mpn/generic/mul_fft.c

949   mp_size_t pl, pl2, pl3, l;
957      We must have pl3 = 3/2 * pl2, with pl2 a multiple of 2^k2, and
958      pl3 a multiple of 2^k3. Since k3 >= k2, both are multiples of 2^k2,
959      and pl2 must be an even multiple of 2^k2. Thus (pl2,pl3) =
972       pl3 = 3 * pl2 / 2; /* since k>=FFT_FIRST_K=4, pl2 is a multiple of 2^4,
974       k3 = mpn_fft_best_k (pl3, sqr);
976   while (mpn_fft_next_size (pl3, k3) != pl3);
978   TRACE (printf ("mpn_mul_fft_full nl=%ld ml=%ld -> pl2=%ld pl3=%ld k=%d\n",
979 		 nl, ml, pl2, pl3, k2));
981   ASSERT_ALWAYS(pl3 <= pl);
982   cc = mpn_mul_fft (op, pl3, n, nl, m, ml, k3);     /* mu */
989   l = pl3 - pl2; /* l = pl2 / 2 since pl3 = 3/2 * pl2 */
1037   /* since pl2+pl3 >= pl, necessary the extra limbs (including cc) are zero */
1038   MPN_COPY (op + pl3, pad_op, pl - pl3);
1039   ASSERT_MPN_ZERO_P (pad_op + pl - pl3, pl2 + pl3 - pl);