指向设备对象成员的指针

Question

I know, that I should not dereference pointer to device. But is the following an dereferencation?

struct example{
 int a;
 float b;
};

void test(){
 example* p_e; 
 cudamalloc(&p_e,sizeof(example));
 float* db = &p_e->b; // is this line valid?
}

It works on my computer, but is this valid?

edit:

With it works, I meant, that it holds in my testrun that:

reinterpret_cast<void*>(&p_e->b) == reinterpret_cast<void*>(reinterpret_cast<unsigned char*>(p_e)+sizeof(int))

Answer 1

Regarding this:

float db = &p_e->b;

I can't imagine a situation where the address of an entity in C or C++ is a float value. So from a language perspective, it makes no sense. If we modify that to:

float *db = &p_e->b;

That is a sensible construct, although whether it is "useful" depends on how you intend to use it. Taking the address of the location p_e->b requires taking the value of the pointer contained in p_e (a location in host memory, so no device dereferencing going on there) and adding to it the offset to the location b within the object. That offset doesn't involve any dereferencing. It can be computed simply by inspection of the class/structure definition. So none of that requires any dereferencing of pointers.

The pointer that is produced ( db ) is now pointing to a location in device memory, so that pointer should only be used (dereferenced) in device code. But you could pass that pointer by value to a CUDA kernel and make "use" of it, sensibly.

If you actually intended to populate the value db with the float contents pointed to by the pointer p_e->b , first of all your code does not do that, and second it would not generally be possible in host code to directly retrieve that data from the device, when the underlying base structure pointer ( b_e ) is allocated using cudaMalloc .