我想计算'out = alpha * px + beta * py'，'px'和'py'是数组。*
我有一个简单的内核：
__global__
void saxpyGPU2( float *out, const float *px, const float *py, size_t N, float alpha,float beta )
{
size_t i = blockDim.x*blockIdx.x + threadIdx.x;
while (i < N)
{
out[i] = alpha * px[i] + beta * py[i];
i += blockDim.x*gridDim.x;
}
}
它有效，所以我想循环展开。
cuda-handbook中的代码是：
template<const int n>
__device__
void saxpy_unrolled(float *out, const float *px, const float *py, size_t N, float alpha,float beta)
{
float x[n], y[n];
size_t i;
for ( i = n*blockIdx.x*blockDim.x+threadIdx.x; i < N-n*blockDim.x*gridDim.x; i += n*blockDim.x*gridDim.x ) {
for ( int j = 0; j < n; j++ ) {
size_t index = i+j*blockDim.x;
x[j] = px[index];
y[j] = py[index];
}
for ( int j = 0; j < n; j++ ) {
size_t index = i+j*blockDim.x;
out[index] = alpha*x[j]+beta* y[j];
}
}
// to avoid the (index<N) conditional in the inner loop,
// we left off some work at the end
for ( int j = 0; j < n; j++ ) {
for ( int j = 0; j < n; j++ ) {
size_t index = i+j*blockDim.x;
if ( index<N ) {
x[j] = px[index];
y[j] = py[index];
}
}
for ( int j = 0; j < n; j++ ) {
size_t index = i+j*blockDim.x;
if ( index<N ) out[index] = alpha*x[j]+beta* y[j];
}
}
}
__global__
void saxpyGPU( float *out, const float *px, const float *py, size_t N, float alpha,float beta )
{
saxpy_unrolled<4>( out, px, py, N, alpha ,beta);
}
当我＆gt;我在第二个分支时不明白N-正* blockDim.x * gridDim.x。为什么要使用外循环
for ( int j = 0; j < n; j++ ) {
for ( int j = 0; j < n; j++ )....}
我测试了这两个内核，第一个是好的，但是我从书中复制的第二个是不正确的。
我初始化了两个数组while(i<1024) a[i] = i; b[i] = 10*i;i++，我想计算c = alpha * a + beta * b使用上面的两个内核，但是循环展开的内核中的结果对于c中的所有元素都是4.3e8
这是我的测试代码：
int main(){
int arraySize = 1024;
float* a =new float[arraySize];
float* b =new float[arraySize];
float* c =new float[arraySize];
for (int i =0;i<arraySize;i++)
{
a[i] = 1.0* i;
b[i] = 10.0*i;
c[i] = 0.0;
}
float* d_a;
float* d_b;
float* d_c;
cudaMalloc((void**)&d_a,sizeof(float)*arraySize);
cudaMemcpy(d_a,a,sizeof(float)*arraySize,cudaMemcpyHostToDevice);
cudaMalloc((void**)&d_b,sizeof(float)*arraySize);
cudaMemcpy(d_b,b,sizeof(float)*arraySize,cudaMemcpyHostToDevice);
cudaMalloc((void**)&d_c,sizeof(float)*arraySize);
for (int i=0;i<arraySize;i++)
{
c[i] = a[i] + b[i];
}
dim3 block_size(256,1,1);
dim3 grid_size((arraySize -1)/block_size.x+1,1,1);
float alpha = 1.0;
float beta = 1.0;
bool flag = true;
if(flag)
{
saxpyGPU<<<grid_size,block_size>>>(d_c,d_a,d_b,arraySize,alpha,beta);
float* temp = new float[arraySize];
cudaMemcpy(temp,d_c,arraySize*sizeof(float),cudaMemcpyDeviceToHost);
for (int i = 0;i<arraySize;i++)
{
cout<<(temp[i] - c[i])<<",";
}
}
else
{
saxpyGPU2<<<grid_size,block_size>>>(d_c,d_a,d_b,arraySize,alpha,beta);
cudaMemcpy(temp,d_c,arraySize*sizeof(float),cudaMemcpyDeviceToHost);
for (int i = 0;i<arraySize;i++)
{
cout<<(temp[i] - c[i])<<",";
}
这两个内核显示不同的结果
答案 0 :(得分：2)
您发布的内核代码完全正确并产生预期结果。这可以使用以下代码演示：
#include <thrust/random.h>
