operator overloading

**dontoo** · 02-19-2010

I overloaded << operator to format output in this fashion
"(" variable1 ", " variable2 ", " variable3 ")";
Lets say that my Vector class has another methods and I want to output those methods in normal way. How to undone overloading operator <<? What to write and where in the code?

Code:

#include <d3dx10.h>
#include <iostream>
using namespace std;

// Overload the  "<<" operators so that we can use cout to 
// output D3DXVECTOR3 objects.

ostream& operator<<(ostream& os, D3DXVECTOR3& v)
{
	os << "(" << v.x << ", " << v.y << ", " << v.z << ")";
	return os;
}

int main()
{
	// Using constructor, D3DXVECTOR3(FLOAT x, FLOAT y, FLOAT z);
	D3DXVECTOR3 u(1.0f, 2.0f, 3.0f);

	// Using constructor, D3DXVECTOR3(CONST FLOAT *);
	float x[3] = {-2.0f, 1.0f, -3.0f};
	D3DXVECTOR3 v(x);

	// Using constructor, D3DXVECTOR3() {};
	D3DXVECTOR3 a, b, c, d, e;  

	// Vector addition: D3DXVECTOR3 operator + 
	a = u + v;

	// Vector subtraction: D3DXVECTOR3 operator - 
	b = u - v;

	// Scalar multiplication: D3DXVECTOR3 operator * 
	c = u * 10;

	// ||u||
	float L = D3DXVec3Length(&u);

	// d = u / ||u||
	D3DXVec3Normalize(&d, &u);

	// s = u dot v
	float s = D3DXVec3Dot(&u, &v);

	// e = u x v
	D3DXVec3Cross(&e, &u, &v);

	cout << "u             = " << u << endl;
	cout << "v             = " << v << endl;
	cout << "a = u + v     = " << a << endl;
	cout << "b = u - v     = " << b << endl;
	cout << "c = u * 10    = " << c << endl;
	cout << "d = u / ||u|| = " << d << endl;
	cout << "e = u x v     = " << e << endl;
	cout << "L = ||u||     = " << L << endl;
	cout << "s = u.v       = " << s << endl;

	return 0;
}

**laserlight** · 02-19-2010

I do not really understand your question, but it sounds like you should just write another function to print in the exact way that you want.

**hk_mp5kpdw** · 02-19-2010

Originally Posted by dontoo

I overloaded << operator to format output in this fashion
"(" variable1 ", " variable2 ", " variable3 ")";
Lets say that my Vector class has another methods and I want to output those methods in normal way. How to undone overloading operator <<? What to write and where in the code?

How do you output a method? Do you mean produce output from within a method? What's the normal way? Either you handle the output in a different way within those methods on an individual basis or you can do something that I've done using a static data member and member function that sets and stores the current output type so that the print routine can do something different depending on what the value is:

Code:

#include <iostream>

class foo
{
public:
    enum out_type {TYPE1,TYPE2,TYPE3};
    int data;
    foo(int _data = 0) : data(_data)
    {
    }
    std::ostream& print(std::ostream& os) const
    {
        if(output == TYPE1)
            return os << "This is format 1: " << data;
        else if(output==TYPE2)
            return os << "This is format 2: \"" << data << '\"';
        else if(output==TYPE3)
            return os << "This is format 3: ***" << data << "***";
        else return os;
    }
    static void SetOutputType(out_type typ)
    {
        output = typ;
    }
private:
    static out_type output;
};

foo::out_type foo::output;

std::ostream& operator<<(std::ostream& os, const foo& rhs)
{
    return rhs.print(os);
}

int main()
{
    foo bar1, bar2(10);

    foo::SetOutputType(foo::TYPE1);
    std::cout << bar1 << '\n' << bar2 << '\n' << std::endl;

    foo::SetOutputType(foo::TYPE2);
    std::cout << bar1 << '\n' << bar2 << '\n' << std::endl;

    foo::SetOutputType(foo::TYPE3);
    std::cout << bar1 << '\n' << bar2 << '\n' << std::endl;

    return 0;
}

Output is as follows:

Code:

This is format 1: 0
This is format 1: 10

This is format 2: "0"
This is format 2: "10"

This is format 3: ***0***
This is format 3: ***10***

**dontoo** · 02-19-2010

That will work, thx

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