This is an unofficial snapshot of the ISO/IEC JTC1 SC22 WG21 Core Issues List revision 113d. See http://www.open-std.org/jtc1/sc22/wg21/ for the official list.
2024-04-05
[Voted into the WP at the September, 2008 meeting.]
I believe that the committee has neglected to take into account one of the differences between C and C++ when defining sequence points. As an example, consider
(a += b) += c;
where a, b, and c all have type int. I believe that this expression has undefined behavior, even though it is well-formed. It is not well-formed in C, because += returns an rvalue there. The reason for the undefined behavior is that it modifies the value of `a' twice between sequence points.
Expressions such as this one are sometimes genuinely useful. Of course, we could write this particular example as
a += b; a += c;
but what about
void scale(double* p, int n, double x, double y) {
for (int i = 0; i < n; ++i) {
(p[i] *= x) += y;
}
}
All of the potential rewrites involve multiply-evaluating p[i] or unobvious circumlocations like creating references to the array element.
One way to deal with this issue would be to include built-in operators in the rule that puts a sequence point between evaluating a function's arguments and evaluating the function itself. However, that might be overkill: I see no reason to require that in
x[i++] = y;
the contents of `i' must be incremented before the assignment.
A less stringent alternative might be to say that when a built-in operator yields an lvalue, the implementation shall not subsequently change the value of that object as a consequence of that operator.
I find it hard to imagine an implementation that does not do this already. Am I wrong? Is there any implementation out there that does not `do the right thing' already for (a += b) += c?
7.6.19 [expr.ass] paragraph 1 says,
The result of the assignment operation is the value stored in the left operand after the assignment has taken place; the result is an lvalue.
What is the normative effect of the words "after the assignment has taken place"? I think that phrase ought to mean that in addition to whatever constraints the rules about sequence points might impose on the implementation, assignment operators on built-in types have the additional constraint that they must store the left-hand side's new value before returning a reference to that object as their result.
One could argue that as the C++ standard currently stands, the effect of x = y = 0; is undefined. The reason is that it both fetches and stores the value of y, and does not fetch the value of y in order to compute its new value.
I'm suggesting that the phrase "after the assignment has taken place" should be read as constraining the implementation to set y to 0 before yielding the value of y as the result of the subexpression y = 0.
Francis Glassborow:
My understanding is that for a single variable:
It is the 3) that is often ignored because in practice the compiler hardly ever codes for the read because it already has that value but in complicated evaluations with a shortage of registers, that is not always the case. Without getting too close to the hardware, I think we both know that a read too close to a write can be problematical on some hardware.
So, in x = y = 0;, the implementation must NOT fetch a value from y, instead it has to "know" what that value will be (easy because it has just computed that in order to know what it must, at some time, store in y). From this I deduce that computing the lvalue (to know where to store) and the rvalue to know what is stored are two entirely independent actions that can occur in any order commensurate with the overall requirements that both operands for an operator be evaluated before the operator is.
Erwin Unruh:
C distinguishes between the resulting value of an assignment and putting the value in store. So in C a compiler might implement the statement x=y=0; either as x=0;y=0; or as y=0;x=0; In C the statement (x += 5) += 7; is not allowed because the first += yields an rvalue which is not allowed as left operand to +=. So in C an assignment is not a sequence of write/read because the result is not really "read".
In C++ we decided to make the result of assignment an lvalue. In this case we do not have the option to specify the "value" of the result. That is just the variable itself (or its address in a different view). So in C++, strictly speaking, the statement x=y=0; must be implemented as y=0;x=y; which makes a big difference if y is declared volatile.
Furthermore, I think undefined behaviour should not be the result of a single mentioning of a variable within an expression. So the statement (x +=5) += 7; should NOT have undefined behaviour.
In my view the semantics could be:
Jerry Schwarz:
My recollection is different from Erwin's. I am confident that the intention when we decided to make assignments lvalues was not to change the semantics of evaluation of assignments. The semantics was supposed to remain the same as C's.
