device function pointers

Question

I need a device version of the following host code:

double (**func)(double x);

double func1(double x)
{
 return x+1.;
}

double func2(double x)
{
 return x+         


        

Answer 1

function pointers are allowed on Fermi. This is how you could do it:

typedef double (*func)(double x);

__device__ double func1(double x)
{
return x+1.0f;
}

__device__ double func2(double x)
{
return x+2.0f;
}

__device__ double func3(double x)
{
return x+3.0f;
}

__device__ func pfunc1 = func1;
__device__ func pfunc2 = func2;
__device__ func pfunc3 = func3;

__global__ void test_kernel(func* f, int n)
{
  double x;

  for(int i=0;i<n;++i){
   x=f[i](2.0);
   printf("%g\n",x);
  }
}

int main(void)
{
  int N = 5;
  func* h_f;
  func* d_f;
  h_f = (func*)malloc(N*sizeof(func));
  cudaMalloc((void**)&d_f,N*sizeof(func));

  cudaMemcpyFromSymbol( &h_f[0], pfunc1, sizeof(func));
  cudaMemcpyFromSymbol( &h_f[1], pfunc1, sizeof(func));
  cudaMemcpyFromSymbol( &h_f[2], pfunc2, sizeof(func));
  cudaMemcpyFromSymbol( &h_f[3], pfunc3, sizeof(func));
  cudaMemcpyFromSymbol( &h_f[4], pfunc3, sizeof(func));

  cudaMemcpy(d_f,h_f,N*sizeof(func),cudaMemcpyHostToDevice);

  test_kernel<<<1,1>>>(d_f,N);

  cudaFree(d_f);
  free(h_f);

  return 0;
}