Host to kernel sending information issue - parallel-processing

Question

The kernel below in C language changes a cell of an array:
__global__ void test(int *mt[matrix_size])
{
mt[0][0]=12;
}
The code below copies kernel results to host but it doesn't send the array to host correctly:
int *matrix[matrix_size],*d_matrix[matrix_size];
for(int i=0;i<matrix_size;i++)
matrix[i] = (int *)malloc(n*n*sizeof(int));
for(int i=0;i<matrix_size;i++)
cudaMalloc((void**)&d_matrix[i],sizeof(int));
test<<<1,1>>>(d_matrix);
cudaMemcpy(*matrix,*d_matrix,n*n*sizeof(int),cudaMemcpyDeviceToHost);
printf("\n\n %d \n\n",matrix[0][0]); //the result is zero instead of 12
How can I fix the problem?

Answer 1

You have gotten a lot wrong here.
The root cause is that d_matrix is in host memory and can't be passed directly to a kernel. If you check runtime errors you will see that firstly the first cudaMemcpy call fails because of the wrong direction argument, then when you fix that, the kernel fails with a invalid address error.
To fix this, you need to allocate a copy of d_matrix on the GPU and copy d_matrix to that copy. This is because the array you are passing to the kernel decays to a pointer and is not passed by value.
Something like this:
#include <cstdio>
const int n = 9;
const int matrix_size = 16;
__global__
void test(int *mt[matrix_size])
{
mt[threadIdx.x][0] = 12 + threadIdx.x;
}
int main()
{
int *matrix[matrix_size],*d_matrix[matrix_size];
for(int i=0;i<matrix_size;i++) {
matrix[i] = (int *)malloc(n * n * sizeof(int));
cudaMalloc((void**)&d_matrix[i], n * n * sizeof(int));
}
int **dd_matrix;
cudaMalloc(&dd_matrix, matrix_size * sizeof(int*));
cudaMemcpy(dd_matrix, d_matrix, matrix_size * sizeof(int *), cudaMemcpyHostToDevice);
test<<<1,matrix_size>>>(dd_matrix);
for(int i=0;i<matrix_size;i++) {
cudaMemcpy(matrix[i], d_matrix[i], n*n*sizeof(int), cudaMemcpyDeviceToHost);
printf("%d = %d \n", i, matrix[i][0]);
}
return 0;
}
Which when run gives this:
$ nvcc -g -G -arch=sm_52 -o bozocu bozocu.cu
$ ./bozocu
0 = 12
1 = 13
2 = 14
3 = 15
4 = 16
5 = 17
6 = 18
7 = 19
8 = 20
9 = 21
10 = 22
11 = 23
12 = 24
13 = 25
14 = 26
15 = 27
is, I believe, more in line with what you were expecting.