Re: i = v[i++] results in undefined behavior. Can't understand why.
On Sep 28, 12:03 pm, litb <litb...@googlemail.com> wrote:
On 28 Sep., 09:34, James Kanze <james.ka...@gmail.com> wrote:
On Sep 27, 6:50 pm, "Johannes Schaub (litb)" <schaub-johan...@web.de>
wrote:
James Kanze wrote:
On Sep 26, 6:57 pm, "Johannes Schaub (litb)" <schaub-johan...@web.d=
e>
wrote:
Armen Tsirunyan wrote:
Also, is
(++i)++;
defined?
I guess not, am I right?
++i is
i = i + 1
The Standard says for postfix i++:
"The value computation of the ++ expression is sequenced before th=
e
modification of the operand object."
And it says for i = x
"In all cases, the assignment is sequenced after the value
computation of the right and left operands, and before the
value computation of the assignment expression."
Together this means that all side effects are sequenced and
all side effects are sequenced relative to all value
computations. Thus the behavior of "(++i)++" is defined.
Where do you get that from? The sentences you quote only speak
of sequencing the *value* computation. Nothing about side
effects.
It says "The value computation of the ++ expression is
sequenced before the modification of the operand object". The
latter is the side effect. So you have the modification of
"++i" before value computation of that expression,
The passage you quote says just the opposite. And logically, it
couldn't be any other way: you can't have the side effect (the
modification of i) before having calculated its value.
No. It says that the modification in "i++" happens after value
computation of "i++".
In the expression "i++", the side effect is sequenced after the
value computation. Which is in direct contradiction to the last
sentence of yours to which I was responding.
What I referred to is the modification in "++i"
to be sequenced before value computation of "++i".
The side effects of "++i" are not sequenced before anything in
any of the expressions in this thread.
I should probably have started another paragraph to make that
clear, but it seemed obvious to me.
and that value computation before value computation of the
"<lval>++" expression (because value computation of operands
of an operator is sequenced before value computation of their
result), which in turn is sequenced before modification of
"<lval>".
So since "sequenced before" is a transitive relation, both
side effects are sequenced with respect to each other and with
respect to value computations.
Where do you see any sequencing of the two side effects?
I see it in my above description.
Yes, but on what do you base your description? Where is
there text in the standard supporting it? (There may be,
but I've not seen it.) Where do you find anything
concerning the sequencing of side effects (as opposed to the
value computation) of "i++"? (The only thing I find is the
very generic text in =A71.9/14-15, which guarantees that the
side effects are sequenced before the end of the full
expression and before function calls. There's additional
text in =A75.14, =A75.15, =A75.16 and =A75.18 which guarantees the
sequencing of side effects where the &&, ||, ?: and comma
operator are involved. But that's all I find. (In fact,
the places where side effects are sequenced seem to
correspond exactly to the places where there are sequence
points in C++03. This isn't by accident; the intent is that
in a single threaded environment, the rules don't change.)
Please show us where you don't see it. I.e where the
sequence breaks for you, so we can work it out.
Otherwise, I'm just going to copy/paste the above
description again.
Your "description" is a vacuous claim. I'm asking about
evidence. As far as I can tell, the current rules do not
change anything with regards to what is or is not legal in
a single threaded program. (They do make it a lot clearer.)
--
James Kanze