c++ - How to determine what is 'sequenced before' others?

Question

I went through this excellent answer regarding Undefined Behaviour and Sequenced [Before/After] relations in C++11. I understand the binary relation concepts, but am missing what the new rules governing sequencing are.

For these familiar examples, how do the new sequencing rules apply?

1. i = ++i;
2. a[++i] = i;

More specifically, what are the new C++11 sequencing rules?

I am looking for some rules like (this one is completely made up)

The lhs of an '=' statement is always sequenced before the rhs, and is thus evaluated first.

In case these are available in the standard itself, can someone quote the same here?

Answer 1

The sequenced-before relationship, and the rules concerning it are a "tidying up" of the prior rules on sequence points, defined in a consistent way with the other memory model relationships such as happens-before and synchronizes-with so that it can be precisely specified which operations and effects are visible under which circumstances.

The consequences of the rules are unchanged for simple single-threaded code.

Let's start with your examples:

1. i = ++i;

If i is a built-in type such as int then there are no function calls involved, everything is a built-in operator. There are thus 4 things that happen:

(a) The value computation of ++i, which is original-value-of-i +1

(b) The side effect of ++i, which stores original-value-of-i +1 back into i

(c) The value computation of the assignment, which is just the value stored, in this case the result of the value computation of ++i

(d) The side effect of the assignment, which stores the new value into i

All of these things are sequenced-before the following full expression. (i.e. they are all complete by the final semicolon of the statement)

Since ++i is equivalent to i+=1, the side effect of storing the value is sequenced-before the value computation of ++i, so (b) is sequenced-before (a).

The value computation of both operands of an assignment is sequenced-before the value computation of the assignment itself, and that is in turn sequenced-before the side effect of storing the value. Therefore (a) is sequenced before (c), and (c) is sequenced-before (d).

We therefore have (b) -> (a) -> (c) -> (d), and this is thus OK under the new rules, whereas it was not OK under C++98.

If i was a class, then the expression would be i.operator=(i.operator++()), or i.operator=(operator++(i)), and all effects of the operator++ call are sequenced-before the call to operator=.

2. a[++i] = i;

If a is an array type, and i is an int, then again the expression has several parts:

(a) The value computation of i

(b) The value computation of ++i

(c) The side effect of ++i, which stores the new value back into i

(d) The value computation of a[++i], which returns an lvalue for the element of a indexed by the value computation of ++i

(e) The value computation of the assignment, which is just the value stored, in this case the result of the value computation of i

(f) The side effect of the assignment, which stores the new value into the array element a[++i]

Again, all of these things are sequenced-before the following full expression. (i.e. they are all complete by the final semicolon of the statement)

Again, since ++i is equivalent to i+=1, the side effect of storing the value is sequenced-before the value computation of ++i, so (c) is sequenced-before (b).

The value computation of the array index ++i is *sequenced-before` the value computation of the element selection, so (b) is sequenced-before (d).

The value computation of both operands of an assignment is sequenced-before the value computation of the assignment itself, and that is in turn sequenced-before the side effect of storing the value. Therefore (a) and (d) are sequenced before (e), and (e) is sequenced-before (f).

We therefore have two sequences: (a) -> (d) -> (e) -> (f) and (c) -> (b) -> (d) -> (e) -> (f).

Unfortunately, there is no ordering between (a) and (c). Thus a side effect which stores to i is unsequenced with respect to a value computation on i, and the code exhibits undefined behaviour. This is again given by 1.9p15 of the C++11 standard.

As above, if i is of class type then everything is fine, because the operators become function calls, which impose sequencing.

The rules

The rules are relatively straightforward:

1. The value computations of the arguments of a built-in operator are sequenced-before the value computation of the operator itself.

2. The side effects of a built-in assignment operator or preincrement operator are sequenced-before the value computation of the result.

3. The value computation of any other built-in operator is sequenced-before the side effects of that operator.

4. The value computation and side-effects of the left-hand side of the built-in comma operator are sequenced-before the value computation and side-effects of the right-hand side.

5. All value computations and side effects of a full expression are sequenced-before the next full expression.

6. The value computation and side effects of the arguments of a function call are sequenced before the first full expression in the function.

7. The value computation and side effects of everything inside a function are sequenced-before the value computation of the result.

8. For any two function calls in the full expression, either the value computation of the result of one is sequenced-before the call to the other, or vice-versa. If no other rule specifies the ordering, the compiler may choose.

   Thus in a()+b(), either a() is sequenced-before b(), or b() is sequenced-before a(), but there is no rule to specify which.

9. If there are two side effects that modify the same variable, and neither is sequenced-before the other, the code has undefined behaviour.

10. If there is a side effect that modifies a variable, and a value computation that reads that variable, and neither is sequenced-before the other, the code has undefined behaviour.