Overloading normal functionality ???

[sarathy]

Hi all,
I just wanted to know if it is possible to overload the
original meaning of operators. I read that it is not possible to
change the operator precedence by using operator overloading. Other
than that i think that the operator functionality can be overloaded.
But in the following code, the operator + has been overloaded on type
int. But it does'nt seem to work as the operator fn was not invoked.

In the following code,

# include <iostream>
using namespace std;

namespace blackbox
{
class A
{
public:
int a;
int operator *(int a);
};

int A::operator * (int b)
{
return a+b;
}
}

int main()
{
using namespace blackbox;

int a=10;
int b=20;
int c=a*b;
cout << c << endl;

return 0;
}

OUTPUT:
---------------
200
Thanks and regards,
Sarathy

[Marcus Kwok]

xiaohuamao <ca************ **@gmail.comwro te:

To overload the operator * for two ints, you
should use int operator * (const int &, const int &) instead.

You cannot overload operators for primitive types.

--
Marcus Kwok
Replace 'invalid' with 'net' to reply