Paolo Bonzini <[EMAIL PROTECTED]> writes:
>> Or you can do, since elsewhere in the code you compute time_t_max:
>> for (j = 1; j <= time_t_max / 2 + 1; j *= 2)
>
> No, this does not work. It would work to have:
>
> for (j = 1;;)
> {
> if (j > time_t_max / 2)
> break;
> j *= 2;
> }
>
> Oops.
"Oops" is my thought as well. Even the second version of your code is
incorrect in general, since it assumes that 'int' and 'time_t' are the
same width.
It's surprisingly tricky to get right. This underscores why -O2's
changes in this area are so worrisome.
What I eventually ended up doing -- and this was before reading the
above-quoted email -- was this:
for (j = 1; ; j <<= 1)
if (! bigtime_test (j))
return 1;
else if (INT_MAX / 2 < j)
break;
This is portable and should work on all real platforms.
But that was the easy code! Here's the harder stuff, the
code that computes time_t_max:
static time_t time_t_max;
static time_t time_t_min;
for (time_t_max = 1; 0 < time_t_max; time_t_max *= 2)
continue;
time_t_max--;
if ((time_t) -1 < 0)
for (time_t_min = -1; (time_t) (time_t_min * 2) < 0; time_t_min *= 2)
continue;
Obviously this code is buggy, at least in theory, due to the signed
integer overflows. But rewriting it is not so easy, since we have no
INT_MAX to rescue us as we did in the bigtime_test loop. Here's what
I eventually came up with:
for (;;)
{
time_t t = (time_t_max << 1) + 1;
if (t <= time_t_max)
break;
time_t_max = t;
}
time_t_min = - ((time_t) ~ (time_t) 0 == (time_t) -1) - time_t_max;
This isn't guaranteed to be portable by C99 either, of course; among
other things, left-shift has undefined behavior on signed integer
overflow. I am relying on your heuristic advice to use left shift
rather than multiplication by 2, so that GCC won't mess up here. But
it's a weak heuristic and I'm afraid it doesn't inspire a whole lot of
confidence.
