使用 SSE 缩放复杂向量
Scaling of a complex vector using SSE
我想将 SSE 指令应用于包含复数的向量。没有 SSE 指令,我可以用下面的代码来完成。但是,当我应用 SSE 指令时,我不知道如何将计算出的实部和虚部返回到数组中。我该如何解决?
complex double * complexScaling(complex double *input, double c, int length)
{
for(int i=0; i<length; i++) {
input[i] = c*input[i];
}
return input;
}
complex double * complexScalingSSE(complex double *input, double c, int length)
{
__m128d multiplier,real,imag;
multiplier = _mm_set1_pd(c);
for(int i=0; i<length; i+=2) {
real = _mm_set_pd(creal(input[i]),creal(input[i+1]));
real = _mm_mul_pd(real, multiplier);
imag = _mm_set_pd(cimag(input[i]),cimag(input[i+1]));
imag = _mm_mul_pd(imag, multiplier);
}
return input;
}
您可能想要这样的东西:
complex double * complexScalingSSE(complex double *input, double c, int length)
{
const __m128d vc = _mm_set1_pd(c);
for (int i = 0; i < length; i++)
{
__m128d v = _mm_loadu_pd((double *)&input[i]); // load one complex double
v = _mm_mul_pd(v, vc); // scale it
_mm_storeu_pd((double *)&input[i], v); // store it
}
return input;
}
您可能想尝试以 2 倍展开循环(但请注意,对于大向量,此例程很可能 I/O 绑定):
complex double * complexScalingSSE(complex double *input, double c, int length)
{
const __m128d vc = _mm_set1_pd(c);
int i;
for (i = 0; i <= length - 2; i += 2)
{
__m128d v1 = _mm_loadu_pd((double *)&input[i]); // load two complex doubles
__m128d v2 = _mm_loadu_pd((double *)&input[i + 1]);
v1 = _mm_mul_pd(v1, vc); // scale them
v2 = _mm_mul_pd(v2, vc);
_mm_storeu_pd((double *)&input[i], v1); // store them
_mm_storeu_pd((double *)&input[i + 1], v2);
}
if (i < length) // handle odd element at end
{
__m128d v = _mm_loadu_pd((double *)&input[i]); // load one complex double
v = _mm_mul_pd(v, vc); // scale it
_mm_storeu_pd((double *)&input[i], v); // store it
}
return input;
}
另请注意,一个体面的编译器应该自动矢量化您的原始例程,在这种情况下,您不会从手工编码的 SSE 中看到任何好处。
我想将 SSE 指令应用于包含复数的向量。没有 SSE 指令,我可以用下面的代码来完成。但是,当我应用 SSE 指令时,我不知道如何将计算出的实部和虚部返回到数组中。我该如何解决?
complex double * complexScaling(complex double *input, double c, int length)
{
for(int i=0; i<length; i++) {
input[i] = c*input[i];
}
return input;
}
complex double * complexScalingSSE(complex double *input, double c, int length)
{
__m128d multiplier,real,imag;
multiplier = _mm_set1_pd(c);
for(int i=0; i<length; i+=2) {
real = _mm_set_pd(creal(input[i]),creal(input[i+1]));
real = _mm_mul_pd(real, multiplier);
imag = _mm_set_pd(cimag(input[i]),cimag(input[i+1]));
imag = _mm_mul_pd(imag, multiplier);
}
return input;
}
您可能想要这样的东西:
complex double * complexScalingSSE(complex double *input, double c, int length)
{
const __m128d vc = _mm_set1_pd(c);
for (int i = 0; i < length; i++)
{
__m128d v = _mm_loadu_pd((double *)&input[i]); // load one complex double
v = _mm_mul_pd(v, vc); // scale it
_mm_storeu_pd((double *)&input[i], v); // store it
}
return input;
}
您可能想尝试以 2 倍展开循环(但请注意,对于大向量,此例程很可能 I/O 绑定):
complex double * complexScalingSSE(complex double *input, double c, int length)
{
const __m128d vc = _mm_set1_pd(c);
int i;
for (i = 0; i <= length - 2; i += 2)
{
__m128d v1 = _mm_loadu_pd((double *)&input[i]); // load two complex doubles
__m128d v2 = _mm_loadu_pd((double *)&input[i + 1]);
v1 = _mm_mul_pd(v1, vc); // scale them
v2 = _mm_mul_pd(v2, vc);
_mm_storeu_pd((double *)&input[i], v1); // store them
_mm_storeu_pd((double *)&input[i + 1], v2);
}
if (i < length) // handle odd element at end
{
__m128d v = _mm_loadu_pd((double *)&input[i]); // load one complex double
v = _mm_mul_pd(v, vc); // scale it
_mm_storeu_pd((double *)&input[i], v); // store it
}
return input;
}
另请注意,一个体面的编译器应该自动矢量化您的原始例程,在这种情况下,您不会从手工编码的 SSE 中看到任何好处。