Multiple Assignment Evaluation Debate

[lawrence.jones]

Now, I re-read what 6.5.16 has to say, and yes, I know examples
are not normative, but to me Example 2 (6.5.16.1p5) seems to
support the following interpretation of the above statement:

1) `q` is evaluated
2) its value is assigned to `p->next`
3) that value becomes the value of the rightmost assignment
4) that value is assigned to `p`

It's a bit more complicated than that. First, the top-level assignment
expression (p = p->next = q) is evaluated. That requires evaluating the
left side (p) to determine the location to store into and evaluating the
right side (p->next = q) to determine the value to store. Those
evaluations can occur in either order. Evaluating the left side is
trivial, but the right side is again an assignment expression, so it is
necessary to evaluate the left side (p->next) to determine a location to
store into and the right side (q) to determine the value to be stored,
again in any order. This time the right side is trivial, but evaluating
the left side requires evaluating p. So, there are *lots* of
evaluations, which are only loosely ordered.

-Larry Jones

Something COULD happen today. And if anything DOES,
by golly, I'm going to be ready for it! -- Calvin

 

[Vladimir S. Oka]

(e-mail address removed) opined:

It's a bit more complicated than that. First, the top-level
assignment expression (p = p->next = q) is evaluated. That requires
evaluating the left side (p) to determine the location to store into
and evaluating the right side (p->next = q) to determine the value to
store. Those evaluations can occur in either order.

I admit that the normative text of the standard does allow (but not
mandate) such behaviour. However, as I have pointed earlier, the
Example 2 (non-normative, as examples are) suggests that the order of
evaluation is right-to-left.

Evaluating the left side is trivial, but the right side is again an
assignment expression, so it is necessary to evaluate the left side
(p->next) to determine a location to store into and the right side
(q) to determine the value to be stored, again in any order. This
time the right side is trivial, but evaluating the left side requires
evaluating p. So, there are *lots* of evaluations, which are only
loosely ordered.

I know that modifying a value multiple times between sequence points
results in UB, I don't think evaluating something multiple times
(without modifying it multiple times) is a problem.

--
BR, Vladimir

Technological progress has merely provided us
with more efficient means for going backwards.
-- Aldous Huxley

 

[lawrence.jones]

(e-mail address removed) opined:

I admit that the normative text of the standard does allow (but not
mandate) such behaviour. However, as I have pointed earlier, the
Example 2 (non-normative, as examples are) suggests that the order of
evaluation is right-to-left.

No, it most certainly does not. Example 2 is primarily concerned with
the result type of assignment expressions, not order of evaluation.
About the only thing that it implies about order of evaluation is that
an operator's operands must be evaluated before the final result of the
operation is known, but it says nothing whatsoever about the order in
which the operands are evaluated. Compiler can, and do, evaluate
operands in various orders. In fact, a common strategy is to evaluate
the most complex operand first.

I know that modifying a value multiple times between sequence points
results in UB, I don't think evaluating something multiple times
(without modifying it multiple times) is a problem.

It is if there is at least one modification and the other evaluations
aren't "read only to determine the value to be stored" (6.5p2).

-Larry Jones

There's a connection here, I just know it. -- Calvin

 

[Vladimir S. Oka]

(e-mail address removed) opined:

No, it most certainly does not. Example 2 is primarily concerned
with the result type of assignment expressions, not order of
evaluation. About the only thing that it implies about order of
evaluation is that an operator's operands must be evaluated before
the final result of the operation is known, but it says nothing
whatsoever about the order in
which the operands are evaluated.

So, according to what you just said, in order to execute:

p = p->next = q;

the right assignment has to be done first. Otherwise, the final result
is not known and the left assignment cannot happen. Hence, the right
one must be executed first, QED. Executing the right assignment does
not modify `p`, so there's no multiple modifications between sequence
points, and no UB (`p` is modified once, and `p->next` is modified
once -- that's it).

Compiler can, and do, evaluate
operands in various orders. In fact, a common strategy is to
evaluate the most complex operand first.


It is if there is at least one modification and the other evaluations
aren't "read only to determine the value to be stored" (6.5p2).

