question about k&r2

[CBFalconer]

Hardly, the latter causes undefined behaviour. Without an
intervening sequence point, p is both written, and read for
a purpose other than to determine the value to be written.
(That purpose is in fact to determine the location to write
another value).

p is NOT written. p->next is. That does not affect the value of
p. Well defined.

 

[Keith Thompson]

p is NOT written. p->next is. That does not affect the value of
p. Well defined.

Yes, p is written; it's the target of an assignment.

<SEQUENCE_POINT> p = p->next = q; <SEQUENCE_POINT>

C99 6.5p2:

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.

Between the two sequence points, p is modified once (when the result
of "p->next = q" is assigned to it), *and* p is read for a purpose
other than to determine the value to be stored (in the evaluation of
the lvalue p->next, to determine the location to which to assign the
value of q).

 

[Old Wolf]

p is NOT written. p->next is. That does not affect the value of
p. Well defined.

p most certainly is written; it is by itself on the left of an
assignment operator!

This code straight-out violates C99 6.5p2 (which Keith was kind
enough to quote). Ben Pfaff says there has been some debate over
this; if so then perhaps the debate should be about whether to
change the standard..?

 

[CBFalconer]

Yes, p is written; it's the target of an assignment.

I was referring to the interior portion of the statement. I wasn't
clear.

<SEQUENCE_POINT> p = p->next = q; <SEQUENCE_POINT>

Agreed so far

C99 6.5p2:

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.

Between the two sequence points, p is modified once (when the result
of "p->next = q" is assigned to it), *and* p is read for a purpose
other than to determine the value to be stored (in the evaluation of
the lvalue p->next, to determine the location to which to assign the
value of q).

And I disagree here, although I see your point. I think this
should head for the standards group, so cross-posted.

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[pete]

No, but I still disagree. The left assignment is of the value of
the statement "p->next = q".
To evaluate that statement

If (p->next = q) was a statement, then you'd be right.
To evaluate that "expression" ...

you have to evaluate the assignment "= q".

The value of the expresssion (p->next = q), is equal to (q).

There never is an assignment "p = q" in the chain.
This is because, in C, assignment statements have a value,

Statements don't have values in C.

and that statement has to be performed before that
value is available.

You're mixing your terms "statements" and "expressions"
and that's the problem.
You're seeing sequence points where there are none.
Statements are sequence points.
Assignment isn't a sequence point.


(p = p->next = q) associates like (p = (p->next = q))

The value of (p->next = q) is the same as (q)
and that's why the value of q can be assigned to p.