How can I use cout << myclass

it's very easy, just implement :

std::ostream & operator<<(std::ostream & os, const myclass & foo)
{
os << foo.var;
return os;
}

You need to return a reference to os in order to chain the outpout (cout << foo << 42 << endl)

Typically by overloading operator<< for your class:

struct myclass {
int i;
};


std::ostream &operator<<(std::ostream &os, myclass const &m) {
return os << m.i;
}


int main() {
myclass x(10);


std::cout << x;
return 0;
}

You need to overload the << operator,

std::ostream& operator<<(std::ostream& os, const myclass& obj)
{
os << obj.somevalue;
return os;
}

Then when you do cout << x (where x is of type myclass in your case), it would output whatever you've told it to in the method. In the case of the example above it would be the x.somevalue member.

If the type of the member can't be added directly to an ostream, then you would need to overload the << operator for that type also, using the same method as above.

Alternative:

struct myclass {
int i;
inline operator int() const
{
return i;
}
};

Even though other answer provide correct code, it is also recommended to use a hidden friend function to implement the operator<<. Hidden friend functions has a more limited scope, therefore results in a faster compilation. Since there is less overloads cluttering the namespace scope, the compiler has less lookup to do.

struct myclass {
int i;


friend auto operator<<(std::ostream& os, myclass const& m) -> std::ostream& {
return os << m.i;
}
};


int main() {
auto const x = myclass{10};
std::cout << x;


return 0;
}