CUDA surf3Dread overloaded types, only one works? - cuda

Question

While testing if I know how to allocate surface objects, I was designing a dummy kernel to read a single value.
This kernel was failing at compile time because
"no instance of overloaded function "surf3Dread" matches the argument list"
__global__ void test_surface(cudaSurfaceObject_t surfImg,int x, int y, int z){
float test = surf3Dread(surfImg , (int)(x*sizeof(float)) , y , z ,cudaBoundaryModeZero);
printf("%f \n",test);
}
it works when I do this instead:
__global__ void test_surface(cudaSurfaceObject_t surfImg,int x, int y, int z){
float test;
surf3Dread(&test,surfImg , (int)(x*sizeof(float)) , y , z ,cudaBoundaryModeZero);
printf("%f \n",test);
}
This is nor a problem really, but I was doing the first because the documentation of surf3Dread states that you this function is defined as:
template<class T>
T surf3Dread(cudaSurfaceObject_t surfObj,
int x, int y, int z,
boundaryMode = cudaBoundaryModeTrap);
template<class T>
void surf3Dread(T* data,
cudaSurfaceObject_t surfObj,
int x, int y, int z,
boundaryMode = cudaBoundaryModeTrap);
Maybe I am not understanding the documentation correctly, but I'd say that the first kernel here corresponds to the first documented way of calling the function and the second kernel to the second. Why does only one work? If I misunderstood the first function in the documentation, how do you call that version?
I am using CUDA 10.2

Answer 1

In the first instance, the compiler cannot deduce the template instantiation from the supplied function arguments. You need to specify the type explicitly to the compiler. This:
#include <cstdio>
__global__ void test_surface(cudaSurfaceObject_t surfImg,int x, int y, int z){
float test = surf3Dread<float>(surfImg, (int)(x*sizeof(float)), y, z, cudaBoundaryModeZero);
printf("%f \n",test);
}
will work where your version will not.