Optimize symmetry_bd with stride-based fast paths

This commit is contained in:
2026-03-01 15:50:56 +08:00
parent 03416a7b28
commit 16013081e0

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@@ -5,6 +5,7 @@
#include <stddef.h>
#include <math.h>
#include <stdio.h>
#include <string.h>
/* 主网格0-based -> 1D */
static inline size_t idx_ex(int i0, int j0, int k0, const int ex[3]) {
const int ex1 = ex[0], ex2 = ex[1];
@@ -87,60 +88,159 @@ static inline size_t idx_funcc_F(int iF, int jF, int kF, int ord, const int extc
* funcc(:,:,-i) = funcc(:,:,i+1)*SoA(3)
* enddo
*/
static inline void symmetry_bd_impl(int ord,
int shift,
const int extc[3],
const double *__restrict func,
double *__restrict funcc,
const double SoA[3])
{
const int extc1 = extc[0], extc2 = extc[1], extc3 = extc[2];
const int nx = extc1 + ord;
const int ny = extc2 + ord;
const size_t snx = (size_t)nx;
const size_t splane = (size_t)nx * (size_t)ny;
const size_t interior_i = (size_t)shift + 1u; /* iF = 1 */
const size_t interior_j = ((size_t)shift + 1u) * snx; /* jF = 1 */
const size_t interior_k = ((size_t)shift + 1u) * splane; /* kF = 1 */
const size_t interior0 = interior_k + interior_j + interior_i;
/* 1) funcc(1:extc1,1:extc2,1:extc3) = func */
for (int k0 = 0; k0 < extc3; ++k0) {
const double *src_k = func + (size_t)k0 * (size_t)extc2 * (size_t)extc1;
const size_t dst_k0 = interior0 + (size_t)k0 * splane;
for (int j0 = 0; j0 < extc2; ++j0) {
const double *src = src_k + (size_t)j0 * (size_t)extc1;
double *dst = funcc + dst_k0 + (size_t)j0 * snx;
memcpy(dst, src, (size_t)extc1 * sizeof(double));
}
}
/* 2) funcc(-i,1:extc2,1:extc3) = funcc(i+1,1:extc2,1:extc3)*SoA(1) */
const double s1 = SoA[0];
if (s1 == 1.0) {
for (int ii = 0; ii < ord; ++ii) {
const size_t dst_i = (size_t)(shift - ii);
const size_t src_i = (size_t)(shift + ii + 1);
for (int k0 = 0; k0 < extc3; ++k0) {
const size_t kbase = interior_k + (size_t)k0 * splane + interior_j;
for (int j0 = 0; j0 < extc2; ++j0) {
const size_t off = kbase + (size_t)j0 * snx;
funcc[off + dst_i] = funcc[off + src_i];
}
}
}
} else if (s1 == -1.0) {
for (int ii = 0; ii < ord; ++ii) {
const size_t dst_i = (size_t)(shift - ii);
const size_t src_i = (size_t)(shift + ii + 1);
for (int k0 = 0; k0 < extc3; ++k0) {
const size_t kbase = interior_k + (size_t)k0 * splane + interior_j;
for (int j0 = 0; j0 < extc2; ++j0) {
const size_t off = kbase + (size_t)j0 * snx;
funcc[off + dst_i] = -funcc[off + src_i];
}
}
}
} else {
for (int ii = 0; ii < ord; ++ii) {
const size_t dst_i = (size_t)(shift - ii);
const size_t src_i = (size_t)(shift + ii + 1);
for (int k0 = 0; k0 < extc3; ++k0) {
const size_t kbase = interior_k + (size_t)k0 * splane + interior_j;
for (int j0 = 0; j0 < extc2; ++j0) {
const size_t off = kbase + (size_t)j0 * snx;
funcc[off + dst_i] = funcc[off + src_i] * s1;
}
}
}
}
/* 3) funcc(:,-j,1:extc3) = funcc(:,j+1,1:extc3)*SoA(2) */
const double s2 = SoA[1];
if (s2 == 1.0) {
for (int jj = 0; jj < ord; ++jj) {
const size_t dst_j = (size_t)(shift - jj) * snx;
const size_t src_j = (size_t)(shift + jj + 1) * snx;
for (int k0 = 0; k0 < extc3; ++k0) {
const size_t kbase = interior_k + (size_t)k0 * splane;
double *dst = funcc + kbase + dst_j;
const double *src = funcc + kbase + src_j;
for (int i = 0; i < nx; ++i) dst[i] = src[i];
}
}
} else if (s2 == -1.0) {
for (int jj = 0; jj < ord; ++jj) {
const size_t dst_j = (size_t)(shift - jj) * snx;
const size_t src_j = (size_t)(shift + jj + 1) * snx;
for (int k0 = 0; k0 < extc3; ++k0) {
