By Ben Jackson

2013-03-01 21:08:04 8 Comments

If I write f(x)->g(args, ...) can I rely on a sequence point after f(x) before the evaluation of args, ...? I can see arguments both ways:

  • §1.9.17 "When calling a function (whether or not the function is inline), there is a sequence point after the evaluation of all function arguments (if any) which takes place before execution of any expressions or statements in the function body. There is also a sequence point after the copying of a returned value and before the execution of any expressions outside the function."
  • On the other hand, the object pointer is implicitly a hidden argument this as if I'd written g(f(x), args, ...) which suggests it's like an argument, and thus unspecified.

The -> operator is not a normal binary operator, since clearly g(...) cannot be evaluated before f(x) like it could if I wrote f(x) + g(...). I'm surprised I can't find some specific statement about it.


@Jonathan Wakely 2013-03-01 22:36:53

The answer depends on which version of the C++ standard you are using (or your compiler is using).

C++ 2003 5.2.2 p8 said:


This means there is not a sequence point between evaluating f(x) and args.

In C++ 2011 the whole concept of sequence points has been replaced (see N1944), and that wording is now just a note:

[ Note: The evaluations of the postfix expression and of the argument expressions are all unsequenced relative to one another. All side effects of argument expression evaluations are sequenced before the function is entered (see 1.9). — end note ]

and 1.9 p15 says


This says the expression f(x) and the expression args are sequenced before everything in the body of g, but that they are unsequenced relative to each other, which is the same as the C++03 rules but worded differently.

C++14 has the same rules as C++11, but as noted in the comment below, the rules changed in C++17.

C++ 2017 8.2.2 [] p5 says:


This means for your example the following steps happen in order:

  • f is evaluated.
  • x is evaluated and the parameters of f are initialized.
  • The function call f(x) is evaluated.
  • f(x)->g is evaluated.
  • args and the other arguments to g are evaluated and the parameters of g are initialized (in an unspecified order).
  • Finally, the function call f(x)->g(args, ...) is evaluated.

@Gabriel Kerneis 2018-10-22 13:35:57

This has changed in C++17: "14) In a function-call expression, the expression that names the function is sequenced before every argument expression and every default argument." So now, f(x) will evaluate before the args.

@Dave S 2013-03-01 22:00:30

Note, I think that you're asking one question in your title, and another in the body of your question.

Well, it's not really contradictory. To evaluate your function, the following things have to happen (not necessarily in this order).

  • x is evaluated (A)
  • args is evaluated (B)
  • ... is evaluated (C)
  • f(x) is called (D)
  • the return value of f(x) is copied (E)
  • return->g(args, ...) is called (F)

Now, the rules you've quoted indicate that

  1. (A) has to happen before (D), since there is a sequence point of evaluating the arguments to a function prior to evaluating.
  2. (D) happens before (E), since the copy can't be made until the function runs.
  3. (F) happens after (E), since the implicit pointer is necessary to invoke g(args) *
  4. (B) and (C) happen before (F), since they are arguments.

However, what is unsequenced is the relationship between (A), (B), and (C), or in your question between (B) and (C) and (D), since they aren't arguments to (F), they could be evaluated afterwards. OR, they could be evaluated prior.

* Interesting question. What happens if g(args, ...) is a static member function. In this case, since the returned pointer from f(x) isn't actually passed in, can it be sequenced earlier? But that's a separate question.

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