Are multiple mutations within initializer lists undefined behavior?

Question

I am curious about initializer lists and sequence points. I read a while ago that the order of evaluation in initializer lists is left to right. If that is so, then there must be some kind of sequence point between the points of evaluation, am I wrong? So with that said is the following valid code? Is there anything that causes undefined behavior in it?

int i = 0;

struct S {
    S(...) {} 
    operator int() { return i; }
};

int main() {
    i = S{++i, ++i};
}

Any and all responses are appreciated.

Answer 1

Yes, the code is valid and does not have undefined behavior. Expressions in an initizalizer list are evaluated left-to-right and sequenced before the evaluation of the constructor of S . Therefore, your program should consistently assign value 2 to variable i .

Quoting § 8.5.4 of the C++ Standard:

"Within the initializer-list of a braced-init-list, the initializer-clauses, including any that result from pack expansions (14.5.3), 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."

Thus, what happens is:

1. ++i gets evaluated, yielding i = 1 (first argument of S 's constructor);
2. ++i gets evaluated, yielding i = 2 (second argument of S 's constructor);
3. S 's constructor is executed;
4. S 's conversion operator is executed, returning value 2 ;
5. value 2 is assigned to i (which already had value 2 ).

Another relevant paragraph of the Standard is § 1.9/15, which also mentions similar examples that do have undefined behavior:

i = v[i++]; // the behavior is undefined
i = i++ + 1; // the behavior is undefined

However, the same paragraph says:

" Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced. [...] When calling a function (whether or not the function is inline), every value computation and side effect associated with any argument expression, or with the postfix expression designating the called function, is sequenced before execution of every expression or statement in the body of the called function. "

Since 1) the evaluation of the expressions in the initializer list is sequenced left-to-right, 2) the execution of the constructor of S is sequenced after the evaluation of all expressions in the initializer list, and 3) the assignment to i is sequenced after the execution of the constructor of S (and its conversion operator), the behavior is well-defined.

Answer 2

Yes indeed you have case of undefined behavior.

Below are examples of situations causing undefined behavior:

• A variable is changed several times within one sequence point . As a canonical example, the i=i++ expression is often cited where assignment of the i variable and its increment are performed at the same time. To learn more about this kind of errors, read the section "sequence points" .
• Using a variable before initializing it. Undefined behavior occurs when trying to use the variable.
• Memory allocation using the new [] operator and subsequent release using the delete operator. For example: T *p = new T[10]; delete p;. The correct code is: T *p = new T[10]; delete [] p;.

EDITED

Also your code S{++i, ++i}; is not compiled for VS2012. May be you mean S(++i, ++i);? If you use "()" then undefined behavour is present. In other case your source code is incorrect.