--
BR, Vladimir

Bubble Memory, n.:
A derogatory term, usually referring to a person's
intelligence. See also "vacuum tube".

 

[Jordan Abel]

(e-mail address removed) opined:


So, according to what you just said, in order to execute:

p = p->next = q;

the right assignment has to be done first. Otherwise, the final result
is not known and the left assignment cannot happen. Hence, the right
one must be executed first, QED. Executing the right assignment does
not modify `p`, so there's no multiple modifications between sequence
points, and no UB (`p` is modified once, and `p->next` is modified
once -- that's it).

That sounds right.

[No-one has claimed that "p = p->next" alone is UB, and I don't see what
this adds that would be]

 

[ena8t8si]

(e-mail address removed) opined:

(e-mail address removed) opined:


Vladimir S. Oka wrote:
Vladimir S. Oka wrote:
On Friday 17 March 2006 04:48, (e-mail address removed) opined
(in
<[email protected]>):


pete wrote:
(e-mail address removed) wrote:

Micah Cowan wrote:

Richard G. Riley schrieb:
On 2006-03-12, pete <[email protected]>
wrote:

I'm coming around to thinking that 6.5 [#2] is
relevant, and that p = p->next = q
is undefined, and not just unspecified.
but
p=(p->next=q);
is fine? I hope .

No. Whether you write
a = b = c;
or
a = (b = c);
does not change anything -- apart from clarifying
your intent. You still are modifying p twice
between sequence points

Well, no he's not.
But he's reading its prior value for purposes other
than to determine the value stored, so same deal.

Look again. Reading p is necessary to evaluate p->q
= next,

No, it isn't.
The value of (p->q = next) is (next)

The value of p->q = next is next. But in order to get
the value, the assignment p->q = next must _have been
evaluated_, which needed both operands. Evaluation
includes all the actions in the Semantics paragraph,
which includes starting the side effect of storing the
value.

I think pete is right. The compiler _knows_ that the
result of (p->q = next) _will_be_ next before it even
produces any code, and can use that knowledge once it does
get to producing some.

You've fallen into the trap of arguing based on what a
compiler might be capable of. Regardless of what the
compiler knows, it's still obliged to produce code that
behaves according to how the Semantics paragraphs and
everything else in the standard says it must.

But it still does not evaluate p->q in order to assign next
to it. What is evaluated is next, and its value is /assigned/
to p->q. Next, the assignment is evaluated and assigned to p.
The assignment evaluates to next, but I don't think that
compiler is obliged to do that in any particular way.

Sorry, the compiler _is_ obliged to do the evaluation as per
the Semantics description. The Semantics paragraphs in 6.5.16
specify _both_ that the store happens and what the value is.
(Yes, the side effect of the store may happen later.) Where
in the standard is there any statement that says an expression
(_any_ expression) may yield a value _before_ it has been
evaluated?

Ok, now my head starts to spin a bit, but bear with me...

Is it right that the original problem statement was:

p = p->next = q;
Yes.

Assuming it was: `p` is not modified twice, it is read once to
get `next`, and modified once to assign value to it in the
leftmost assignment. Assigning value to `next` does not modify
`p`, right?

Now, I re-read what 6.5.16 has to say, and yes, I know examples
are not normative, but to me Example 2 (6.5.16.1p5) seems to
support the following interpretation of the above statement:

1) `q` is evaluated
2) its value is assigned to `p->next`
3) that value becomes the value of the rightmost assignment
4) that value is assigned to `p`

I read that it happens in the order I give above. Also, `p` is
modified only once (4), and read only once (2).

Yes. Also the side effect of updating the value of p->next
can happen any time after (2), and the side effect of updating
the value of p can happen any time after (4).

I agree.

However, what is the side effect of updating `p->next`? It's just plain
storing a value into a variable. Again, `p` is not modified, AFAICT.
It's the struct member `next` that is, and `p` is just read to get its
address (`next`'s address).

Whether there's any side effect of updating `p` after (4) is, IMO,
irrelevant to the discussion (again, I can't see any side effect of
storing a value into variable).