#include <thrust/device_vector.h>
#include <thrust/transform.h>
#include <thrust/copy.h>
#include <thrust/iterator/counting_iterator.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
template<const int n>
__device__
void saxpy_unrolled(float *out, const float *px, const float *py,
size_t N, float alpha,float beta) {
float x[n], y[n];
size_t i;
for ( i = n*blockIdx.x*blockDim.x+threadIdx.x;
i < N-n*blockDim.x*gridDim.x;
i += n*blockDim.x*gridDim.x ) {
for ( int j = 0; j < n; j++ ) {
size_t index = i+j*blockDim.x;
x[j] = px[index];
y[j] = py[index];
}
for ( int j = 0; j < n; j++ ) {
size_t index = i+j*blockDim.x;
out[index] = alpha*x[j]+beta* y[j];
}
}
for ( int j = 0; j < n; j++ ) {
for ( int j = 0; j < n; j++ ) {
size_t index = i+j*blockDim.x;
if ( index<N ) {
x[j] = px[index];
y[j] = py[index];
}
}
for ( int j = 0; j < n; j++ ) {
size_t index = i+j*blockDim.x;
if ( index<N ) {
out[index] = alpha*x[j] + beta*y[j];
}
}
}
}
__global__
void saxpyGPU( float *out, const float *px, const float *py,
size_t N, float alpha,float beta ) {
saxpy_unrolled<4>( out, px, py, N, alpha ,beta);
}
struct prg {
float a, b;
__host__ __device__
prg(float _a=0.f, float _b=1.f) : a(_a), b(_b) {};
__host__ __device__
float operator()(const unsigned int n) const {
thrust::default_random_engine rng;
thrust::uniform_real_distribution<float> dist(a, b);
rng.discard(n);
return dist(rng);
}
};
int main(void) {
const int N = 100000;
const float alpha = 0.12345f, beta = 0.9876f;
prg gen(1.f, 2.f);
thrust::device_vector<float> x(N), y(N), z(N);
thrust::counting_iterator<unsigned int> iseqx(0);
thrust::counting_iterator<unsigned int> iseqy(N);
thrust::transform(iseqx, iseqx + N, x.begin(), gen);
thrust::transform(iseqy, iseqy + N, y.begin(), gen);
float *xp = thrust::raw_pointer_cast(&x[0]);
float *yp = thrust::raw_pointer_cast(&y[0]);
float *zp = thrust::raw_pointer_cast(&z[0]);
dim3 blockdim(128);
dim3 griddim(16);
saxpyGPU<<<griddim, blockdim>>>(zp, xp, yp, N, alpha, beta);
cudaDeviceSynchronize();
std::vector<float> xh(N), yh(N), zh(N);
thrust::copy(x.begin(), x.end(), xh.begin());
thrust::copy(y.begin(), y.end(), yh.begin());
thrust::copy(z.begin(), z.end(), zh.begin());
float maxabserr = -1.f, maxrelerr = -1.f;
for(int i=0; i<N; i++) {
float saxpyval = alpha * xh[i] + beta * yh[i];
float abserr = fabs(zh[i]-saxpyval);
float relerr = abserr / fmaxf(fabs(zh[i]), fabs(saxpyval));
maxabserr = fmaxf(abserr, maxabserr);
maxrelerr = fmaxf(relerr, maxrelerr);
}
std::cout.precision(10);
std::cout << "Maximum absolute error = " << maxabserr << std::endl;
std::cout << "Maximum relative error = " << maxrelerr << std::endl;
return 0;
}
给了我以下内容：
$ nvcc -arch=sm_30 -o unrolled_saxpy unrolled_saxpy.cu
$ ./unrolled_saxpy
Maximum absolute error = 2.384185791e-07
Maximum relative error = 1.1920676e-07
如果您（仍然）不理解为什么内核按原样编写，请按照我在上一个问题中向您展示的内容并手动展开saxpy函数。从n = 1开始并确认它在功能上与展开的等效物相同，然后尝试n = 2，n = 4等，以查看循环展开的动作是什么。