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
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