Why did device code can access host variable in cuda like this?

Question

In the code below, I have use the thrust library, what I don't understand is in the device

code I can access v , which I think is in host memory, because I direct give it the value

4 in the constructor.

struct op{
    float v;
    op(float iv):v(iv){}
    __device__
    void operator()(float v2)
    {
         printf("%f\n", v+v2);
    }
};
int main()
{
    device_vector<float> data(4,1);
    for_each(data.begin(), data.end(), op(4));
    return 0;
}

The output of the code above is 5.0 5.0 5.0 5.0 .

So, can you give some hints about this behavior? I have searched through the official

manual, but didn't get any information about this behavior.

---

Sorry for my express, just like Ben Voigt said, my problem is " what I don't understand is in the device code I can access v, which I think is in host memory "

Answer 1

v is not in host memory, at least not if the functor object is instantiated in device code. (And your struct, whose () operator only has a __device__ decoration, would only "work" in device code.)

iv starts out in host memory, but the process of copying from iv (host) to v (device, at least if the op object is used in a thrust device operation) is handled by thrust prior to computation of the for_each function.

The copying of the op() initialization parameter (4, in your case) is really no different than what needs to happen with the 1 parameter here:

device_vector<float> data(4,1);
                            ^

That 1 parameter exists in host code . However the thrust constructor for device_vector initiates the necessary host->device memory copies (and/or kernel calls, depending on the specific thrust function) to initialize the device_vector storage, which exists on the device .

This is handled at runtime, not compile-time. You can replace your 4 constant with a variable, and still get valid results, just as you could replace your 1 initializer for device_vector with a variable.

That is how I would explain it. Probably talonmies or JaredHoberock (or others) can correct my terminology, but I think the gist is correct.