c++17 evaluation order with operator overloading f

Regarding this question

What are the evaluation order guarantees introduced by C++17?

With this specification

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0145r3.pdf

And this text from the specification

Furthermore, we suggest the following additional rule: the order of evaluation of an expression involving an overloaded operator is determined by the order associated with the corresponding built-in operator, not the rules for function calls.

Does this mean that these two expressions are no longer equivalent?

a << b;
operator<<(a, b);

As the second one looks like a function call, hence there is no guaranteed evaluation order in the parameters?

"As the second one looks like a function call, hence there is no guaranteed evaluation order in the parameters?"

Indeed. [expr.call]/5 contains an example specifically covering the difference between the two cases covered in your question [emphasis mine]:

The postfix-expression is sequenced before each expression in the expression-list and any default argument. The initialization of a parameter, including every associated value computation and side effect, is indeterminately sequenced with respect to that of any other parameter.

...

Note: If an operator function is invoked using operator notation, argument evaluation is sequenced as specified for the built-in operator; see [over.match.oper]. [ Example:
struct S {
  S(int);
};
int operator<<(S, int);
int i, j;
int x = S(i=1) << (i=2);
int y = operator<<(S(j=1), j=2);
After performing the initializations, the value of i is 2 (see [expr.shift]), but it is unspecified whether the value of j is 1 or 2.

— end example ]