operator int()

the book im reading has an exercise to make a class Int and overload the basic operators. there is 2 things i dont understand. it asks for bounds checking, but i dont know why the answer in the back of the book needs to convert it to long double. the other thing i dont get is 'operator int()'. when does this come in play in the class? its a conversion from Int to type int, i know that much. but when is this operator called. thanks in advance for responses :)

Code:

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//////////////////////////class Int/////////////////////////
class Int
{
private:
        int i;
public:
        Int(): i(0)  //set default value (constructor)
        {        }

        Int(int integer):        i(integer)          //1-arg constructor
        {        }
       
        void put_int()                        //display int
        { cout << i << endl; }

        void get_int()        //get int
        { cin >> i; }

        operator int()  //conversion operator from Int to int
        {        return i;        }

        Int operator + (Int i2)
        {        return limit_check(long double(i) + long double(i2));        }

        Int operator - (Int i2)
        {        return limit_check(long double(i) - long double(i2));        }

        Int operator * (Int i2)
        {        return limit_check(long double(i) * long double(i2));        }

        Int operator / (Int i2)
        {        return limit_check(long double(i) / long double(i2));        }

        Int limit_check(long double num)
        {
                if(num > 2147482648.0L || num < -2147482648.0L)
                {
                        cout << "\nOverflow error:  Exiting." << endl;
                        exit(1);
                }

                return Int(int(num));
        }

};

/////////////////////////////////////////////////////////////

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Quote:

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when is this operator called

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It is called when a variable of type int is expected. For example:

Code:

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void func(int nParam)
{
}

int main()
{
    Int a=5;

    func(a);
}

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In this case 'a' will call operator int() to resolve itself to type 'int'. Quite ingenious really.

>>Int a=5;
int a=5;

Geez.

Maybe you misunderstand, XSquared. Int is a class.

Whoops. You're right. Name your clases differently, geez! :p

Quote:

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Originally Posted by bennyandthejets

In this case 'a' will call operator int() to resolve itself to type 'int'. Quite ingenious really.

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thanks for the reply.:D
why is it required to convert it to long double and check the bounds?

Quote:

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Name your clases differently, geez!

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I agree. I would have called it CInt.

Quote:

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why is it required to convert it to long double and check the bounds?

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I don't know, it does seem a bit pointless doesn't it.