put_time into stream weird behavior - c++

Question

It was my understanding that overloaded operators are translated by the compiler into method calls, for example :
stream << "test";
Is the same as:
stream.operator<<("test");
If that is indeed the case, then why does this lead into a compilation error:
#include <iostream>
#include <iomanip>
#include <ctime>
int main()
{
std::ostream s {std::cout.rdbuf()};
auto t = std::time(nullptr);
auto tm = *std::localtime(&t);
s.operator<<(std::put_time(&tm, "%H:%M:%S"));//doesn't work
return 0;
}
While changing the doesn't work line to:
s << std::put_time(&tm, "%H:%M:%S");
Makes the code compile.
Both GCC and MSVC 2017 on C++14 seem to produce an error. Is this a compiler bug or are those two statements not equal?

Answer 1

Here's your code, but reduced:
struct X { friend void operator+(X, X); };
int main() {
X var;
var + var; // ok
var.operator+(var); // fail
}
The thing is, by explicitly calling the operator<<, you are relying on the fact that it is indeed an operator of the class (like T& T::operator<<(const U&);) and not an operator defined outside the class (because if the operator is not a member, you can't write var.operator<<(/*...*/)).
If you write var << something, then in addition to the operators defined in the class, lookup (called ADL) finds the operators defined outside of the class as well.
The operator can be defined outside of the class, because std::put_time's return type is specified to ([ext.manip]:
An object of unspecified type such that if out is an object of type basic_­ostream<charT, traits> then the expression out << put_­time(tmb, fmt) behaves as if it called f(out, tmb, fmt), where the function f is defined as: [...]
Because the requirement is of the form var << var2, nothing prohibits standard library implementations to define it outside of the class so that it is found using ADL.