operator const char* overwrites (?) my another variable in strange way

Question

#include <iostream>
#include <sstream>

class Vector
{
    double _x;
    double _y;
public:
    Vector(double x, double y) : _x(x), _y(y) {}
    double getX() { return _x; }
    double getY() { return _y; }

    operator const char*()
    {
        std::ostringstream os;
        os << "Vector(" << getX() << "," << getY() << ")";
        return os.str().c_str();
    }
};
int main()
{
    Vector w1(1.1,2.2);
    Vector w2(3.3,4.4);
    std::cout << "Vector w1(" << w1.getX() << ","<< w1.getY() << ")"<< std::endl;
    std::cout << "Vector w2(" << w2.getX() << ","<< w2.getY() << ")"<< std::endl;

    const char* n1 = w1;
    const char* n2 = w2;

    std::cout << n1 << std::endl;
    std::cout << n2 << std::endl;
}

Output of this program:

$ ./a.out 
Vector w1(1.1,2.2)
Vector w2(3.3,4.4)
Vector(3.3,4.4)
Vector(3.3,4.4)

I don't understand why I get the output. It seems that "const char* n2 = w2;" overwrites n1 and then I get two times "Vector(3.3,4.4)". Could anybody explain me this phenomena?

Try strdup (or the best C++ way to do it) when you return the char. — Maxime Chéramy, Nov 19 '13 at 13:01

score 13 · Accepted Answer · edited Nov 19 '13 at 13:19

It's undefined behavior that sometimes works (by luck), sometimes doesn't.

You're returning a pointer to a temporary local object. The pointer to a temporary local object being the internals of the string object obtained by calling os.str().c_str().

If you want print those objects easily by cout, you can overload operator << for output streams. Like:

ostream& operator<<(ostream& out, const Vector &a)
{
   std::ostringstream os;
   os << "Vector(" << a.getX() << "," << a.getY() << ")";
   out << os.str();

   return out;
}

and then

std::cout << w1 << std::endl;
std::cout << w2 << std::endl;

operator const char* overwrites (?) my another variable in strange way

1 Answers1