transposer pour les 8 registres d'éléments 16 bits sur SSE2 / SSSE3
https://stackoverflow.com/questions/2517584
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(je suis un débutant à SSE / asm, excuses si cela est évident ou redondant)
Y at-il une meilleure façon de transposer 8 registres SSE contenant des valeurs 16 bits que l'exécution 24 [lh] unpck ps et 8/16 + battages et en utilisant 8 registres supplémentaires? (Note à l'aide jusqu'à 3 SSSE instructions, Intel Merom, manquant aka MELANGE * de SSE4.)
Supposons que vous avez des registres v [0-7] et utiliser t0-t7 comme aux registres. En pseudo code intrinsics:
/* Phase 1: process lower parts of the registers */
/* Level 1: work first part of the vectors */
/* v[0] A0 A1 A2 A3 A4 A5 A6 A7
** v[1] B0 B1 B2 B3 B4 B5 B6 B7
** v[2] C0 C1 C2 C3 C4 C5 C6 C7
** v[3] D0 D1 D2 D3 D4 D5 D6 D7
** v[4] E0 E1 E2 E3 E4 E5 E6 E7
** v[5] F0 F1 F2 F3 F4 F5 F6 F7
** v[6] G0 G1 G2 G3 G4 G5 G6 G7
** v[7] H0 H1 H2 H3 H4 H5 H6 H7 */
t0 = unpcklps (v[0], v[1]); /* Extract first half interleaving */
t1 = unpcklps (v[2], v[3]); /* Extract first half interleaving */
t2 = unpcklps (v[4], v[5]); /* Extract first half interleaving */
t3 = unpcklps (v[6], v[7]); /* Extract first half interleaving */
t0 = pshufhw (t0, 0xD8); /* Flip middle 2 high */
t0 = pshuflw (t0, 0xD8); /* Flip middle 2 low */
t1 = pshufhw (t1, 0xD8); /* Flip middle 2 high */
t1 = pshuflw (t1, 0xD8); /* Flip middle 2 low */
t2 = pshufhw (t2, 0xD8); /* Flip middle 2 high */
t2 = pshuflw (t2, 0xD8); /* Flip middle 2 low */
t3 = pshufhw (t3, 0xD8); /* Flip middle 2 high */
t3 = pshuflw (t3, 0xD8); /* Flip middle 2 low */
/* t0 A0 B0 A1 B1 A2 B2 A3 B3 (A B - 0 1 2 3)
** t1 C0 D0 C1 D1 C2 D2 C3 D3 (C D - 0 1 2 3)
** t2 E0 F0 E1 F1 E2 F2 E3 F3 (E F - 0 1 2 3)
** t3 G0 H0 G1 H1 G2 H2 G3 H3 (G H - 0 1 2 3) */
/* L2 */
t4 = unpcklps (t0, t1);
t5 = unpcklps (t2, t3);
t6 = unpckhps (t0, t1);
t7 = unpckhps (t2, t3);
/* t4 A0 B0 C0 D0 A1 B1 C1 D1 (A B C D - 0 1)
** t5 E0 F0 G0 H0 E1 F1 G1 H1 (E F G H - 0 1)
** t6 A2 B2 C2 D2 A3 B3 C3 D3 (A B C D - 2 3)
** t7 E2 F2 G2 H2 E3 F3 G3 H3 (E F G H - 2 3) */
/* Phase 2: same with higher parts of the registers */
/* A A0 A1 A2 A3 A4 A5 A6 A7
** B B0 B1 B2 B3 B4 B5 B6 B7
** C C0 C1 C2 C3 C4 C5 C6 C7
** D D0 D1 D2 D3 D4 D5 D6 D7
** E E0 E1 E2 E3 E4 E5 E6 E7
** F F0 F1 F2 F3 F4 F5 F6 F7
** G G0 G1 G2 G3 G4 G5 G6 G7
** H H0 H1 H2 H3 H4 H5 H6 H7 */
t0 = unpckhps (v[0], v[1]);
