我正在循环一些数据,计算一些 double 和每 2 个 __m128d
操作,我想将数据存储在 __m128
float 上。
所以 64+64 + 64+64 (2 __m128d
) 存入 1 32+32+32+32 __m128
.
我做了这样的事情:
__m128d v_result;
__m128 v_result_float;
...
// some operations on v_result
// store the first two "slot" on float
v_result_float = _mm_cvtpd_ps(v_result);
// some operations on v_result
// I need to store the last two "slot" on float
v_result_float = _mm_cvtpd_ps(v_result); ?!?
但它每次都会(显然)覆盖前 2 个 float “插槽”。
我如何“间隔”_mm_cvtpd_ps
以开始向 3° 和 4°“插槽”插入值,第二次?
完整代码如下:
__m128d v_pA;
__m128d v_pB;
__m128d v_result;
__m128 v_result_float;
float *pCEnd = pTest + roundintup8(blockSize);
for (; pTest < pCEnd; pA += 8, pB += 8, pTest += 8) {
v_pA = _mm_load_pd(pA);
v_pB = _mm_load_pd(pB);
v_result = _mm_add_pd(v_pA, v_pB);
v_result = _mm_max_pd(v_boundLower, v_result);
v_result = _mm_min_pd(v_boundUpper, v_result);
v_result = _mm_mul_pd(v_rangeLn2per12, v_result);
v_result = _mm_add_pd(v_minLn2per12, v_result);
// two double processed: store in 1° and 2° float slot
v_result_float = _mm_cvtpd_ps(v_result);
v_pA = _mm_load_pd(pA + 2);
v_pB = _mm_load_pd(pB + 2);
v_result = _mm_add_pd(v_pA, v_pB);
v_result = _mm_max_pd(v_boundLower, v_result);
v_result = _mm_min_pd(v_boundUpper, v_result);
v_result = _mm_mul_pd(v_rangeLn2per12, v_result);
v_result = _mm_add_pd(v_minLn2per12, v_result);
// another two double processed: store in 3° and 4° float slot
v_result_float = _mm_cvtpd_ps(v_result); // fail
v_result_float = someFunction(v_result_float);
_mm_store_ps(pTest, v_result_float);
v_pA = _mm_load_pd(pA + 4);
v_pB = _mm_load_pd(pB + 4);
v_result = _mm_add_pd(v_pA, v_pB);
v_result = _mm_max_pd(v_boundLower, v_result);
v_result = _mm_min_pd(v_boundUpper, v_result);
v_result = _mm_mul_pd(v_rangeLn2per12, v_result);
v_result = _mm_add_pd(v_minLn2per12, v_result);
// two double processed: store in 1° and 2° float slot
v_result_float = _mm_cvtpd_ps(v_result);
v_pA = _mm_load_pd(pA + 6);
v_pB = _mm_load_pd(pB + 6);
v_result = _mm_add_pd(v_pA, v_pB);
v_result = _mm_max_pd(v_boundLower, v_result);
v_result = _mm_min_pd(v_boundUpper, v_result);
v_result = _mm_mul_pd(v_rangeLn2per12, v_result);
v_result = _mm_add_pd(v_minLn2per12, v_result);
// another two double processed: store in 3° and 4° float slot
v_result_float = _mm_cvtpd_ps(v_result); // fail
v_result_float = someFunction(v_result_float);
_mm_store_ps(pTest + 4, v_result_float);
}
最佳答案
您需要使用movlhps
(_mm_movelh_ps
) 将第二次转换的低位字移动到第一次转换结果的高位字。简化示例:
#include <immintrin.h>
__m128d some_double_operation(__m128d);
__m128 some_float_operation(__m128);
void foo(double const* input, float* output, int size)
{
// assuming everything is already nicely aligned ...
for(int i=0; i<size; i+=4, input+=4, output+=4)
{
__m128d res_lo = some_double_operation(_mm_load_pd(input));
__m128d res_hi = some_double_operation(_mm_load_pd(input+2));
__m128 res_float = _mm_movelh_ps(_mm_cvtpd_ps(res_lo), _mm_cvtpd_ps(res_hi));
__m128 res_final = some_float_operation(res_float);
_mm_store_ps(output, res_final);
}
}
Godbolt-演示:https://godbolt.org/z/wgKjxN .
如果 some_double_operation
是内联的,编译器可能会将第一个 double 运算的结果保存在一个寄存器中,而该寄存器不会被第二次函数调用使用,因此不需要将任何内容存储到内存中。
关于c++ - 如何将两个_pd 转换为一个_ps?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54517648/