Ervin seems to assume that because assignments are lvalues, an assignment's value must be determined by a read of the location. But that was definitely not our intention. As he notes this has a significant impact on the semantics of assignment to a volatile variable. If Erwin's interpretation were correct we would have no way to write a volatile variable without also reading it.
Lawrence Crowl:
For x=y=0, lvalue semantics implies an lvalue to rvalue conversion on the result of y=0, which in turn implies a read. If y is volatile, lvalue semantics implies both a read and a write on y.
The standard apparently doesn't state whether there is a value dependence of the lvalue result on the completion of the assignment. Such a statement in the standard would solve the non-volatile C compatibility issue, and would be consistent with a user-implemented operator=.
Another possible approach is to state that primitive assignment operators have two results, an lvalue and a corresponding "after-store" rvalue. The rvalue result would be used when an rvalue is required, while the lvalue result would be used when an lvalue is required. However, this semantics is unsupportable for user-defined assignment operators, or at least inconsistent with all implementations that I know of. I would not enjoy trying to write such two-faced semantics.
Erwin Unruh:
The intent was for assignments to behave the same as in C. Unfortunately the change of the result to lvalue did not keep that. An "lvalue of type int" has no "int" value! So there is a difference between intent and the standard's wording.
So we have one of several choices:
I think the last one has the least impact on existing programs, but it is an ugly solution.
Andrew Koenig:
Whatever we may have intended, I do not think that there is any clean way of making
volatile int v;
int i;
i = v = 42;
have the same semantics in C++ as it does in C. Like it or not, the
subexpression v = 42 has the type ``reference to volatile int,''
so if this statement has any meaning at all, the meaning must be to store 42
in v and then fetch the value of v to assign it to i.
Indeed, if v is volatile, I cannot imagine a conscientious programmer writing a statement such as this one. Instead, I would expect to see
v = 42;
i = v;
if the intent is to store 42 in v and then fetch the (possibly
changed) value of v, or
v = 42;
i = 42;
if the intent is to store 42 in both v and i.
What I do want is to ensure that expressions such as ``i = v = 42'' have well-defined semantics, as well as expressions such as (i = v) = 42 or, more realistically, (i += v) += 42 .
I wonder if the following resolution is sufficient:
Append to 7.6.19 [expr.ass] paragraph 1:
There is a sequence point between assigning the new value to the left operand and yielding the result of the assignment expression.
I believe that this proposal achieves my desired effect of not constraining when j is incremented in x[j++] = y, because I don't think there is a constraint on the relative order of incrementing j and executing the assignment. However, I do think it allows expressions such as (i += v) += 42, although with different semantics from C if v is volatile.
Notes on 10/01 meeting:
There was agreement that adding a sequence point is probably the right solution.
Notes from the 4/02 meeting:
The working group reaffirmed the sequence-point solution, but we will look for any counter-examples where efficiency would be harmed.
For drafting, we note that ++x is defined in 7.6.2.3 [expr.pre.incr] as equivalent to x+=1 and is therefore affected by this change. x++ is not affected. Also, we should update any list of all sequence points.
Notes from October 2004 meeting:
Discussion centered around whether a sequence point “between assigning the new value to the left operand and yielding the result of the expression” would require completion of all side effects of the operand expressions before the value of the assignment expression was used in another expression. The consensus opinion was that it would, that this is the definition of a sequence point. Jason Merrill pointed out that adding a sequence point after the assignment is essentially the same as rewriting
b += a
as
b += a, b
Clark Nelson expressed a desire for something like a “weak” sequence point that would force the assignment to occur but that would leave the side effects of the operands unconstrained. In support of this position, he cited the following expression:
j = (i = j++)
With the proposed addition of a full sequence point after the assignment to i, the net effect is no change to j. However, both g++ and MSVC++ behave differently: if the previous value of j is 5, the value of the expression is 5 but j gets the value 6.
Clark Nelson will investigate alternative approaches and report back to the working group.
Proposed resolution (March, 2008):
This issue is resolved by the adoption of the sequencing rules and the resolution of issue 637.