what does operator<< in std namespace do?

Question

Of course following code works (it calls std::cout::operator<<):

cout << 1 << '1' << "1" << endl;

Happened to find there is also std::operator<<, and it seems it only works for char or char* arguments:

operator<<(cout, '1'); // ok
operator<<(cout, "1"); // ok
operator<<(cout, 1);   // error

So why do we need this operator and how to use it?

Thanks.

Answer 1

operator<<(cout, '1'); // ok
operator<<(cout, "1"); // ok
operator<<(cout, 1);   // error

The first two works because they invoke non-member functions taking two arguments. The functions which takes char and char const* as argument are defined as non-member (free) functions.

However, the function which takes int as argument is defined as member function, which means the third one needs to invoke a member function. If you invoke it as non-member function, then int would have to be converted into some type for which there exists a non-member function. So when this conversion is considered, it results in ambiguity because there are many possible conversions equally good.

As said, this should work:

cout.operator<<(1); //should work

As to why some functions are defined as members and others as non-members, I don't know the answer. It requires lots of study of the proposals and the decisions that led to this design of the library.