const size_t kbase = interior_k + (size_t)k0 * splane;
double *dst = funcc + kbase + dst_j;
const double *src = funcc + kbase + src_j;
for (int i = 0; i < nx; ++i) dst[i] = -src[i];
}
}
} else {
for (int jj = 0; jj < ord; ++jj) {
const size_t dst_j = (size_t)(shift - jj) * snx;
const size_t src_j = (size_t)(shift + jj + 1) * snx;
for (int k0 = 0; k0 < extc3; ++k0) {
const size_t kbase = interior_k + (size_t)k0 * splane;
double *dst = funcc + kbase + dst_j;
const double *src = funcc + kbase + src_j;
for (int i = 0; i < nx; ++i) dst[i] = src[i] * s2;
}
}
}
/* 4) funcc(:,:,-k) = funcc(:,:,k+1)*SoA(3) */
const double s3 = SoA[2];
if (s3 == 1.0) {
for (int kk = 0; kk < ord; ++kk) {
const size_t dst_k = (size_t)(shift - kk) * splane;
const size_t src_k = (size_t)(shift + kk + 1) * splane;
double *dst = funcc + dst_k;
const double *src = funcc + src_k;
for (size_t p = 0; p < splane; ++p) dst[p] = src[p];
}
} else if (s3 == -1.0) {
for (int kk = 0; kk < ord; ++kk) {
const size_t dst_k = (size_t)(shift - kk) * splane;
const size_t src_k = (size_t)(shift + kk + 1) * splane;
double *dst = funcc + dst_k;
const double *src = funcc + src_k;
for (size_t p = 0; p < splane; ++p) dst[p] = -src[p];
}
} else {
for (int kk = 0; kk < ord; ++kk) {
const size_t dst_k = (size_t)(shift - kk) * splane;
const size_t src_k = (size_t)(shift + kk + 1) * splane;
double *dst = funcc + dst_k;
const double *src = funcc + src_k;
for (size_t p = 0; p < splane; ++p) dst[p] = src[p] * s3;
}
}
}
static inline void symmetry_bd(int ord,
const int extc[3],
const double *func,
double *funcc,
const double SoA[3])
{
const int extc1 = extc[0], extc2 = extc[1], extc3 = extc[2];
if (ord <= 0) return;
// 1) funcc(1:extc1,1:extc2,1:extc3) = func
// Fortran 的 (iF=1..extc1) 对应 C 的 func(i0=0..extc1-1)
for (int k0 = 0; k0 < extc3; ++k0) {
for (int j0 = 0; j0 < extc2; ++j0) {
for (int i0 = 0; i0 < extc1; ++i0) {
const int iF = i0 + 1, jF = j0 + 1, kF = k0 + 1;
funcc[idx_funcc_F(iF, jF, kF, ord, extc)] = func[idx_func0(i0, j0, k0, extc)];
}
/* Fast paths used by current C kernels: ord=2 (derivs), ord=3 (lopsided/KO). */
if (ord == 2) {
symmetry_bd_impl(2, 1, extc, func, funcc, SoA);
return;
}
if (ord == 3) {
symmetry_bd_impl(3, 2, extc, func, funcc, SoA);
return;
}
// 2) do i=0..ord-1: funcc(-i, 1:extc2, 1:extc3) = funcc(i+1, ...)*SoA(1)
for (int ii = 0; ii <= ord - 1; ++ii) {
const int iF_dst = -ii; // 0, -1, -2, ...
const int iF_src = ii + 1; // 1, 2, 3, ...
for (int kF = 1; kF <= extc3; ++kF) {
for (int jF = 1; jF <= extc2; ++jF) {
funcc[idx_funcc_F(iF_dst, jF, kF, ord, extc)] =
funcc[idx_funcc_F(iF_src, jF, kF, ord, extc)] * SoA[0];
}
}
}
// 3) do i=0..ord-1: funcc(:,-i, 1:extc3) = funcc(:, i+1, 1:extc3)*SoA(2)
// 注意 Fortran 这里的 ":" 表示 iF 从 (-ord+1..extc1) 全覆盖
for (int jj = 0; jj <= ord - 1; ++jj) {
const int jF_dst = -jj;
const int jF_src = jj + 1;
for (int kF = 1; kF <= extc3; ++kF) {
for (int iF = -ord + 1; iF <= extc1; ++iF) {
funcc[idx_funcc_F(iF, jF_dst, kF, ord, extc)] =
funcc[idx_funcc_F(iF, jF_src, kF, ord, extc)] * SoA[1];
}
}
}
// 4) do i=0..ord-1: funcc(:,:,-i) = funcc(:,:, i+1)*SoA(3)
for (int kk = 0; kk <= ord - 1; ++kk) {
const int kF_dst = -kk;
const int kF_src = kk + 1;
for (int jF = -ord + 1; jF <= extc2; ++jF) {
for (int iF = -ord + 1; iF <= extc1; ++iF) {
funcc[idx_funcc_F(iF, jF, kF_dst, ord, extc)] =
funcc[idx_funcc_F(iF, jF, kF_src, ord, extc)] * SoA[2];
}
}
}
symmetry_bd_impl(ord, ord - 1, extc, func, funcc, SoA);
}
#endif