Is f(++i, ++i) undefined?

Question

I seem to recall that in C++11, they made some changes to the sequencing behaviour and that now i++ and ++i have different sequencing requirements.

Is f(++i, ++i) still undefined behaviour? What is the difference between f(i++, i++) and f(++i, ++i)?

Since the example in [intro.execution]/15 explicitly calls out `f(i = -1, i = -1);` as UB, the answer to the question (for scalar `i`s) should be obvious. — T.C., Mar 28 '15 at 11:38

score 7 · Accepted Answer · answered Mar 28 '15 at 11:33

It's undefined behaviour unless i is a class type. From C++11 1.9/15:

Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced.

followed by a note to clarify that this does apply to function arguments:

[ Note: Value computations and side effects associated with different argument expressions are unsequenced. —end note ]

Your code modifies the same object twice without sequencing, so by the same paragraph:

If a side effect on a scalar object is unsequenced relative to either another side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined.

If i were a class type, then ++ would call a function, and function calls are always sequenced with respect to each other. So any modifications of scalar objects would be indeterminately sequenced; there's no undefined behaviour, but the result is unspecified.

score 3 · Answer 2 · answered Mar 28 '15 at 11:33

It's still undefined behaviour:

If a side effect on a scalar object is unsequenced relative to either another side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined.

§1.9 [intro.execution]

and the order of evaluation of function parameters is unsequenced relative to each other.

score 1 · Answer 3 · edited Jun 20 '20 at 09:12

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In C++17, it's not undefined. http://en.cppreference.com/w/cpp/language/eval_order#Undefined_behavior

f(++i, ++i); // undefined behavior until C++17, unspecified after C++17

In a function call, value computations and side effects of the initialization of every parameter are indeterminately sequenced with respect to value computations and side effects of any other parameter.

edited Jun 20 '20 at 09:12

Community

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answered Sep 06 '17 at 19:25

Tuff Contender

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To add to this answer: code `int i = 0; f(++i, ++i)` would result in `i == 2` and `f` called with any of the following combination of parameters: (1, 1), (1, 2), (2, 1), (2, 2) – Revolver_Ocelot Sep 06 '17 at 19:38
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[@Revolver_Ocelot](https://stackoverflow.com/users/3410396/revolver-ocelot), I think only the calling of `f(1, 2)` and `f(2, 1)` conform to the C++17 standard, as value computations and side effects of the initialization of two parameters should not overlap. – Tuff Contender Sep 06 '17 at 21:22

Is f(++i, ++i) undefined?

3 Answers3