Add vector with CUDA C++ : don't get the expected result

Question

I recently started CUDA programming. I followed tutorials and guides about it, and made a first program copied from the original example ( here pg. 25 ), which basically use the GPU to add two vectors. I can compile it but the result I get isn't the one I expected.

Here is my code :

#include <iostream>
#include <random>

using namespace std;

//Size of vector
#define N 16


//Kernel for adding vector
__global__ void add(int* a, int* b, int* c){
  c[blockIdx.x] = a[blockIdx.x] + b[blockIdx.x];
}


//Functions
void printArray(int* a){
  cout << "[ ";
  for(int i = 0; i < N; i++){
    cout << *(a+i) << " ";
  } cout << "]" << endl;
}

void fillRandomValue(int* a, int size, int range){
  for(int i = 0; i < size; i++){
    *(a+i) = rand() % range;
  }
}


int main(void){
  //Declaration of vector for host and device : a for host ; d_a for device a ;
  int* a,* b,* c;
  int* d_a,* d_b,*d_c;
  int size = N * sizeof(int);

  //Allocate memory into the device
  cudaMalloc((void** )&d_a, size);
  cudaMalloc((void** )&d_b, size);
  cudaMalloc((void** )&d_c, size);

  //Declaring 3 array
  a = new int[size];
  b = new int[size];
  c = new int[size];

  //Randomn array
  //Initialize randomn seed
  srand (time(NULL));
  fillRandomValue(a, N, 10);
  fillRandomValue(b, N, 10);

  //Copy host to device
  cudaMemcpy(d_a, &a, size, cudaMemcpyHostToDevice);
  cudaMemcpy(d_b, &b, size, cudaMemcpyHostToDevice);

  //Adding vector
  add<<<N,1>>>(d_a, d_b, d_c);

  //Get back the result
  cudaMemcpy(c, d_c, size, cudaMemcpyDeviceToHost);

  //Output to console
  cout << "a = "; printArray(a); cout << endl;
  cout << "b = "; printArray(b); cout << endl;
  cout << "a + b = "; printArray(c); cout << endl;

  //Free memory
  free(a); free(b); free(c);
  cudaFree(d_a); cudaFree(d_b); cudaFree(d_c);

  return 0;

}

The problem is I get the following result:

a = [ 5 5 7 2 9 5 3 5 4 5 2 9 6 7 4 9 ]

b = [ 9 6 9 8 9 7 5 6 6 6 2 8 7 3 1 2 ]

a + b = [ -971240544 44196 -2117203120 54617 1031799296 65038 1031800320 65038 -2117199280 54617 -1009853744 44196 -1048817168 44196 -1307968288 54865 ]

I tried to see if those weren't address, but no. I also tried to copy the straight code from the link above but I got no luck.

I can reproduce this behavior as much as I want .

This is incorrect: `cudaMemcpy(d_a, &a, size, cudaMemcpyHostToDevice);` get rid of the ampersand (&). And likewise for the next line. You did it correctly on the copy operation after the kernel (where there is no ampersand in front of `c`). You should also use [proper CUDA error checking](https://stackoverflow.com/questions/14038589/what-is-the-canonical-way-to-check-for-errors-using-the-cuda-runtime-api). And just to be clear, although you may have "copied" something from that presentation you linked, you changed it also (compared to what is on slide 25). Those changes matter. — Robert Crovella, Jul 23 '20 at 14:31
FYI, there is no C/C++ languages. C and C++ are distinct languages. For example, C++ has `std::vector` and C doesn't. C++ has templates, inheritance and function overriding, the C language doesn't. Please update you post title and language tags as appropriate. — Thomas Matthews, Jul 23 '20 at 14:35
Thank you @RobertCrovella ! I didn't see it ( more didn't think of it if it english ). I'm going to mark the question solved thank you. — Hugo Chittaro, Jul 23 '20 at 14:35

score 2 · Accepted Answer · answered Jul 23 '20 at 14:38

2

As said in the comment, the problem was on the cudeMemcpy(d_a, &a, size, cudaMemcpyHostToDevice) The ampersand was not supposed to be here.

answered Jul 23 '20 at 14:38

Hugo Chittaro

111
10

Add vector with CUDA C++ : don't get the expected result

1 Answers1