我的第一个 CUDA 应用程序有问题。基本上它应该生成 N 个线性微分方程并使用一阶方法对它们进行数值求解。变量t（时间）从 0 迭代到T步长 = TAU= 0.0001。如果T足够小（比如 0.001），一切正常，但如果T== 0.1 或更大，内核似乎什么也不做。我如何检查这种情况？
N- 方程数，TAU- 时间步长，TN- 每个块的线程数，T- 结束时间
变量r不做任何事情。我用它来验证内核是否做了任何事情。因此，如果T== 0.0001，则r== 283，但如果T== 0.1，则r== 0。
#include <stdio.h>
#include <math.h>
#include <time.h>
#define N 4096
#define TAU 0.0001f
#define TN 2
#define T 0.1f
#define PI 3.141592f
__global__ void kern(float* v, float* m, float* r)
{
*r = 283;
__syncthreads();
int tid = blockIdx.x*TN + threadIdx.x;
for(float t = 0; t <= T; t += TAU)
{
float f = 0;
__syncthreads();
for(int k = 0; k < N; ++k)
f += m[N*tid + k]*v[k];
f *= TAU;
f += v[tid];
__syncthreads();
v[tid] = f;
}
}
int main()
{
float* v = new float[N];
float* m = new float[N*N];
for(int i = 0; i < N; ++i)
v[i] = sin(2*PI*i/N); //setting initial conditions
for(int i = 0; i < N*N; ++i)
m[i] = cos(2*PI*i/(N*N)); //coefficients in right hand part of the equations
// printing some of the values (total: 8 values) to compare with result
for(int i = 0; i < N*N; i += N*N / 8) printf("%f ", m[i]); printf("\n\n");
for(int i = 0; i < N; i += N / 8) printf("%f ", v[i]); printf("\n");
float* cv;
float* cm;
float* cr;
cudaMalloc((void**)&cv, N*sizeof(float));
cudaMalloc((void**)&cm, N*N*sizeof(float));
cudaMalloc((void**)&cr, sizeof(float));
cudaMemcpy(cv, v, N*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(cm, m, N*N*sizeof(float), cudaMemcpyHostToDevice);
dim3 blocks(N / TN);
dim3 threads(TN);
time_t ts = time(0);
printf("starting kernel\n");
kern<<<blocks, threads>>>(cv, cm, cr);
printf("kernel stopped\n");
time_t ts_end = time(0);
cudaMemcpy(v, cv, N*sizeof(float), cudaMemcpyDeviceToHost);
float r;
cudaMemcpy(&r, cr, sizeof(float), cudaMemcpyDeviceToHost);
for(int i = 0; i < N; i += N / 8) printf("%f ", v[i]); printf("\n");
printf("%d\n", ts_end - ts);
printf("result: %f\n", r);
delete[] m;
delete[] v;
cudaFree(cv);
cudaFree(cm);
cudaFree(cr);
}