+= Operator

I have a main file rational.cxx and a header file rational.h.
Everything seems to be working except i can't get the +=, *=, -=, and /= operators to work in the rational.cxx file.
I have a function in rational.h so I can use it which is below.

Code:

---

int operator += (const rational& lhs, const rational& rhs)
{
    return lhs += rhs;
}

---

For example when I get to the point in the program where += is located in the rational.cxx file I get a Segmentation Fault. I can't find why it's doing this I'm hoping it's just a silly mistake I made. I forgot to say I'm using a class in the header file which is why I need to do this. So any help would be appreciated.

You are getting a seg fault most likely because your function is never ending recursion.

Also why is your += returning a bool instead of a rational reference?

I'm still new to this so I haven't really figured out what to use yet, I mean to type int though.

Unless you have a specific reason to do otherwise your math operators return a reference or a new object of the type of thing you are doing the operatioin on.

Also as a general rule your += -= *= /= and %= (if they are defined) are member functions.

The problem I see from the little bit of code is that you defined the += operator in terms of itself. So the compiler doesn't know how to do += on rational so it calls your function which sees the += again and class the function again. This keeps going until you run out of memory and try to access some that doesn't belong to you

Silly me, thanks for that I realized one of my problems. I have to take the numerator of lhs and rhs and add the two. Now I just have to figure out how to do that. Any ideas? lol

Quote:

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Originally Posted by StarOrbs

Silly me, thanks for that I realized one of my problems. I have to take the numerator of lhs and rhs and add the two. Now I just have to figure out how to do that. Any ideas? lol

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[edit]

Code:

---

int operator += (const rational& lhs, const rational& rhs)
{
    return lhs += rhs;
}

---

You don't list both objects as parameters. The lhs object is the one calling the operator+= method, so it should not be listed. You can access the lhs object's members directly by just using their names.

[/edit]

Presumably your objects lhs and rhs are defined something like this:

Code:

---

class myObject
{
    int x;
    int y;

[edit]don't need get/set methods-->methods of a class,
 like operator+=, can access private members directly[/edit]
....
}

---

So, you would do something like this inside operator+=:

[edit]
x += rhs.x;
[/edit]

and as Thantos pointed out, you want to return a reference to an object, which in this case I think should be the lhs object. That is so you can chain operations together like this:

myObject1 += myObject2 += myObject3;

Otherwise, I don't think you would need to return anything: your lhs object would already have the appropriate changes made inside the operator+= method. Also, you should be able to see that returning an int probably isn't correct. If you do this:

myObject1 += myObject2;

the return value is not assigned to a variable so it's discarded. If you wanted to assign the return value to a variable, you would need to do this:

int result = (myObject1 += myObject2);

Is that something you want to be doing? If not, the only time the return value would be important is in a chain:

myObject1 += (myObject2 += myObject3);

If the operator+= returns an int, then you are left with:

myObject += someInt;

and you probably don't even have an operator+= defined for an int type on the rhs. What you want to be left with is:

myObject += Object2;

I got the += working but for some reason when I got to *= I got snagged again. I'm trying to multiply, I'll just use your example.. x times a integer.. For example..
x * 8 ..
And in the .cxx file it would say x *= 8

I hope I said that right.

Think about what x * 8 means.

Its simply:
x + x + x + x + x + x + x + x

Don't you already have an operator that defines addition?

I tried to figure out what you mean but it won't let me multiply something that is part of an object by an integer.

Maybe this example will help you out:

Code:

---

#include <iostream>
using namespace std;

class Apple
{
        int size;

public:

        Apple(int s)
        {
                size = s;
        }
        Apple()
        {
                size = 0;
        }

        int getsize(){return size;}

        Apple& operator+=(const Apple& rhs)
        {
                size += rhs.size;
                return *this; 
                //'this' is a pointer to the object that is calling this function, i.e.
                //the object on the lhs, and dereferencing it gives you the object
        }

        Apple& operator*=(const Apple& rhs)
        {
                size *= rhs.size;
                return *this;
        }

};

int main()
{
        Apple A(5);
        Apple B(2);

        A += B;
        cout<<A.getsize()<<endl;

        A *= B;
        cout<<A.getsize()<<endl;

        return 0;
}

---

I understand what your typed but all that program does it ouput 0 to the screen. This is driving me nuts.

Let me post the code, maybe you guys will see where I'm going wrong. Don't worry about the rest of the code, I just need help with the operators.. += and such

Code:

---

#include <sstream>
`
class rational
{
  private:
        int num;
        int den;
 
  public:
        rational(int rn, int rd);
        rational();
        string tostring() const;
       
        int operator += (rational& rhs);
        int operator -= (rational& rhs);
        int operator *= (int& rhs);
        int operator /= (rational& rhs);
       
          int compare (const rational& rhs) const;
          int setden(int denom);
          int setnum(int numer);
        int denominator() const;
        int numerator() const;