I still think the operation in this case has to be equivalent to:

p->next = q;
p = p->next;

Right. Except the last part is better written as

t = q;
p->next = t;
p = t;

 

[ena8t8si]

It's a bit more complicated than that. First, the top-level assignment
expression (p = p->next = q) is evaluated. That requires evaluating the
left side (p) to determine the location to store into and evaluating the
right side (p->next = q) to determine the value to store. Those
evaluations can occur in either order. ...

Why do you say the top level expression is evaluated
first? It makes more sense to say that subexpressions
(including simple operands) are evaluated first, and a
larger containing expression is evaluated only after
its operand subexpressions have been evaluated.

 

[ena8t8si]

No, it most certainly does not. Example 2 is primarily concerned with
the result type of assignment expressions, not order of evaluation.
About the only thing that it implies about order of evaluation is that
an operator's operands must be evaluated before the final result of the
operation is known, but it says nothing whatsoever about the order in
which the operands are evaluated. Compiler can, and do, evaluate
operands in various orders. In fact, a common strategy is to evaluate
the most complex operand first.

When asked about evaluation order by a beginner,
try saying this:

Precedence _does_ determine order of evaluation
for operATORS, but _doesn't_ determine order of
evaluation for operANDS.

Of course, it isn't completely accurate (because
of &&, etc), but usually after hearing that one
sentence the lights will go on.

 

[Flash Gordon]

(e-mail address removed) opined:

So, according to what you just said, in order to execute:

p = p->next = q;

the right assignment has to be done first. Otherwise, the final result
is not known and the left assignment cannot happen. Hence, the right
one must be executed first, QED. Executing the right assignment does
not modify `p`, so there's no multiple modifications between sequence
points, and no UB (`p` is modified once, and `p->next` is modified
once -- that's it).

That sounds right.

[No-one has claimed that "p = p->next" alone is UB, and I don't see what
this adds that would be]

The problem with "p = p->next = q" is that wither the new or the old
value of p could be used to decide where to write q to. Regardless of
whether that is a real problem, there is also section 6.5 para 2 which says:
| Between the previous and next sequence point an object shall have its
| stored value modified at most once by the evaluation of an expression.
| Furthermore, the prior value shall be read only to determine the value
| to be stored.

Note the second sentence. In "p = p->next = q" p is read to determine
where to store q, and this violates the shall in the second sentence.
Unless, of course, evaluating p->next is part of determining the value
to be assigned to p...

Personally I don't like the line even if it is defined behaviour.

 

[Vladimir S. Oka]

Flash Gordon opined:

(e-mail address removed) opined:

(e-mail address removed) opined:
It's a bit more complicated than that. First, the top-level
assignment expression (p = p->next = q) is evaluated. That
requires evaluating the left side (p) to determine the location
to store into and evaluating the right side (p->next = q) to
determine the value to
store. Those evaluations can occur in either order.
I admit that the normative text of the standard does allow (but
not mandate) such behaviour. However, as I have pointed earlier,
the Example 2 (non-normative, as examples are) suggests that the
order of evaluation is right-to-left.
No, it most certainly does not. Example 2 is primarily concerned
with the result type of assignment expressions, not order of
evaluation. About the only thing that it implies about order of
evaluation is that an operator's operands must be evaluated before
the final result of the operation is known, but it says nothing
whatsoever about the order in
which the operands are evaluated.
So, according to what you just said, in order to execute:

p = p->next = q;

the right assignment has to be done first. Otherwise, the final
result is not known and the left assignment cannot happen. Hence,
the right one must be executed first, QED. Executing the right
assignment does not modify `p`, so there's no multiple
modifications between sequence points, and no UB (`p` is modified
once, and `p->next` is modified once -- that's it).

That sounds right.

[No-one has claimed that "p = p->next" alone is UB, and I don't see
[what
this adds that would be]

The problem with "p = p->next = q" is that wither the new or the old
value of p could be used to decide where to write q to. Regardless of
whether that is a real problem, there is also section 6.5 para 2
which says:
| Between the previous and next sequence point an object shall have
| its stored value modified at most once by the evaluation of an
| expression. Furthermore, the prior value shall be read only to
| determine the value to be stored.

