c++17 evaluation order with operator overloading functions

Question

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?

Worl

That's correct, the two different syntaxes have different semantics now. The operator syntax has its peculiar semantics, the function call syntax has the (new!) semantics of function calls. – Robbynrobe

@KerrekSB This is precisely why this proposal: open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0431r0.htm was written but I don't believe it gained any traction. – Leftwards

@ForeverLearning: It is certainly a subject that inspires hearts, minds and proposals! – Robbynrobe

Answer 1

"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 ]

Oeildeboeuf

Look no further than: open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0431r0.htm Proposal was DOA though. – Leftwards

@Leftwards thanks for the related link, that's interesting. – Oeildeboeuf