我写了一个简单的程序来实现SSE内在函数来计算两个大(100000或更多元素)向量的内积。该程序比较两种内部产品的执行时间,以传统方式计算内部产品和使用内在函数。一切都很好,直到我在计算内积的语句之前插入(仅为了它的乐趣)一个内循环。在我走得更远之前,这是代码:
//this is a sample Intrinsics program to compute inner product of two vectors and compare Intrinsics with traditional method of doing things.
#include <iostream>
#include <iomanip>
#include <xmmintrin.h>
#include <stdio.h>
#include <time.h>
#include <stdlib.h>
using namespace std;
typedef float v4sf __attribute__ ((vector_size(16)));
double innerProduct(float* arr1, int len1, float* arr2, int len2) { //assume len1 = len2.
float result = 0.0;
for(int i = 0; i < len1; i++) {
for(int j = 0; j < len1; j++) {
result += (arr1[i] * arr2[i]);
}
}
//float y = 1.23e+09;
//cout << "y = " << y << endl;
return result;
}
double sse_v4sf_innerProduct(float* arr1, int len1, float* arr2, int len2) { //assume that len1 = len2.
if(len1 != len2) {
cout << "Lengths not equal." << endl;
exit(1);
}
/*steps:
* 1. load a long-type (4 float) into a v4sf type data from both arrays.
* 2. multiply the two.
* 3. multiply the same and store result.
* 4. add this to previous results.
*/
v4sf arr1Data, arr2Data, prevSums, multVal, xyz;
//__builtin_ia32_xorps(prevSums, prevSums); //making it equal zero.
//can explicitly load 0 into prevSums using loadps or storeps (Check).
float temp[4] = {0.0, 0.0, 0.0, 0.0};
prevSums = __builtin_ia32_loadups(temp);
float result = 0.0;
for(int i = 0; i < (len1 - 3); i += 4) {
for(int j = 0; j < len1; j++) {
arr1Data = __builtin_ia32_loadups(&arr1[i]);
arr2Data = __builtin_ia32_loadups(&arr2[i]); //store the contents of two arrays.
multVal = __builtin_ia32_mulps(arr1Data, arr2Data); //multiply.
xyz = __builtin_ia32_addps(multVal, prevSums);
prevSums = xyz;
}
}
//prevSums will hold the sums of 4 32-bit floating point values taken at a time. Individual entries in prevSums also need to be added.
__builtin_ia32_storeups(temp, prevSums); //store prevSums into temp.
cout << "Values of temp:" << endl;
for(int i = 0; i < 4; i++)
cout << temp[i] << endl;
result += temp[0] + temp[1] + temp[2] + temp[3];
return result;
}
int main() {
clock_t begin, end;
int length = 100000;
float *arr1, *arr2;
double result_Conventional, result_Intrinsic;
// printStats("Allocating memory.");
arr1 = new float[length];
arr2 = new float[length];
// printStats("End allocation.");
srand(time(NULL)); //init random seed.
// printStats("Initializing array1 and array2");
begin = clock();
for(int i = 0; i < length; i++) {
// for(int j = 0; j < length; j++) {
// arr1[i] = rand() % 10 + 1;
arr1[i] = 2.5;
// arr2[i] = rand() % 10 - 1;
arr2[i] = 2.5;
// }
}
end = clock();
cout << "Time to initialize array1 and array2 = " << ((double) (end - begin)) / CLOCKS_PER_SEC << endl;
// printStats("Finished initialization.");
// printStats("Begin inner product conventionally.");
begin = clock();
result_Conventional = innerProduct(arr1, length, arr2, length);
end = clock();
cout << "Time to compute inner product conventionally = " << ((double) (end - begin)) / CLOCKS_PER_SEC << endl;
// printStats("End inner product conventionally.");
// printStats("Begin inner product using Intrinsics.");
begin = clock();
result_Intrinsic = sse_v4sf_innerProduct(arr1, length, arr2, length);
end = clock();
cout << "Time to compute inner product with intrinsics = " << ((double) (end - begin)) / CLOCKS_PER_SEC << endl;
//printStats("End inner product using Intrinsics.");
cout << "Results: " << endl;
cout << " result_Conventional = " << result_Conventional << endl;
cout << " result_Intrinsics = " << result_Intrinsic << endl;
return 0;
}
我使用以下g ++调用来构建它:
g++ -W -Wall -O2 -pedantic -march=i386 -msse intrinsics_SSE_innerProduct.C -o innerProduct
在两个函数中,上面的每个循环总共运行N ^ 2次。但是,假设arr1和arr2(两个浮点向量)加载值为2.5,则数组的长度为100,000,两种情况下的结果应为6.25e + 10。我得到的结果是:
结果:
result_Conventional = 6.25e + 10
result_Intrinsics = 5.36871e + 08
这不是全部。似乎从使用内在函数的函数返回的值在上面的值“饱和”。我尝试为数组的元素和不同的大小添加其他值。但似乎数组内容的任何大于1.0的值和1000以上的任何大小都符合我们上面看到的相同值。
最初,我认为这可能是因为SSE中的所有操作都处于浮点状态,但浮点应该能够存储大约e + 08的数字。
我试图看看我哪里可能出错但似乎无法弄明白。我使用的是g ++版本:g ++(GCC)4.4.1 20090725(Red Hat 4.4.1-2)。
欢迎任何帮助。
谢谢,
斯利拉姆。
答案 0 :(得分:5)
你遇到的问题是虽然float
可以存储6.25e + 10,但它只有几个有效位数。
这意味着当您通过一次添加许多小数字来构建大数字时,您会到达较小数字小于较大数字中最低精度数字的点,因此将其相加没有效果。
至于为什么你没有在非内在版本中得到这种行为,很可能{1}}变量被保存在一个寄存器中,该寄存器使用更高的精度来实现浮动的实际存储,所以它是在循环的每次迭代中都不会被截断为result
的精度。您必须查看生成的汇编程序代码才能确定。