On 02/25/2011 06:10 PM, Jonathan M Davis wrote:
On Friday, February 25, 2011 17:31:36 Ali Çehreli wrote:
On 02/25/2011 05:09 PM, bearophile wrote:
> int j;
> int[2] y;
> y[j] = j = 1;
I think that's undefined behavior in C and C++. It is not defined
whether j's previous or past value is used in y[j].
I would expect the situation be the same in D.
No, that should be perfectly defined. What's undefined is when you do something
like func(j, y[j]). The evaluation order of the function arguments is undefined.
However, the evaluation order when dealing with an assignment should be
defined.
I _could_ be wrong about that, but there's no question that the assignments
themselves are guaranteed to be done in right-to-left order.
- Jonathan M Davis
Standard texts are very difficult to read. I found a 2005 draft of the
C++ standard. 5 Expressions, paragraph 4:
<quote>
Except where noted, the order of evaluation of operands of individual
operators and subexpressions of individual expres-
sions, and the order in which side effects take place, is
unspecified.58) Between the previous and next sequence point a
scalar object shall have its stored value modified at most once by the
evaluation of an expression. Furthermore, the prior
value shall be accessed only to determine the value to be stored. The
requirements of this paragraph shall be met for
each allowable ordering of the subexpressions of a full expression;
otherwise the behavior is undefined. [ Example:
i = v [ i ++]; / / the behavior is undefined
i = 7 , i ++ , i ++; / / i becomes 9
i = ++ i + 1; / / the behavior is undefined
i = i + 1; / / the value of i is incremented
— end example ]
</quote>
To complete, footnote 58 is:
<quote>
58)
The precedence of operators is not directly specified, but it can be
derived from the syntax.
</quote>
The section for the assignment operator does not mention anything to the
contrary. According to my current understanding and my now-hazy
recollections of old discussions on C++ news groups, in the following
statement
a() = b() = c();
the value of c() is assigned to b(), and the value of that expression is
assigned to a(); but the order in which the three expressions are
evaluated are unspecified.
Hence, if the code behaves contrary to 5.1 above, it has undefined behavior.
This was bearophile's statement:
y[j] = j = 1;
The assignment operators do not introduce sequence points; there are
only two: before and after the whole line above. The code does not obey
"the prior value shall be accessed only to determine the value to be
stored". Above, the prior value is used to determine which element of y
is being assigned to.
Ali
