如何将两个_pd转换成一个_ps？

Question

我循环的一些数据，计算一些双，每2个__m128d操作，我想将数据存储在一个__m128浮动。

所以64 + 64 + 64 + 64（2 __m128d）存储到1 32 + 32 + 32 + 32 __m128。

我喜欢这样出头：

__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); ?!?

但它覆盖（明显）的前2浮动“槽”每次。

我的“空间”的_mm_cvtpd_ps如何开始插入值到3°和4°“时隙”，第二时间？

下面是完整的代码：

__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

你需要使用movlhps（_mm_movelh_ps）的第二次转换的低字移动到第一个转换结果的高字。简单的例子：

#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被内联，编译器可能会保持不使用第二次调用功能寄存器中的第一双操作的结果，因此不需要对任何存储内存。