cuda copy data which dynamic malloc in kernel from device memory - cuda

Question

I met a problem about using cudaMemcpy with cudaMemcpyDeviceToHost.
There is a struct which have a pointer int* a, It will malloc in the kernel function.
And then I need copy this int* a to host memory.
My question is: I didn't know how it can not work by using cudaMemcpy.
There my codes:
#include <cuda_runtime.h>
#include <stdio.h>
typedef struct { int n, m; int *a; } myst;
__global__ void xthread(myst *st)
{
unsigned int idx = blockIdx.x*blockDim.x + threadIdx.x;
myst *mst = &st[idx];
mst->n = idx;
mst->m = idx+1;
mst->a = (int *)malloc((mst->m)*sizeof(int));
mst->a[0] = idx;
}
int main(int argc,char **argv)
{
dim3 dimGrid(1);
dim3 dimBlock(2);
myst *mst = NULL;
myst *hst = (myst *)malloc(2 * sizeof(myst));
cudaMalloc(&mst, 2 * sizeof(myst));
xthread<<<dimGrid, dimBlock>>>(mst);
cudaDeviceSynchronize();
cudaMemcpy(&hst[0],&mst[0],sizeof(myst),cudaMemcpyDeviceToHost);
cudaMemcpy(&hst[1],&mst[1],sizeof(myst),cudaMemcpyDeviceToHost);
int *pInt1 = (int *)malloc((hst[0].m)*sizeof(int)) ;
int *pInt2 = (int *)malloc((hst[1].m)*sizeof(int)) ;
cudaMemcpy(pInt1, hst[0].a, (hst[0].m)*sizeof(int), cudaMemcpyDeviceToHost);
cudaMemcpy(pInt2, hst[1].a, (hst[1].m)*sizeof(int), cudaMemcpyDeviceToHost);
printf("%d\t%d\t%d\n",hst[0].n,hst[0].m, pInt1[0]);
printf("%d\t%d\t%d\n",hst[1].n,hst[1].m, pInt2[0]);
free(pInt1);
free(pInt2);
return 0;
}
The codes will go warning about "Cuda API error detected: cudaMemcpy returned (0xb)"
I saw a similar question : copy data which is allocated in device from device to host
But it seem that can not solve my problem.
Thx.

Answer 1

Alright, I work it out with a stupid way (-.-!!).
While return form the kernel function, I count how many space I have to malloc in Host and Device, and cudaMalloc again a big space . Next, in other kernel function named ythread, copy the data which in the Heap to the big space.
typedef struct { int n, m; int *a; } myst;
__global__ void xthread(myst *st) {
unsigned int idx = blockIdx.x * blockDim.x + threadIdx.x;
myst *mst = &st[idx];
mst->n = idx;
mst->m = idx + 1;
mst->a = (int *) malloc((mst->m) * sizeof(int));
for (int i = 0; i < mst->m; i++) {
mst->a[i] = idx + 900 + i * 10;
}
}
__global__ void ythread(myst *st, int *total_a) {
unsigned int idx = blockIdx.x*blockDim.x + threadIdx.x;
myst *mst = &st[idx];
int offset=0;
for(int i=0; i<idx; i++) {
offset += st[i].m;
}
for(int i=0; i<mst->m; i++) {
total_a[offset+i] = mst->a[i];
}
}
int main(int argc,char **argv) {
dim3 dimGrid(1);
dim3 dimBlock(2);
myst *mst = NULL;
cudaMalloc((void**)&mst, dimBlock.x * sizeof(myst));
xthread<<<dimGrid, dimBlock>>>(mst);
cudaDeviceSynchronize();
myst *hst = (myst *)malloc(dimBlock.x * sizeof(myst));
cudaMemcpy(hst, mst, dimBlock.x*sizeof(myst),cudaMemcpyDeviceToHost);
int t_size = 0;
for(int i=0; i<dimBlock.x; i++) {
t_size += hst[i].m;
}
printf("t_size:%d\n", t_size);
int * t_a_h = (int *)malloc(t_size*sizeof(int));
int * t_a_d = NULL;
cudaMalloc((void**)&t_a_d, t_size*sizeof(int));
ythread<<<dimGrid, dimBlock>>>(mst, t_a_d);
cudaDeviceSynchronize();
cudaMemcpy(t_a_h, t_a_d, t_size*sizeof(int),cudaMemcpyDeviceToHost);
for(int i=0; i<t_size; i++) {
printf("t_a_h[%d]:%d\n", i, t_a_h[i]);
}
free(t_a_h);
cudaFree(mst);
cudaFree(t_a_d);
return 0;
}
Emmmmmm, it work, but I think there is a better way to solve this problem.