

• For an operator to be a method of class Foo, it must take a Foo as its left-hand argument.
• It would only take a single explicit argument—the other, implicit, left-hand argument would be *this, the current object.
• In the case of cout << f, the left-hand argument is an ostream.
• You can’t add methods to class ostream.
  • It’s not yours.
  • That class is finished.
• Therefore, in this case, operator<< is not a method; it is, instead, a free function.

Mutation

Consider this surprising result:

int main() {
    char now[] = "21:40:57";
    cout << now+3 << '\n';
}

40:57

• now is an array of char.
• An array name (now) is the same as a pointer to the first element, or a char *.
• Adding 3 to a pointer moves it forward three items.

I don’t like this. I want it to simply append a "3" to the string. Let’s redefine it!

Futility

string operator+(char *p, int n) {
    return p + to_string(n);
}

int main() {
    char now[] = "21:40:57";
    cout << now+3 << '\n';
}

c.cc:1: error: ‘std::string operator+(char*, int)’ must have an argument of 
   class or enumerated type

That failed. I can’t mutate the language. char * + int is already defined, and I can’t change it.

Besides, at least one of the arguments has to be a non-built-in type. It wouldn’t work if I tried to define operator%(char *p, int n)

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