Question

Hey. Is it possible to overload operator<< for primitive types? Fx lets say that I want to write a std::endl each time want to write a int. Can I overload operator<< for int, so that it automatic puts a std::endl to the output? I have tried with this,

std::ostream& operator<<(std::ostream& strm, int & i) {    strm << i << std::endl;    return strm; } 

but it doesn’t work. I cant recall the compiler error message, but I think that I’m getting operator overloading all wrong any ways. I try to call the above overloaded operator<< in this way,

int main() {    int i = 2;    std::out<<'Here is an int ' << i;     return 0; } 

But it doesn’t work at all. Maybe I can’t overload POD types?

Answer 1

As zabzonk said, the standard library provides an (ostream&, int) overload so you can’t define another.

To simulate what you were doing (though it is completely pointless in its present form 🙂 :

class EndlinedInteger { public:     EndlinedInteger(int i) : i(i) { }     friend ostream& operator<<(ostream&, EndlinedInteger const&); private:     int i; };  ostream& operator<<(ostream& out, EndlinedInteger const& ei) {     out << ei.i << endl;     return out; }  int main() {    EndlinedInteger i = 2;    std::cout<<'Here is an int ' << i; }