Question

我正在循环一些数据，计算两次double，每2 __m128d个操作，我想将数据存储在__m128浮点数上。

因此将64 + 64 + 64 + 64（2 __m128d）存储到1 32 + 32 + 32 + 32 __m128中。

我做这样的事情：

__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); ?!?

但是它每次都会覆盖（显然）前两个浮点数“ slots”。

如何第二次“间隔” _mm_cvtpd_ps以开始将值插入3°和4°的“槽”？

这是完整的代码：

__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);
}

Answer 1

您需要使用movlhps（_mm_movelh_ps）将第二次转换的低位词移至第一次转换结果的高位词。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-Demo：https://godbolt.org/z/wgKjxN。

如果内联some_double_operation，则编译器可能会将第一个double操作的结果保存在该函数的第二个调用未使用的寄存器中，因此不需要将任何内容存储到内存中。

如何将两个_pd转换为一个_ps？

1 个答案: