गुरुवार, 13 अक्तूबर 2011
बुधवार, 12 अक्तूबर 2011
An Example of Operator Overloading
An Example of Operator Overloading
Complex a(1.2,1.3); //this class is used to represent complex numbers
Complex b(2.1,3); //notice the construction taking 2 parameters for the real and imaginary part
Complex c = a+b; //for this to work the addition operator must be overloaded
The addition without having overloaded operator + could look like this:
Complex c = a.Add(b);
This piece of code is not as readable as the first example though--we're
dealing with numbers, so doing addition should be natural. (In contrast
to cases when programmers abuse this technique, when the concept
represented by the class is not related to the operator--ike using + and
- to add and remove elements from a data structure. In this cases
operator overloading is a bad idea, creating confusion.)In order to allow operations like Complex c = a+b, in above code we overload the "+" operator. The overloading syntax is quite simple, similar to function overloading, the keyword operator must be followed by the operator we want to overload:
class Complex
{
public:
Complex(double re,double im)
:real(re),imag(im)
{};
Complex operator+(const Complex& other);
Complex operator=(const Complex& other);
private:
double real;
double imag;
};
Complex Complex::operator+(const Complex& other)
{
double result_real = real + other.real;
double result_imaginary = imag + other.imag;
return Complex( result_real, result_imaginary );
}
The assignment operator can be overloaded similarly. Notice that we did
not
have to call any accessor functions in order to get the real and
imaginary
parts from the parameter other since the overloaded operator is a member
of the
class and has full access to all private data.
Alternatively, we could have defined the addition operator globally and
called a member to do the actual work. In that case, we'd also have to
make the method a friend of the class, or use an accessor method to get
at the private data:
friend Complex operator+(Complex);
Complex operator+(const Complex &num1, const Complex &num2)
{
double result_real = num1.real + num2.real;
double result_imaginary = num1.imag + num2.imag;
return Complex( result_real, result_imaginary );
}
Why would you do this? when the operator is a class member, the first object in the
expression must be of that particular type. It's as if you were writing:
Complex a( 1, 2 );
Complex a( 2, 2 );
Complex c = a.operator=( b );
when it's a global function, the implicit or user-defined conversion can allow the operator to act even if the
first operand is not exactly of the same type:
Complex c = 2+b; //if the integer 2 can be converted by the Complex class, this expression is valid
By the way, the number of operands to a function is fixed; that is, a binary
operator takes two operands, a unary only one, and you can't change it. The
same is true for the precedence of operators too; for example the multiplication operator
is called before addition. There are some operators that need the first
operand to be assignable, such as : operator=, operator(), operator[] and operator->, so
their use is restricted just as member functions(non-static), they can't be
overloaded globally. The operator=, operator& and operator, (sequencing) have
already defined meanings by default for all objects, but their meanings can be
changed by overloading or erased by making them private.
Another intuitive meaning of the "+" operator from the STL string class which is overloaded to do concatenation:
string prefix("de");
string word("composed");
string composed = prefix+word;
Using "+" to concatenate is also allowed in Java, but note that this is
not extensible to other classes, and it's not a user defined behavior.
Almost all operators can be overloaded in C++:
+ - * / % ^ & |
~ ! , = < > <= >=
++ -- << >> == != && ||
+= -= /= %= ^= & = |= *=
<<= >>= [ ] ( ) -> ->* new delete
The only operators that can't be overloaded are the operators for scope
resolution (::), member selection (.), and member selection through a
pointer to a function(.*). Overloading assumes you specify a behavior
for an operator that acts on a user defined type and it can't be used
just with general pointers. The standard behavior of operators for
built-in (primitive) types cannot be changed by overloading, that is,
you can't overload operator+(int,int). The logic(boolean) operators have by the default a short-circuiting way of acting in expressions with multiple boolean operations. This means that the expression:
if(a && b && c)
will not evaluate all three operations and will stop after a false one
is found. This behavior does not apply to operators that are overloaded
by the programmer.
Even the simplest C++ application, like a "hello world" program, is using overloaded operators. This is due to the use of this technique almost everywhere in the standard library (STL). Actually the most basic operations in C++ are done with overloaded operators, the IO(input/output) operators are overloaded versions of shift operators(<<, >>). Their use comes naturally to many beginning programmers, but their implementation is not straightforward. However a general format for overloading the input/output operators must be known by any C++ developer. We will apply this general form to manage the input/output for our Complex class:
friend ostream &operator<<(ostream &out, Complex c) //output
{
out<<"real part: "<>(istream &in, Complex &c) //input
{
cout<<"enter real part:\n";
in>>c.real;
cout<<"enter imag part: \n";
in>>c.imag;
return in;
}
Notice the use of the friend keyword in order to access the private
members in the above implementations. The main distinction between them
is that the operator>> may encounter unexpected errors for
incorrect input, which will make it fail sometimes because we haven't
handled the errors correctly. A important trick that can be seen in this
general way of overloading IO is the returning reference for
istream/ostream which is needed in order to use them in a recursive
manner:
Complex a(2,3);
Complex b(5.3,6);
cout<
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