CUDA function pointers

Question

I was trying to make somtehing like this (actually I need to write some integration functions) in CUDA

#include <iostream>
using namespace std;

float f1(float x) {
    return x * x;
}

float f2(float x) {
    return x;
}

void tabulate(float p_f(float)) {
    for (int i = 0; i != 10; ++i) {
        std::cout << p_f(i) << ' ';
    }
    std::cout << std::endl;
}

int main() {
    tabulate(f1);
    tabulate(f2);
    return 0;
}

output:

0 1 4 9 16 25 36 49 64 81
0 1 2 3 4 5 6 7 8 9

---

I tried the following but only got the error

Error: Function pointers and function template parameters are not supported in sm_1x.

float f1(float x) {
    return x;
}

__global__ void tabulate(float lower, float upper, float p_function(float), float* result) {
    for (lower; lower < upper; lower++) {
        *result = *result + p_function(lower);
    }
}

int main() {
    float res;
    float* dev_res;

    cudaMalloc( (void**)&dev_res, sizeof(float) ) ;

    tabulate<<<1,1>>>(0.0, 5.0, f1, dev_res);
    cudaMemcpy(&res, dev_res, sizeof(float), cudaMemcpyDeviceToHost);

    printf("%f\n", res);
    /************************************************************************/
    scanf("%s");

    return 0;
}

Answer 1

To get rid of your compile error, you'll have to use -gencode arch=compute_20,code=sm_20 as a compiler argument when compiling your code. But then you'll likely have some runtime problems:

Taken from the CUDA Programming Guide http://docs.nvidia.com/cuda/cuda-c-programming-guide/index.html#functions

Function pointers to __global__ functions are supported in host code, but not in device code. Function pointers to __device__ functions are only supported in device code compiled for devices of compute capability 2.x and higher.

It is not allowed to take the address of a __device__ function in host code.

so you can have something like this (adapted from the "FunctionPointers" sample):

//your function pointer type - returns unsigned char, takes parameters of type unsigned char and float
typedef unsigned char(*pointFunction_t)(unsigned char, float);

//some device function to be pointed to
__device__ unsigned char
Threshold(unsigned char in, float thresh)
{
   ...
}

//pComputeThreshold is a device-side function pointer to your __device__ function
__device__ pointFunction_t pComputeThreshold = Threshold;
//the host-side function pointer to your __device__ function
pointFunction_t h_pointFunction;

//in host code: copy the function pointers to their host equivalent
cudaMemcpyFromSymbol(&h_pointFunction, pComputeThreshold, sizeof(pointFunction_t))

You can then pass the h_pointFunction as a parameter to your kernel, which can use it to call your __device__ function.

//your kernel taking your __device__ function pointer as a parameter
__global__ void kernel(pointFunction_t pPointOperation)
{
    unsigned char tmp;
    ...
    tmp = (*pPointOperation)(tmp, 150.0)
    ...
}

//invoke the kernel in host code, passing in your host-side __device__ function pointer
kernel<<<...>>>(h_pointFunction);

Hopefully that made some sense. In all, it looks like you would have to change your f1 function to be a __device__ function and follow a similar procedure (the typedefs aren't necessary, but they do make the code nicer) to get it as a valid function pointer on the host-side to pass to your kernel. I'd also advise giving the FunctionPointers CUDA sample a look over

Answer 2

Even though you may be able to compile this code (see @Robert Crovella's answer) this code will not work. You cannot pass function pointers from host code as the host compiler has no way of figuring out the function address.

Answer 3

Here is a simple class for function pointers that are callable from within a kernel I wrote based on this question:

template <typename T>
struct cudaCallableFunctionPointer
{
public:
  cudaCallableFunctionPointer(T* f_)
  {
    T* host_ptr = (T*)malloc(sizeof(T));
    cudaMalloc((void**)&ptr, sizeof(T));

    cudaMemcpyFromSymbol(host_ptr, *f_, sizeof(T));
    cudaMemcpy(ptr, host_ptr, sizeof(T), cudaMemcpyHostToDevice);
    
    cudaFree(host_ptr)
  }

  ~cudaCallableFunctionPointer()
  {
    cudaFree(ptr);
  }

  T* ptr;
};

you could use it like this:

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

typedef double (*func)(double x);
__device__ func f_ = func1;



__global__ void test_kernel(func* f)
{
    double x = (*f)(2.0);
    printf("%g\n", x);
}



int main()
{
    cudaCallableFunctionPointer<func> f(&f_);

    test_kernel << < 1, 1 >> > (f.ptr);
}

output:

3