silly cuda function question (modifyable parameters)

Accelerated Computing CUDA CUDA Programming and Performance
1

simply put:

__device test(int x)
{
x++;
}

//kernel code:

r1 = 0;
test(r1);

// does r1 = 1 or r1 = 0 after this call?

2

Seriously? The standard C argument model applies, of course.

3

I’m not sure, but i think if you need to do something like that you need to use a int pointer. Do something like this:

simply put:

__device test(int *x)
{
*x++;
}

//kernel code:

int r1 = 0;
test(&r1);

4

The compiler understands pass by reference. You can make the parameter int &x if you want the parameter to be modifiable, but without the ugly pointer dereferencing.

5

So if I understand correctly the right away of doing this operation is:

__device test(int x)

{

x++;

}

//kernel code:

int r1 = 0;

test(&r1);

Is that it?

6

No. C++ introduced the concept of references so that pointer dereferencing is not required for modifiable arguments. So the function definition becomes

__device test(int & x)

{

x++;

}

Any good reference book on C++ (I like Stroustrup’s “The C++ Programming Language”) has lots of detail on the semantics of references.

7

I have problems with call-by-reference in CUDA 2.2 (look here: forum entry).
Can anybody please try to confirm this?

Thx,
Marcel