Question

代码编译时没有错误，但调试时会出现访问冲突消息。任何人都可以指出下面的代码有什么问题吗？

我的代码实际上运行了1000次相同等式的1000次迭代它是一个递归非线性方程。目的只是为了欣赏这种能力并行运行多个（迭代）方程式。

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <conio.h>
#include <cuda.h>
#include <cutil.h>
#include <time.h> 

#define TOTAL_THREADS 1024
#define THREADS_PER_BLOCK 256
#define TOTAL_BLOCKS 4
#define VALUES_PER_THREAD 1000
#define THETA_VALUES_PER_THREAD 15

__global__ void my_compute(float *y_d, float *theta_d, float *u_d) 
{
   int offset = ((blockIdx.x * blockDim.x) + threadIdx.x) * VALUES_PER_THREAD;
   int theta_offset = ((blockIdx.x * blockDim.x) + threadIdx.x) * THETA_VALUES_PER_THREAD;

   for (int i = 7; i < 1000; i++) {
      y_d[offset + i] = theta_d[theta_offset + 0] * y_d[offset + i - 1] +
                        theta_d[theta_offset + 1] * y_d[offset + i - 3] +
                        theta_d[theta_offset + 2] * u_d[offset + i - 5] * u_d[offset + i - 4] +
                        theta_d[theta_offset + 3] + 
                        theta_d[theta_offset + 4] * u_d[offset + i - 6] +
                        theta_d[theta_offset + 5] * u_d[offset + i - 4] * y_d[offset + i - 6] +
                        theta_d[theta_offset + 6] * u_d[offset + i - 7] +
                        theta_d[theta_offset + 7] * u_d[offset + i - 7] * u_d[offset + i - 6] +
                        theta_d[theta_offset + 8] * y_d[offset + i - 4] +
                        theta_d[theta_offset + 9] * y_d[offset + i - 5] +
                        theta_d[theta_offset + 10] * u_d[offset + i - 4] * y_d[offset + i - 5] +
                        theta_d[theta_offset + 11] * u_d[offset + i - 4] * y_d[offset + i - 2] +
                        theta_d[theta_offset + 12] * u_d[offset + i - 7] * u_d[offset + i - 3] +
                        theta_d[theta_offset + 13] * u_d[offset + i - 5] +
                        theta_d[theta_offset + 14] * u_d[offset + i - 4];
   }
}

int main(void) 
{
   float y[1000000];
   FILE * fpoo;
   FILE * u;
   float theta[15000];
   float u_data[1000000];
   float *y_d;
   float *theta_d;
   float *u_d;
   cudaEvent_t start, stop;
   float time;

   cudaEventCreate(&start);
   cudaEventCreate(&stop);

   //memory allocation
   cudaMalloc((void **) &y_d, 1000000 * sizeof(float));
   cudaMalloc((void **) &theta_d, 15000 * sizeof(float));
   cudaMalloc((void **) &u_d, 1000000 * sizeof(float));
   cudaEventRecord(start, 0);

   // importing data for theta and input of model//
   fpoo = fopen("c:\\Fly_theta.txt", "r");
   u = fopen("c:\\Fly_u.txt", "r");

   for (int i = 0; i < 1000; i++) {
      for (int j = 0; j < 15; j++)
         fscanf(fpoo, "%f\n", &theta[15 * i + j]);
   } 
   for (int i = 0; i < 1000; i++) {
      for (int j = 0; j < 1000; j++)
         fscanf(u, "%f\n", &u_data[1000 * i + j]);
   }
   //initialising past input with the value of zero//
   for (int i = 0; i < 1000; i++) {
      for (int j = 0; j < 8; j++)
         y[8 * i + j] = 0;
   }
   cudaMemcpy(y_d, y, 1000000 * sizeof(float), cudaMemcpyHostToDevice);

   cudaMemcpy(theta_d, theta, 15000 * sizeof(float), cudaMemcpyHostToDevice);
   cudaMemcpy(u_d, u_data, 1000000 * sizeof(float), cudaMemcpyHostToDevice);

   //calling kernel function//
   my_compute <<< 4, 256 >>> (y_d, theta_d, u_d);

   cudaMemcpy(y, y_d, 1000000 * sizeof(float), cudaMemcpyDeviceToHost);

   for (int i = 0; i < 1000; i++) {
      for (int j = 0; j < 1000; j++)
         printf("%f", y[1000 * i + j]);
   } 
   cudaEventRecord(stop, 0);
   cudaEventSynchronize(stop);
   cudaEventElapsedTime(&time, start, stop);
   cudaEventDestroy(start);
   cudaEventDestroy(stop);
   printf("Time to generate:  %3.1f ms \n", time);

   cudaFree(y_d);
   cudaFree(theta_d);
   cudaFree(u_d);

   fclose(u);
   fclose(fpoo);

