Thread: Overloading operators some problems with (*,/,=) operators.

Hello i am doing some exercises with operators to train them, and I have got problem with those three cause they work bad for me
(machine doesnt accept actions with this operators)
I can use only 2 times double.
Here is my code:

Code:

class MayCal
{
 protected:
      double liczba;
      
 public:
 MayCal()
 {
 liczba = 0;
};
 MayCal(double a)
 {
 liczba = a;
}
MayCal operator/ (MayCal const&);
MayCal operator* (MayCal const&);
MayCal & operator =(MayCal const& );
MayCal MayCal::operator/ (MayCal const& ex)
{    
if(ex.liczba == 0)
{
     
   MayCal tmp(liczba = 2012);
   
  return tmp;
     
   
}
else
{
       MayCal tmp(liczba/ex.liczba);
   return tmp;  
}    

MayCal MayCal::operator* (MayCal const& ex)
{    
   MayCal tmp(liczba * ex.liczba);
   return tmp;    
}
MayCal& MayCal::operator= (MayCal const& ex)
{
     liczba = ex.liczba;
}

In operator/ I have got If cause in exercise always when i do divison
MyCal a = 500 , c = 0;
a/c it must give 2012
500/0 = must give 2012.

I think i got something bad with this division operator on my examples it work and operator= completly doesnt work and operator* works too on examples but machine tests both operator* and operator/ so i dont know which is bad or maybe both of them.
On my examples operator* and operator/ works, but machine says it gives bad answer, and operator= totally doesnt work.

When i do in main()

Code:

MayCal a = 50 , b = 30;
cout << "a++ = " << a=b << endl;

I get error [Error] no match for 'operator<<' in 'b << std::endl'

I dont know that operator= is wrong made or my operator << doesnt work for this example.

Could somebody look at those code? I will be very thankful.

I add machine accepts for me operator+, operator- which have got same build as operator* and operator/ i think something is wrong with my division.

A definition of operator*() like this

Code:

MayCal MayCal::operator* (MayCal const& ex)
{   
   MayCal tmp(liczba * ex.liczba);
   return tmp;   
}

has to appear outside the class definition.

Code:

class MayCal
{
         //    the definition in the form above cannot be in here
};

// it can be here

All assignment operators have lower precedence than operator <<. So this

Code:

MayCal a = 50 , b = 30;
cout << "a++ = " << a=b << endl;

is equivalent to

Code:

MayCal a = 50 , b = 30;
cout << "a++ = " << a= (b << endl);

which requires class MyCal to have an operator<<() that accepts a stream manipulator - which yours does not. You probably intended something like

Code:

MayCal a = 50 , b = 30;
cout << "a++ = " << (a= b) << endl;    // note brackets

For that to work, MyCal's assignment operator needs to return a reference to something (otherwise the result of the assignment cannot be streamed to cout). Currently, your operator=() does not have a return statement, so the function returns an invalid reference to the caller. That causes the "cout .... "statement to exhibit undefined behaviour (since, to output something, cout's streaming operator accesses a non-existent object via the reference).

And your MyCal needs to have a friend operator<<() in order to be streamed.

No, not without a fairly liberal interpretation of "something like this".

The variable r in your function is auto (i.e. it ceases to exist when the function returns) and your operator[] returns a reference to it. A function should never return a pointer or reference to an auto variable which is local to that function, since it will cause the caller to exhibit undefined if it uses that pointer or reference.

Generally speaking, operator[] is only needed by classes that contain a collection of objects (e.g. an array) and need to provide caller access to elements of those collections. For example, if MayCal contains a member named x which is an array of 10 float, and the caller needs to access elements of that array, there are various types of operator[] that can be defined.

Although not unheard of, it is unusual for an object to contain a collection of objects of the same type (for example, an apple does not generally contain a collection of apples).


whiteflag's description, incidentally, is only one form of operator[]. There are others. The return value is not required to be a reference, for example (albeit, doing it that way imposes other constraints on the caller).