Separating even and odd numbers in CUDA - cuda

Question

I have an array of numbers as {1,2,3,4,5,6,7,8,9,10} and I want to separate even and odd numbers as:
even = {2,4,6,8}
and:
odd = {1,3,5,7}
I am aware of atomic operations in CUDA, and also aware that the output is not expected to suffer from race conditions. I don't want to use atomic operations. How can I achieve this without using atomic keywords?
CODE:
#include <stdio.h>
#include <cuda.h>
// Kernel that executes on the CUDA device
__global__ void square_array(float *total,float *even,float *odd, int N)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
int a=total[idx];
if ((a%2)==0)
{
for (int i=0;i<=idx;i++)
{
int b = even[i];
if(b==0)
{
even[i] = total[idx];
break;
}
}
}
else
{
for (int i=0;i<idx;i++)
{
int c = odd[i];
odd[i] = total[idx];
break;
}
}
}
// main routine that executes on the host
int main(void)
{
float *total_h,*even_h, *odd_h,*total_d, *even_d,*odd_d; // Pointer to host & device arrays
const int N = 10; // Number of elements in arrays
size_t size = N * sizeof(float);
total_h = (float *)malloc(size); // Allocate array on host
even_h = (float *)malloc(size); // Allocate array on host
odd_h = (float *)malloc(size); // Allocate array on host
cudaMalloc((void **) &total_d, size);
cudaMalloc((void **) &even_d, size);
cudaMemset(even_d,0,size);
cudaMalloc((void **) &odd_d, size); // Allocate array on device
cudaMemset(odd_d,0,size);
// Initialize host array and copy it to CUDA device
for (int i=0; i<N; i++) total_h[i] = (float)i+1;
cudaMemcpy(total_d, total_h, size, cudaMemcpyHostToDevice);
// Do calculation on device:
square_array <<< 1,10 >>> (total_d,even_d,odd_d, N);
// Retrieve result from device and store it in host array
cudaMemcpy(even_h, even_d, sizeof(float)*N, cudaMemcpyDeviceToHost);
cudaMemcpy(odd_h, odd_d, sizeof(float)*N, cudaMemcpyDeviceToHost);
// Print results
printf("total Numbers\n");
for (int i=0; i<N; i++) printf("%f\n",total_h[i]);
printf("EVEN Numbers\n");
for (int i=0; i<N; i++) printf("%f\n",even_h[i]);
printf("ODD Numbers\n");
for (int i=0; i<N; i++) printf("%f\n",odd_h[i]);
// Cleanup
free(total_h);
free(even_h);
free(odd_h);
cudaFree(total_d);
cudaFree(even_d);
cudaFree(odd_d);
}
OUTPUT:

Answer 1

As suggested by Jared Hoberock, it would be much more easy to use the efficient partitioning algorithm available in CUDA Thrust instead of starting the development of a partitioning routine of your own. Below, please find a complete worked example.
#include <thrust\device_vector.h>
#include <thrust\partition.h>
#include <thrust\execution_policy.h>
struct is_even { __host__ __device__ bool operator()(const int &x) { return (x % 2) == 0; } };
void main() {
const int N = 10;
thrust::host_vector<int> h_data(N);
for (int i=0; i<N; i++) h_data[i] = i;
thrust::device_vector<int> d_data(h_data);
thrust::device_vector<int> d_evens(N/2);
thrust::device_vector<int> d_odds(N/2);
thrust::partition_copy(d_data.begin(), d_data.end(), d_evens.begin(), d_odds.begin(), is_even());
printf("Even numbers\n");
for (int i=0; i<N/2; i++) {
int val = d_evens[i];
printf("evens[%i] = %i\n",i,val);
}
printf("Odd numbers\n");
for (int i=0; i<N/2; i++) {
int val = d_odds[i];
printf("odds[%i] = %i\n",i,val);
}
}