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.
Paolo
