assembly - 转置 SSE2/SSSE3 上的 8 个 16 位元素寄存器

Question

(我是 SSE/asm 的新手，如果这很明显或多余，我深表歉意)

有没有比执行 24 个 unpck[lh]ps 和 8/16+ shuffle 并使用 8 个额外寄存器更好的方法来转置包含 16 位值的 8 个 SSE 寄存器？ (注意最多使用 SSSE 3 指令，Intel Merom，也就是缺少 SSE4 中的 BLEND*。)

假设您有寄存器 v[0-7] 并使用 t0-t7 作为辅助寄存器。在伪内在函数代码中:

/* 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 */

每个 unpck* 需要 3 个延迟周期，或 2 个相互吞吐量周期(由 Agner 报告)。这扼杀了使用 SSE(在此代码上)的大部分性能提升，因为这种寄存器舞蹈几乎每个元素需要一个周期。我试图理解 x86 转置的 x264 的 asm 文件，但未能理解宏。

谢谢!

Answer 1

是的，您总共可以使用 24 条指令来完成:

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)

如果您需要更多详细信息，请告诉我，但这很明显。