In article <d16e4f6a-c4f2-4b6f-bf7d-
ea1258da781c@[EMAIL PROTECTED]
>, james.kanze@[EMAIL PROTECTED]
says...
> On 9 mai, 19:21, "Victor Bazarov" <v.Abaza...@[EMAIL PROTECTED]
> wrote:
[ ... ]
> > For example, it's not undefined behaviour to do
>
> > f(i++) + ++i;
>
> Yes it is, since there is no sequence point between the two
> incrementations. Sequence points only define a partial
> ordering: there is a sequence point before calling f, but that
> only establishes and ordering between i++ and the call to f; it
> doesn't establish any ordering between the two incrementations.
I believe it establishes _some_ ordering, but not enough[1]. In
particular, I believe the compiler is required to treat evaluation of a
function argument and calling the function as atomic -- i.e. once the
evaluation of any argument takes place, it must proceed to evaluated the
other arguments (if any) and then call the function.
In the expression above, I don't believe it's allowed for the post-
increment to be evaluated, then the pre-increment, then the function
call. It is, however, allowed for the pre-increment, then the post-
increment, then the function call -- and in this ordering, there is no
sequence point between the pre-increment and the post-increment, so the
result is undefined behavior.
1: Though it's open to some question. $1.9/8 says: "Once the execution
of a function begins, no expressions from the calling function are
evaluated until execution of the called function has completed."
I'm interpreting evaluating the arguments to a function as part of
execution of the function. If you choose to interpret it as a completely
separate act that happens before the function's execution, then you're
right -- no ordering is defined. At least in this case, it doesn't make
any real difference though -- the overall result is undefined behavior
either way.
--
Later,
Jerry.
The universe is a figment of its own imagination.
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