t0 = pshufhw (t0, 0xD8); /* Flip middle 2 high */
t0 = pshuflw (t0, 0xD8); /* Flip middle 2 low */
t1 = unpckhps (v[2], v[3]);
t1 = pshufhw (t1, 0xD8); /* Flip middle 2 high */
t1 = pshuflw (t1, 0xD8); /* Flip middle 2 low */
t2 = unpckhps (v[4], v[5]);
t2 = pshufhw (t2, 0xD8); /* Flip middle 2 high */
t2 = pshuflw (t2, 0xD8); /* Flip middle 2 low */
t3 = unpckhps (v[6], v[7]);
t3 = pshufhw (t3, 0xD8); /* Flip middle 2 high */
t3 = pshuflw (t3, 0xD8); /* Flip middle 2 low */
/* t0 A4 B4 A5 B5 A6 B6 A7 B7 (A B - 4 5 6 7)
** t1 C4 D4 C5 D5 C6 D6 C7 D7 (C D - 4 5 6 7)
** t2 E4 F4 E5 F5 E6 F6 E7 F7 (E F - 4 5 6 7)
** t3 G4 H4 G5 H5 G6 H6 G7 H7 (G H - 4 5 6 7) */
/* Back to first part, v[0-3] can be re-written now */
/* L3 */
v[0] = unpcklpd (t4, t5);
v[1] = unpckhpd (t4, t5);
v[2] = unpcklpd (t6, t7);
v[3] = unpckhpd (t6, t7);
/* v[0] = A0 B0 C0 D0 E0 F0 G0 H0
** v[1] = A1 B1 C1 D1 E1 F1 G1 H1
** v[2] = A2 B2 C2 D2 E2 F2 G2 H2
** v[3] = A3 B3 C3 D3 E3 F3 G3 H3 */
/* Back to second part, t[4-7] can be re-written now... */
/* L2 */
t4 = unpcklps (t0, t1);
t5 = unpcklps (t2, t3);
t6 = unpckhps (t0, t1);
t7 = unpckhps (t2, t3);
/* t4 A4 B4 C4 D4 A5 B5 C5 D5 (A B C D - 4 5)
** t5 E4 F4 G4 H4 E5 F5 G5 H5 (E F G H - 4 5)
** t6 A6 B6 C6 D6 A7 B7 C7 D7 (A B C D - 6 7)
** t7 E6 F6 G6 H6 E7 F7 G7 H7 (E F G H - 6 7) */
/* L3 */
v[4] = unpcklpd (t4, t5);
v[5] = unpckhpd (t4, t5);
v[6] = unpcklpd (t6, t7);
v[7] = unpckhpd (t6, t7);
/* v[4] = A4 B4 C4 D4 E4 F4 G4 H4
** v[5] = A5 B5 C5 D5 E5 F5 G5 H5
** v[6] = A6 B6 C6 D6 E6 F6 G6 H6
** v[7] = A7 B7 C7 D7 E7 F7 G7 H7 */
Chaque unpck * prend 3 cycles de latence ou 2 pour un débit réciproque (rapporté par Agner.) Ce ne tue grande partie des gains de performance de l'utilisation de SSE (sur ce code) car cette danse du registre prend presque un cycle par élément . J'ai essayé de comprendre le fichier asm de x264 x86 pour transposition, mais la compréhension des macros a échoué.
Merci!
La solution
Oui, vous pouvez le faire en 24 instructions au total:
8 x _mm_unpacklo_epi16/_mm_unpackhi_epi16 (PUNPCKLWD/PUNPCKHWD)
8 x _mm_unpacklo_epi32/_mm_unpackhi_epi32 (PUNPCKLDQ/PUNPCKHDQ)
8 x _mm_unpacklo_epi64/_mm_unpackhi_epi64 (PUNPCKLQDQ/PUNPCKHQDQ)
Faites-moi savoir si vous avez besoin de plus de détails, mais il est assez évident.
