如何将两个_pd 转换为一个_ps? [英] How to convert two _pd into one _ps?

问题描述

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 浮点数上.

I'm looping some data, calculating some double and every 2 __m128d operations, I want to store the data on a __m128 float.

所以 64+64 + 64+64 (2 __m128d) 存入 1 32+32+32+32 __m128.

So 64+64 + 64+64 (2 __m128d) stored into 1 32+32+32+32 __m128.

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); ?!?

但它每次都会覆盖(显然)前 2 个浮动插槽".

But it overwrite (obviously) the first 2 float "slots" everytime.

我怎样才能空格" _mm_cvtpd_ps 以开始第二次将值插入 3° 和 4°槽"?

How can I "space" the _mm_cvtpd_ps to start insert values to the 3° and 4° "slot", the second time?

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);
}

推荐答案

需要使用movlhps(<代码>_mm_movelh_ps).简化示例:

You need to move the low words of the second conversion to the high words of the result of the first conversion using movlhps (_mm_movelh_ps). Simplified example:

#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.

Godbolt-Demo: https://godbolt.org/z/wgKjxN.

如果 some_double_operation 是内联的，编译器可能会将第一次双精度操作的结果保存在第二次调用函数未使用的寄存器中，因此不需要将任何内容存储到内存中.

If some_double_operation is inlined, the compiler will likely keep the result of the first double operation in a register not used by the second call to the function, thus not require to store anything to memory.

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