   //fclose();
   _getche();
   return 0;
}

Answer 1

您可能有堆栈溢出。

尝试在数组声明中添加static说明符：

static float y[1000000];
static float theta[15000];
static float u_data[1000000];

Answer 2

我尝试使用集成了cuda-memcheck的Toolkit 5.0中的cuda-gdb进行编码。

它向我展示的是在块（3,0,0）的线程（232,0,0）上，theta_offset是15000 - 并且从我对你的代码的预期看起来它应该永远不会超过14999（ theta_d size是15000，这是可以使用的最大索引）

注意4个块* 256个线程* 15个元素/线程= 15360。

Answer 3

Eugene很受欢迎。问题是theta_d：
的原始分配
theta_d的字节大小应为：

THREADS_PER_BLOCK * TOTAL_BLOCKS * THETA_VALUES_PER_THREAD * ELEMENT_SIZE = 256 * 4 * 15 * 4 = 61440

确保您选择的访问模式可以访问该数组的所有元素。在应用程序中， theta_d声明为：

cudaMalloc((void **) &theta_d, 15000 * sizeof(float));

实际上，theta_d的大小只有60000字节。因此，内核将在超过字节60000的所有访问中遇到错误。

一种解决方案是根据已定义的常量调整分配大小。如果用“THREADS_PER_BLOCK * TOTAL_BLOCKS * THETA_VALUES_PER_THREAD”替换15000，它将正确调整分配大小。尝试以下主要功能：

int main(void) { float y[1000000]; FILE * fpoo; FILE * u; float theta[THREADS_PER_BLOCK * TOTAL_BLOCKS * THETA_VALUES_PER_THREAD]; float u_data[1000000]; float *y_d; float *theta_d; float *u_d; cudaEvent_t start, stop; float time; cudaEventCreate(&start); cudaEventCreate(&stop); //memory allocation cudaMalloc((void **) &y_d, 1000000 * sizeof(float)); cudaMalloc((void **) &theta_d, THREADS_PER_BLOCK * TOTAL_BLOCKS * THETA_VALUES_PER_THREAD* sizeof(float)); cudaMalloc((void **) &u_d, 1000000 * sizeof(float)); cudaEventRecord(start, 0); // importing data for theta and input of model// fpoo = fopen("c:\\Fly_theta.txt", "r"); u = fopen("c:\\Fly_u.txt", "r"); for (int i = 0; i < 1000; i++) { for (int j = 0; j < 15; j++) fscanf(fpoo, "%f\n", &theta[15 * i + j]); } for (int i = 0; i < 1000; i++) { for (int j = 0; j < 1000; j++) fscanf(u, "%f\n", &u_data[1000 * i + j]); } //initialising past input with the value of zero// for (int i = 0; i < 1000; i++) { for (int j = 0; j < 8; j++) y[8 * i + j] = 0; } cudaMemcpy(y_d, y, 1000000 * sizeof(float), cudaMemcpyHostToDevice); cudaMemcpy(theta_d, theta, THREADS_PER_BLOCK * TOTAL_BLOCKS * THETA_VALUES_PER_THREAD* sizeof(float), cudaMemcpyHostToDevice); cudaMemcpy(u_d, u_data, 1000000 * sizeof(float), cudaMemcpyHostToDevice); //calling kernel function// my_compute <<< 4, 256 >>> (y_d, theta_d, u_d); cudaMemcpy(y, y_d, 1000000 * sizeof(float), cudaMemcpyDeviceToHost); for (int i = 0; i < 1000; i++) { for (int j = 0; j < 1000; j++) printf("%f", y[1000 * i + j]); } cudaEventRecord(stop, 0); cudaEventSynchronize(stop); cudaEventElapsedTime(&time, start, stop); cudaEventDestroy(start); cudaEventDestroy(stop); printf("Time to generate: %3.1f ms \n", time); cudaFree(y_d); cudaFree(theta_d); cudaFree(u_d); fclose(u); fclose(fpoo); //fclose(); _getche(); return 0; }

Answer 4

守则确实是正确的。用于从文件读取数据的'fopen for loops'不是按预期读取数据。所以我把它改成了下面的循环。我的代码现在运行得很好。感谢所有贡献的人。

//导入theta的数据和模型的输入

fpoo= fopen("c:\\Fly_theta.txt","r");

u= fopen("c:\\Fly_u.txt","r");


        for(int j=0;j<15;j++)
            {
                fscanf(fpoo,"%f\n",&theta[j]);
            }
         for(int i=1;i<1000;i++)
         {
             for(int j=0;j<15;j++)
             {
                theta[15*i+j]=theta[j];
             }
         }
        for(int j=0;j<1000;j++)
            {
                fscanf(u,"%f\n",&u_data[j]);
            }
        for(int i=1;i<1000;i++)
         {
             for(int j=0;j<1000;j++)
             {
                u_data[1000*i+j]=u_data[j];
             }
        }

C CUDA代码中的访问冲突

4 个答案: