Pointers in structs changed in kernel function - cuda

Question

I am trying access an data from array allocated in CUDA. First step was allocate a struct defined by me. After I pass the allocated struct to a kernel function that change the values from the struct. Finally, I pass the struct and the array to a host variables so read them. But actually I am having a problem to read the vector allocated.
#include <stdio.h>
#include <stdlib.h>
typedef struct x{
float *y;
float v;
}x_t;
__global__ void initTeste(x_t *param){
param->v = 10;
param->y[0] = 10;
param->y[1] = 10;
}
int main(void) {
x_t *hvar;
x_t hvarBackup;
float *temp = (float*)malloc(10*sizeof(float));
cudaError_t result;
cudaMalloc(&hvar , sizeof(x_t) );
cudaMalloc(&hvarBackup.y, 10*sizeof(float) );
cudaMemcpy(hvar, &hvarBackup, sizeof(x_t), cudaMemcpyHostToDevice);
initTeste<<<1,1>>>(hvar);
cudaMemcpy(&hvarBackup, hvar, sizeof(x_t), cudaMemcpyDeviceToHost);
cudaMemcpy(temp, &hvar->y, 10*sizeof(float), cudaMemcpyDeviceToHost);
printf("%f",(hvarBackup.v)); //here ok
printf("%f",(temp[0])); //here's the problem
return 0;
}

Answer 1

You cannot do it like that, because you haven't allocated y for the device, hence it will only give you segmentation fault when copying from y content to host. Aside of that, you have to allocate y for the device with the amount of 10*sizeof(float), and this is a truthfully pain in the a** job, especially when your struct becomes a huge container of arrays (and you should always know, that arrays inside structs always have to be avoided in CUDA).
Here's what you can do with the current code
int main(void) {
x_t *h_hvar = (x_t*)malloc(sizeof(x_t));
x_t *d_hvar;
float *h_y = (float*)malloc(10*sizeof(float));
float *d_y;
cudaMalloc(&d_hvar, sizeof(x_t) );
cudaMalloc(&d_y, 10*sizeof(float) );
// Insert the float pointer you allocated in CUDA
// to the host pointer first, and then copy the whole thing
// to the device area
h_hvar->y = d_y;
cudaMemcpy(d_hvar, h_hvar, sizeof(x_t), cudaMemcpyHostToDevice);
initTeste<<<1,1>>>(d_hvar);
cudaMemcpy(h_hvar, d_hvar, sizeof(x_t), cudaMemcpyDeviceToHost);
cudaMemcpy(h_y, d_y, 10*sizeof(float), cudaMemcpyDeviceToHost);
printf("%f",h_hvar->v);
printf("%f",h_y[0]);
return 0;
}
And that should give you the right value..

Answer 2

cudaMemcpy(temp, &hvar->y, 10*sizeof(float), cudaMemcpyDeviceToHost);
should be
cudaMemcpy(temp, hvar->y, 10*sizeof(float), cudaMemcpyDeviceToHost);
because hvar->y is already a pointer and you don't want to get the pointer to that pointer.