Note the second sentence. In "p = p->next = q" p is read to determine
where to store q, and this violates the shall in the second sentence.
Unless, of course, evaluating p->next is part of determining the
value to be assigned to p...

Which was my point all along, and why I think it's OK.

Personally I don't like the line even if it is defined behaviour.

I do heartily agree with this. It's just that I think it is.

--
BR, Vladimir

Fourth Law of Revision:
It is usually impractical to worry beforehand about
interferences -- if you have none, someone will make one for you.

 

[lawrence.jones]

Executing the right assignment does
not modify `p`, so there's no multiple modifications between sequence
points, and no UB (`p` is modified once, and `p->next` is modified
once -- that's it).

That's not correct. Since p is modified, it can be read "only to
determine the value to be stored". If it is read for any other reason,
the behavior is undefined. So, the crux of the matter is whether the
read of p in the embedded assignment (p->next = q) is "only to determine
the value to be stored" in p by the outer assignment or not. Many
people have argued that it is not (it's being read to determine where to
store q) and that's where the undefined behavior comes from.

For what it's worth, I believe the standard can be read either way. I
also believe that all of the formal sequence point models that have been
proposed make it well defined. I also believe there is at least one
implementation that gets it wrong.

-Larry Jones

I don't see why some people even HAVE cars. -- Calvin

 

[lawrence.jones]

Why do you say the top level expression is evaluated
first?

Because it is.

It makes more sense to say that subexpressions
(including simple operands) are evaluated first, and a
larger containing expression is evaluated only after
its operand subexpressions have been evaluated.

It's the evaluation of the top-level expression that causes the operand
subexpressions to be evaluated. This is particularly important for
cases where certain operands are *not* evaluated, such as in &&, ||, and
?: expressions.

-Larry Jones

I sure wish I could get my hands on some REAL dynamite. -- Calvin

 

[Vladimir S. Oka]

(e-mail address removed) opined:

That's not correct. Since p is modified, it can be read "only to
determine the value to be stored". If it is read for any other
reason, the behavior is undefined. So, the crux of the matter is
whether the read of p in the embedded assignment (p->next = q) is
"only to determine the value to be stored" in p by the outer
assignment or not. Many people have argued that it is not (it's
being read to determine where to store q) and that's where the
undefined behavior comes from.

That sounds reasonable enough. I'd still argue that `p` is read only to
determine the value to be stored. For me, the fact that reading it is
first used to figure out where to store `q` is just part of
determining what to finally store in `p`.

Count me in the (not so many?) other people, and let's agree to
disagree. Especially since you won't see me using such a construct
anyway.

For what it's worth, I believe the standard can be read either way.
I also believe that all of the formal sequence point models that have
been proposed make it well defined. I also believe there is at least
one implementation that gets it wrong.

I actually haven't tried this on any implementations. It's entirely
possible that some behave differently to the others, given differences
in opinions on this one.

 

[CBFalconer]

(e-mail address removed) opined:
.... snip ...


I actually haven't tried this on any implementations. It's
entirely possible that some behave differently to the others,
given differences in opinions on this one.

I think you are talking about "p = p->next = q;" constructs. The
"p = p->next" part is a very standard way of advancing down a
linked list, and the whole statement is a very normal way of
extending the list and advancing.

--
"If you want to post a followup via groups.google.com, don't use
the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the
"Reply" at the bottom of the article headers." - Keith Thompson
More details at: <http://cfaj.freeshell.org/google/>
Also see <http://www.safalra.com/special/googlegroupsreply/>

 

[Vladimir S. Oka]

CBFalconer opined:

I think you are talking about "p = p->next = q;" constructs. The
"p = p->next" part is a very standard way of advancing down a
linked list, and the whole statement is a very normal way of
extending the list and advancing.

Correct. That's what the whole discussion was about.