       
};
ostream& operator << (ostream& os, const rational& r);



rational::rational(int rn , int rd)
  : num(rn), den(rd)
{}
 
rational::rational()
  : num(0), den(1)
{}


int rational::compare (const rational& rhs) const
{
        if(den == rhs.den)
            return 0;
    if(den > rhs.den)
            return 1;
    if(den < rhs.den)
            return -1;
}

int rational::setden(int denom)
{
        den = denom;
}

int rational::setnum(int numer)
{
        num = numer;
}

int rational::denominator() const
{
    return den;
}

int rational::numerator() const
{
    return num;
}

int rational::operator += (rational& rhs)
{
        num += rhs.num;
        den += rhs.den;
}

int rational::operator -= (rational& rhs)
{
    num -= rhs.num;
        den -= rhs.den;
}

int rational::operator *= (int& rhs)
{
    num *= rhs;
        den *= rhs;
}

int rational::operator /= (rational& rhs)
{
    num /= rhs.num;
        den /= rhs.den;
}

string rational::tostring() const
{
    ostringstream out;
    out << num << "/" << den;
    return out.str();
}


ostream& operator << (ostream& os, const rational& r)
{
    os << r.tostring();
    return os;
}

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Well your += doesn't really do what it should.

1/2 + 3/1 != 4/3

What is:
1/2 * 5? Isn't it 5/2?

So you can multiple 1 * 5 and leave the denomitor alone.

Or you can add 1/2 + 1/2 + 1/2 + 1/2 + 1/2

For rational numbers, multiplication is usually easier than addition, so it doesn't really make sense to implement the latter in terms of the former.

For multiplication you just multiply the two numerators and the two denominators.
For addition, you have to multiply the two denominators with each other, the first numerator with the second denominator and vice versa, and then add the numerators.

In both cases you finish by reducing the fraction.

Quote:

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I understand what your typed but all that program does it ouput 0 to the screen.

---

Well, I think that means one of two things:

1) Your compiler is junk.
2) Your copy and paste technique needs work.

I know how to do rationals, my problem is with the actually code. I've been messing around with everyone's suggestions and ideas. Still no luck.

Besides the problems you are having with the mechanics of compiling and running programs, you need to review some basics on how to declare, define, and call functions. Maybe the following will help.

Here is an example of a very simple function:

Code:

---

void display(int n)
{
        cout<<n<<endl;
}

---

All functions require a return type. In the display() function above, the return type is 'void', which stands for 'nada', 'nothing', 'zip'. That means, the function doesn't return anything. Functions also can have things called 'parameters'. The parameter in the function above is 'int n'. That means the function is expecting any integer to be sent to it. The integer value that is sent to the function will be stored in the variable 'n'.

Here is what the function looks like as part of a program:

Code:

---

#include <iostream> //contains cout

using namespace std;
//makes it so you don't have to write 'std::cout'
//instead you can just write 'cout'


//function:
void display(int n)
{
        cout<<n<<endl;
}


int main()
{
        int num = 10;
        display(num);

        return 0;
}

---

In main(), the function is called by this line:

display(num);

That instructs the program to send the value contained in the variable 'num' to the function named display(), and then execute the code in the function. Since num is equal to 10, 10 is sent to the display() function. The value 10 is then stored in the variable 'n', which is the function parameter. Then, the code in the function is executed. After the last line in the function is executed, execution returns to the place in main() where the function was called, and then the statements thereafter in main() are executed.

Here is another example of a function:

Code:

---

double multiplyBy3(int x)
{
        double temp = x * 3.0;
        return temp;
}

---

This time the return type is double(a double is a number with a decimal point like 2.5). The return type has to be specified as 'double' because the function returns a double value in this line:

return temp;

The multiplyBy3() function also has an int parameter, so it must be sent an int.

Adding the multiplyBy3() function to the previous program looks like this:

Code:

---

#include <iostream> //contains cout

using namespace std;
//makes it so you don't have to write 'std::cout'
//instead you can just write 'cout'

double multiplyBy3(int x)
{
        double temp = x * 3.0;
        return temp;
}

void display(int n)
{
        cout<<n<<endl;
}


int main()
{
        int num = 10;
        display(num);

        int number = 5;
        double result = multiplyBy3(number);
       
        cout<<result<<endl;

        return 0;
}

---

The function multiplyBy3() is called in main() by this line:

double result = multiplyBy3(number);

The function is sent the int value stored in number. Since number is equal to 5, 5 is sent to the multiplyBy3 function, and then 5 is stored in the variable 'x', which is the function parameter. Then, the code in the function is executed. When the return statment is executed:

return temp;

the function gets the value in temp, and sends it back to main(). Since temp equals 15.0, 15.0 is sent back to main(). When a function returns a value, the function call is replaced by the return value. So, this line in main():

double result = multiplyBy3(number);

becomes:

double result = 15.0;

Then, the line:

cout<<result<<endl;

displays the return value. Note that you cannot do this to display 'result':

display(result);

The display() function requires an int to be sent to it, not a double, and result is type double. Therefore, you would get an error if you tried that: the types have to match.

Quote:

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Originally Posted by 7stud

Well, I think that means one of two things:

1) Your compiler is junk.
2) Your copy and paste technique needs work.

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My copy paste technique is fine thank you very much... I'm using g++... I've been messing around with that program and all it ever does it output 0. It's like it never gets to int main()

because it doesn't output:
0
0

it outputs just
0

Also with the *= I'm trying to multiply a numerator that is part of the rational class by a normal intger for example 8
So what I'm trying to do is this:
third(rational class) *= 8;

Perhaps you're calling some old program?

I fixed my problem, thanks everyone for your patience and help.