Cuda program does not give the correct output when using a CUDA compatible GPU

问题

I found the following program from http://llpanorama.wordpress.com/2008/05/21/my-first-cuda-program/

Unfortunately I can't copy paste it here because the code becomes messy

It takes as input a vector of numbers and then gives as an output the vector multiplied by itself, I run it on the emulator that I have installed on my computer and it gives the following output:

0 0.000000
1 1.000000
2 4.000000
3 9.000000
4 16.000000
5 25.000000
6 36.000000
7 49.000000
8 64.000000
9 81.000000

however if I decide to run it on a remote computer which runs debian and has cuda compatible gpu by entering

nvcc test.cu -lcudart -o test
./test

it gives me the following output

why does this happen? Thank you in advance!

回答1:

The problem is that code has no error checking, and there is something wrong with the remote computer. Add error checking to that code (it's not hard to do), re-run it, and then see what happens. If you still have trouble, report back.

Here is the code suitably modified with error checking:

// example1.cpp : Defines the entry point for the console application.
//

#include <stdio.h>
#include <cuda.h>

#define cudaCheckErrors(msg) \
    do { \
        cudaError_t __err = cudaGetLastError(); \
        if (__err != cudaSuccess) { \
            fprintf(stderr, "Fatal error: %s (%s at %s:%d)\n", \
                msg, cudaGetErrorString(__err), \
                __FILE__, __LINE__); \
            fprintf(stderr, "*** FAILED - ABORTING\n"); \
            exit(1); \
        } \
    } while (0)

// Kernel that executes on the CUDA device
__global__ void square_array(float *a, int N)
{
  int idx = blockIdx.x * blockDim.x + threadIdx.x;
  if (idx<N) a[idx] = a[idx] * a[idx];
}

// main routine that executes on the host
int main(void)
{
  float *a_h, *a_d;  // Pointer to host & device arrays
  const int N = 10;  // Number of elements in arrays
  size_t size = N * sizeof(float);
  a_h = (float *)malloc(size);        // Allocate array on host
  cudaMalloc((void **) &a_d, size);   // Allocate array on device
  cudaCheckErrors("cudaMalloc fail");
  // Initialize host array and copy it to CUDA device
  for (int i=0; i<N; i++) a_h[i] = (float)i;
  cudaMemcpy(a_d, a_h, size, cudaMemcpyHostToDevice);
  cudaCheckErrors("cudaMemcpy 1 fail");
  // Do calculation on device:
  int block_size = 4;
  int n_blocks = N/block_size + (N%block_size == 0 ? 0:1);
  square_array <<< n_blocks, block_size >>> (a_d, N);
  cudaDeviceSynchronize();
  cudaCheckErrors("kernel fail");
  // Retrieve result from device and store it in host array
  cudaMemcpy(a_h, a_d, sizeof(float)*N, cudaMemcpyDeviceToHost);
  cudaCheckErrors("cudaMemcpy 2 fail");
  // Print results
  for (int i=0; i<N; i++) printf("%d %f\n", i, a_h[i]);
  // Cleanup
  free(a_h); cudaFree(a_d);

}

来源：https://stackoverflow.com/questions/15177019/cuda-program-does-not-give-the-correct-output-when-using-a-cuda-compatible-gpu

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cuda

gpu