The following trial presents my intention, which failed to compile:
__host__ __device__ void f(){}
int main()
{
f<<<1,1>>>();
}
You need to create a CUDA kernel entry point, e.g. __global__ function. Something like:
#include
__host__ __device__ void f() {
#ifdef __CUDA_ARCH__
printf ("Device Thread %d\n", threadIdx.x);
#else
printf ("Host code!\n");
#endif
}
__global__ void kernel() {
f();
}
int main() {
kernel<<<1,1>>>();
if (cudaDeviceSynchronize() != cudaSuccess) {
fprintf (stderr, "Cuda call failed\n");
}
f();
return 0;
}