| 17 | } |
| 18 | |
| 19 | double |
| 20 | difftime(time_t time1, time_t time0) |
| 21 | { |
| 22 | /* |
| 23 | ** If double is large enough, simply convert and subtract |
| 24 | ** (assuming that the larger type has more precision). |
| 25 | */ |
| 26 | if (sizeof(time_t) < sizeof(double)) { |
| 27 | double t1 = time1, t0 = time0; |
| 28 | return t1 - t0; |
| 29 | } |
| 30 | |
| 31 | /* |
| 32 | ** The difference of two unsigned values can't overflow |
| 33 | ** if the minuend is greater than or equal to the subtrahend. |
| 34 | */ |
| 35 | if (!TYPE_SIGNED(time_t)) |
| 36 | return time0 <= time1 ? time1 - time0 : dminus(time0 - time1); |
| 37 | |
| 38 | /* Use uintmax_t if wide enough. */ |
| 39 | if (sizeof(time_t) <= sizeof(uintmax_t)) { |
| 40 | uintmax_t t1 = time1, t0 = time0; |
| 41 | return time0 <= time1 ? t1 - t0 : dminus(t0 - t1); |
| 42 | } |
| 43 | |
| 44 | /* |
| 45 | ** Handle cases where both time1 and time0 have the same sign |
| 46 | ** (meaning that their difference cannot overflow). |
| 47 | */ |
| 48 | if ((time1 < 0) == (time0 < 0)) |
| 49 | return time1 - time0; |
| 50 | |
| 51 | /* |
| 52 | ** The values have opposite signs and uintmax_t is too narrow. |
| 53 | ** This suffers from double rounding; attempt to lessen that |
| 54 | ** by using long double temporaries. |
| 55 | */ |
| 56 | { |
| 57 | long double t1 = time1, t0 = time0; |
| 58 | return t1 - t0; |
| 59 | } |
| 60 | } |