overloading operators

**thracian** · 06-23-2008

this code is frustrating me; it's an example from cplusplus.com(classes II)

i can't differentiate between which members are the constructors, why the overloading function is even called, and why saying, "CVector c;" is even valid. Are all of the members constructors(except for x,y)? what the hell is this, "CVector () {};"? oh, and why is cvector repeated so many times on the overloaded function call?

Code:

#include <iostream>
using namespace std;

class CVector {
  public:
    int x,y;
    CVector () {};
    CVector (int,int);
    CVector operator + (CVector);
};

CVector::CVector (int a, int b) {
  x = a;
  y = b;
}

CVector CVector::operator+ (CVector param) {
  CVector temp;
  temp.x = x + param.x;
  temp.y = y + param.y;
  return (temp);
}

int main () {
  CVector a (3,1);
  CVector b (1,2);
  CVector c;
  c = a + b;
  cout << c.x << "," << c.y;
  return 0;
}

rgetgtrgvtrgvtr

**tabstop** · 06-23-2008

Let's start with the easy one: saying "CVector c;" is valid because you defined what to do when some idiot tried to construct a CVector without two ints: if the parameter list is () -- that is, empty -- do

Code:

{ }

-- that is, nothing.

In the overloaded operator:

Code:

CVector CVector::operator+ (CVector param)

You know how functions work: return type; class to which the function belongs; type of parameter.

**thracian** · 06-23-2008

Originally Posted by tabstop

Let's start with the easy one: saying "CVector c;" is valid because you defined what to do when some idiot tried to construct a CVector without two ints: if the parameter list is () -- that is, empty -- do

Code:

{ }

-- that is, nothing.

In the overloaded operator:

Code:

CVector CVector::operator+ (CVector param)

You know how functions work: return type; class to which the function belongs; type of parameter.

All right, so then what calls the overloaded function?

**tabstop** · 06-23-2008

Well, it's name is "+". Do you have any "+" in your code?

**thracian** · 06-23-2008

Originally Posted by tabstop

Well, it's name is "+". Do you have any "+" in your code?

So then what is the arguement that is sent, or 'param'?

**tabstop** · 06-23-2008

Do you happen to possibly have a CVector after your +, that could be considered to be added to the other vector?

**thracian** · 06-23-2008

Originally Posted by tabstop

Do you happen to possibly have a CVector after your +, that could be considered to be added to the other vector?

the b object comes after the operator, i don't understand what you mean....

The function operator+ of class CVector is the one that is in charge of overloading the addition operator (+). This function can be called either implicitly using the operator, or explicitly using the function name:

Code:

c = a + b;
c = a.operator+ (b);

in the case of the example, the operator+ is called implicitly with the + sign? i still don't understand what arguement is sent..

**tabstop** · 06-23-2008

Well, what's the thing after the + sign (function call)? b, it looks like. So hey. b is the parameter. The two lines you gave are exactly 100% the same. But rather than make you type all that you can lose the word operator, and since the parentheses are redundant, you can lose them here too.

**thracian** · 06-23-2008

When the constructor is called for the a and b objects:
CVector a (3,1);
CVector b (1,2);

a.x is assigned as 3 and a.y=1 as well as for b?

but then when this is called:

Code:

CVector CVector::operator+ (CVector param) {
  CVector temp;
  temp.x = x + param.x;
  temp.y = y + param.y;
  return (temp);
}

it refers to the x and y alone which is what?

**tabstop** · 06-23-2008

Originally Posted by thracian

When the constructor is called for the a and b objects:
CVector a (3,1);
CVector b (1,2);

a.x is assigned as 3 and a.y=1 as well as for b?

Looks like b.x is 1 and b.y is 2.

Originally Posted by thracian

but then when this is called:

Code:

CVector CVector::operator+ (CVector param) {
  CVector temp;
  temp.x = x + param.x;
  temp.y = y + param.y;
  return (temp);
}

it refers to the x and y alone which is what?

As always with member functions, variables by themselves refer to the members of the object from which the function is called. You could think of x as this->x and y as this->y, if you really wanted to. Since this + was called from a (that's the part before the dot, right, the object whose method we are calling) x and y refer to a.x and a.y.

**thracian** · 06-23-2008

Ah, well I guess it makes sense now. Oh, and yeah, that's what I meant for the first part(b.x=1, and b.y=2), sorry for not making it clearer.

still trying to understand why cvector is repeated so many times. the tutorial says the format is this:

Code:

type operator sign (parameters) { /*...*/ }

it just doesn't look right because in previous examples, the name of the constuctor is used with the scope operator instead of repeated like here.

Code:

CVector CVector::operator+ (CVector param)

is operator+ a member? is it just something you include because of the format?

this is my first day learning classes so thanks for bearing with me man

**tabstop** · 06-23-2008

That's because the constructor is a function with no return type (not even void) whose name is the name of the class.

So CVector::CVector(int, int) is the same as before with the return type, the name of the class, and the name of the function.

Here you are defining the function CVector (taking two int parameters) as a member of the class CVector (and since the names are the same, it will be the constructor). Before you were defining the function operator+ (which you can abbreviate as just + when calling it, though not defining it) as a member of the class CVector, and returning a CVector.

Edit: ooh another example:

Code:

int CVector::taxicab_distance(const CVector& other) {
    return abs(x-other.x)+abs(y-other.y)
}

See if you can work out how that would be called.

**thracian** · 06-23-2008

Originally Posted by tabstop

That's because the constructor is a function with no return type (not even void) whose name is the name of the class.

So CVector::CVector(int, int) is the same as before with the return type, the name of the class, and the name of the function.

Here you are defining the function CVector (taking two int parameters) as a member of the class CVector (and since the names are the same, it will be the constructor). Before you were defining the function operator+ (which you can abbreviate as just + when calling it, though not defining it) as a member of the class CVector, and returning a CVector.

Edit: ooh another example:

Code:

int CVector::taxicab_distance(const CVector& other) {
    return abs(x-other.x)+abs(y-other.y)
}

See if you can work out how that would be called.

ohhh! cvector was also the return type, goddamnit, i completely missed that the first time.

int - return type
cvector- class
taxecab_distance - member function

thanks man

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