--
BR, Vladimir

A successful tool is one that was used to do something
undreamed of by its author.
-- S.C. Johnson

 

[lawrence.jones]

(e-mail address removed) opined:

That sounds reasonable enough. I'd still argue that `p` is read only to
determine the value to be stored. For me, the fact that reading it is
first used to figure out where to store `q` is just part of
determining what to finally store in `p`.

For what it's worth, I and, I think, most, if not all, of the other
members of the C Standards committee agree with that interpretation,
which is why the formal models of sequence points all make the behavior
well-defined in this case. I think the intended interpretation is that
any reads must be in the context of evaluating the value to be stored,
not that they must directly affect the value. That would make well-
defined even somewhat ludicrous expressions like:

p = (p->prev = q->prev, p->next = q);

Count me in the (not so many?) other people, and let's agree to
disagree. Especially since you won't see me using such a construct
anyway.

We don't disagree, and I think we're actually in the majority. But
there is a vocal minority(?) that disagrees, and that disagreement isn't
without justification, which I why I felt the need to discuss it.

I actually haven't tried this on any implementations. It's entirely
possible that some behave differently to the others, given differences
in opinions on this one.

Most implementations get it right. I've only seen one report of an
implementation that gets it wrong.

-Larry Jones

No one can prove I did that!! -- Calvin

 

[ena8t8si]

Because it is.

A non-answer answer. Bravo.

It's the evaluation of the top-level expression that causes the operand
subexpressions to be evaluated. This is particularly important for
cases where certain operands are *not* evaluated, such as in &&, ||, and
?: expressions.

The Semantics paragraphs for && and so forth give constraints
on which, whether, and in what order their subexpressions are
evaluated, but don't say anything about causality or whether
subexpression evaluation is done before or after evaluation of
the operator. So it sounds like you're just expressing a
personal viewpoint.

 

[ena8t8si]

That's not correct. Since p is modified, it can be read "only to
determine the value to be stored". ...

That sentence in the standard deserves a Lifetime Achievement
award in the DR Hall of Fame. Besides being vague about what
"only" quantifies and what constitutes an access "to determine
the value to be stored", there is the nonsense about undefined
behavior when an access is done as part of determining the
lvalue of which object to modify. What an absurdity! Like
it's possible for a value to be stored before the calculation
is made for where to store it... For extra hilarity, read Doug
Gwyn's defense of LHS-access causing UB because it's needed for
optimizing compilers on vector processors. The committee
should just admit they made a mistake in kowtowing to
freewheeling compiler vendors who can't be bothered to write
compilers that generate correct code. Clearly it would have
been better to add an advisory ala' restrict that allows the
more aggressive optimizations where the programmer explicitly
says it's okay.

 

[pete]

Trouble is, an enthusiastic reading of 6.5/2 can make all
sorts of things UB, including *p++.

An enthusiastic reading of 6.5/2 should find the footnote:

70)
This paragraph renders undefined statement expressions such as
i = ++i + 1;
a[i++] = i;
while allowing
i = i + 1;
a = i;

Can you see how this footnote helps? I can't

It should clear up that *p++ is defined.
How do you figure that *p++ can be considered to be undefined
according to 6.5/2?

 

[lawrence.jones]

That sentence in the standard deserves a Lifetime Achievement
award in the DR Hall of Fame.

Indeed. The problem is that it's extremely difficult to express the
intent precisely in English, which is why there is so much interest in a
formal model of sequence points that would do so. Unfortunately, there
hasn't been quite enough interest to succeed in doing so yet.

The committee
should just admit they made a mistake in kowtowing to
freewheeling compiler vendors who can't be bothered to write
compilers that generate correct code.

They didn't (kowtow to freewheeling vendors), they just tried to strike
a balance between performance and determinancy. High performance is
very important to most users of C, but getting the wrong answer isn't of
any use to anyone. Like I said before, I think most experienced C
programmers have an intuitive understanding of what kinds of things
should be well defined (and "p = p->next = q;" falls into that category)
and what kinds of things should be undefined (like "j = ++i + i;"), and
the formal models proposed so far, although still works in progress,
seem to match most people's intuition.

-Larry Jones

Even if lives DID hang in the balance, it would depend on whose they were.
-- Calvin