c++argumentslanguage-lawyerinitializer-listsequencing

Do the Requirements Placed on Function Arguments Also Apply to Initializer Lists?


So I've read here: https://stackoverflow.com/a/598150/2642059 that this is illegal:

foo(i++, i++);

But I believe that's because there is not a forced sequence, which I understand is the case for Initializer Lists. So is this legal code?

const int foo[] = { i++, i++ };

Solution

  • Yes, the order of the evaluation of initializer-clauses is guaranteed in braced-init-list.

    From the standard, §11.6.4/4 List-initialization [dcl.init.list]:

    (emphasis mine)

    Within the initializer-list of a braced-init-list, the initializer-clauses, including any that result from pack expansions, are evaluated in the order in which they appear. That is, every value computation and side effect associated with a given initializer-clause is sequenced before every value computation and side effect associated with any initializer-clause that follows it in the comma-separated list of the initializer-list. [ Note: This evaluation ordering holds regardless of the semantics of the initialization; for example, it applies when the elements of the initializer-list are interpreted as arguments of a constructor call, even though ordinarily there are no sequencing constraints on the arguments of a call. — end note ]

    From cppreference.com:

    Every initializer clause is sequenced before any initializer clause that follows it in the braced-init-list. This is in contrast with the arguments of a function call expression, which are unsequenced.

    Example for the note of the standard,

    struct A { A(int, int) {} };
    ...
    int i = 0;
    A a1(i++, i++); // used as the arguments of the constructor; unsequenced
    A a2{i++, i++}; // used as the arguments of the constructor; sequenced, within the initializer-list of a braced-init-list