Autres conseils
Je devais faire moi-même, voici donc mon résultat final. Notez que les instructions de chargement et de stockage que j'ai utilisés sont pour les tableaux alignés 16 octets, qui ont été déclarés à l'aide m128i * = matrice (__m128i *) _mm_malloc (N * sizeof (uint16_t), 16); OU tableau int16_t [N] __ attribut ((alignés (16)));
__m128i *p_input = (__m128i*)array;
__m128i *p_output = (__m128i*)array;
__m128i a = _mm_load_si128(p_input++);
__m128i b = _mm_load_si128(p_input++);
__m128i c = _mm_load_si128(p_input++);
__m128i d = _mm_load_si128(p_input++);
__m128i e = _mm_load_si128(p_input++);
__m128i f = _mm_load_si128(p_input++);
__m128i g = _mm_load_si128(p_input++);
__m128i h = _mm_load_si128(p_input);
__m128i a03b03 = _mm_unpacklo_epi16(a, b);
__m128i c03d03 = _mm_unpacklo_epi16(c, d);
__m128i e03f03 = _mm_unpacklo_epi16(e, f);
__m128i g03h03 = _mm_unpacklo_epi16(g, h);
__m128i a47b47 = _mm_unpackhi_epi16(a, b);
__m128i c47d47 = _mm_unpackhi_epi16(c, d);
__m128i e47f47 = _mm_unpackhi_epi16(e, f);
__m128i g47h47 = _mm_unpackhi_epi16(g, h);
__m128i a01b01c01d01 = _mm_unpacklo_epi32(a03b03, c03d03);
__m128i a23b23c23d23 = _mm_unpackhi_epi32(a03b03, c03d03);
__m128i e01f01g01h01 = _mm_unpacklo_epi32(e03f03, g03h03);
__m128i e23f23g23h23 = _mm_unpackhi_epi32(e03f03, g03h03);
__m128i a45b45c45d45 = _mm_unpacklo_epi32(a47b47, c47d47);
__m128i a67b67c67d67 = _mm_unpackhi_epi32(a47b47, c47d47);
__m128i e45f45g45h45 = _mm_unpacklo_epi32(e47f47, g47h47);
__m128i e67f67g67h67 = _mm_unpackhi_epi32(e47f47, g47h47);
__m128i a0b0c0d0e0f0g0h0 = _mm_unpacklo_epi64(a01b01c01d01, e01f01g01h01);
__m128i a1b1c1d1e1f1g1h1 = _mm_unpackhi_epi64(a01b01c01d01, e01f01g01h01);
__m128i a2b2c2d2e2f2g2h2 = _mm_unpacklo_epi64(a23b23c23d23, e23f23g23h23);
__m128i a3b3c3d3e3f3g3h3 = _mm_unpackhi_epi64(a23b23c23d23, e23f23g23h23);
__m128i a4b4c4d4e4f4g4h4 = _mm_unpacklo_epi64(a45b45c45d45, e45f45g45h45);
__m128i a5b5c5d5e5f5g5h5 = _mm_unpackhi_epi64(a45b45c45d45, e45f45g45h45);
__m128i a6b6c6d6e6f6g6h6 = _mm_unpacklo_epi64(a67b67c67d67, e67f67g67h67);
__m128i a7b7c7d7e7f7g7h7 = _mm_unpackhi_epi64(a67b67c67d67, e67f67g67h67);
_mm_store_si128(p_output++, a0b0c0d0e0f0g0h0);
_mm_store_si128(p_output++, a1b1c1d1e1f1g1h1);
_mm_store_si128(p_output++, a2b2c2d2e2f2g2h2);
_mm_store_si128(p_output++, a3b3c3d3e3f3g3h3);
_mm_store_si128(p_output++, a4b4c4d4e4f4g4h4);
_mm_store_si128(p_output++, a5b5c5d5e5f5g5h5);
_mm_store_si128(p_output++, a6b6c6d6e6f6g6h6);
_mm_store_si128(p_output, a7b7c7d7e7f7g7h7);
Mon idée vient de ce http://www.randombit.net/bitbashing/programming /integer_matrix_transpose_in_sse2.html
je voudrais segmenter l'une 8x8 4x4 à quatre et que l'affaire mentionnée. enfin échanger le bloc (0,1) et le bloc (1,0)
cependant, je ne comprends toujours pas ce truc de Paul R. Paul aurait-vous me donner quelques coups?
