在std命名空间中operator <<是什么做的？

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.

Answer 2

I always understood the reason for char , unsigned char and char* being defined as non-member functions outside the basic_ostream class is so it's easier to overload them.

See, all other operator<< use the char ones as building blocks. So if you want to create an ostream specialized to a specific charT character type and / or traits trait type - you only have to specialize these operator<< functions.

Had they been member functions, you'd have had to specialized the entire class (basically recreating all the class member functions).

I'm not sure that's the only reason, but that's how I've always seen it.