Program generates wrong result

[alexnovice]

Hallo everyone!

I have found a simple program on the internet and tried to run it. Compiling tells me that theres no problem, but when executing the program there is the wrong result. The program should square all values in my vector, but the values did not change at all. I am compiling with “nvcc -o r vecadd.cu” and running it with ./r. I think that he is not even starting a thread or the kernel is not running at all. But shouldn’t he at least tell me that an error occured or that no threads have been started?

include <stdio.h>

include <cuda.h>

global void square_array(int *a, int N) {

 int x = blockIdx.x * blockDim.x + threadIdx.x;

if(x < N)

   a[x] = a[x] * a[x];

}

int main()

{

 const int N = 10;

 int *ar_h, *ar_d;

//Speicher für Host-Array allokieren

 size_t size = N * sizeof(int);

 ar_h = (int *) malloc( size );

//Speicher für Device-Array allokieren

 cudaMalloc((void **) &ar_d, size);

int i;

//Füllen des Arrays

 for(i = 0; i < N; i++) {

   ar_h[i] = i + 1;

   printf("%d ", ar_h[i]);

   cudaMemcpy(ar_h, ar_d, size, cudaMemcpyHostToDevice);

 }

printf(“\n”);

//Initialisierung des Kernels

 int blockSize = N;

 square_array <<<1, blockSize>>>(ar_d, N);

//Wiederholen der Daten

 cudaMemcpy(ar_d, ar_h, sizeof(int) * N, cudaMemcpyDeviceToHost);

//Ausgabe

 for(i = 0; i < N; i++) {

   printf("%d ", ar_h[i]);

 }

printf(“\n”);

//Speicher freigeben

 free( ar_h );

 cudaFree( ar_d );

return 0;

}

Thank you for your help.

Best greets,

Alex

[Nico]

You probably want to take the first cudaMemcpy call out of the for loop. Also, the first parameter to the cudaMemcpy calls is the destination while the second one is the source, not vice versa.

N.