Geek asks and answers

CUDA cudaMemcpy: invalid argument

Here is my code:

struct S {
    int a, b;
    float c, d;
};
class A {
private:
    S* d;
    S h[3];
public:
    A() {
        cutilSafeCall(cudaMalloc((void**)&d, sizeof(S)*3));
    }
void Init();
};

void A::Init() {
    for (int i=0;i<3;i++) {
        h[i].a = 0;
        h[i].b = 1;
        h[i].c = 2;
        h[i].d = 3;
    }
    cutilSafeCall(cudaMemcpy(d, h, 3*sizeof(S), cudaMemcpyHostToDevice));
}

A a;

This is actually a complex program that contains CUDA and OpenGL. When I debug this program, it does not work when running cudaMemcpy with error information

cudaSafeCall () Runtime 11 API error: invalid argument.

Actually, this program is converted from another, which may work correctly. But in this I used two variables S * d and S h [3] in the main function, and not in the class. What is weirder is that I implement this class A in a small program, it works fine. And I updated my driver, the error still exists.

Can someone give me a clue why this is happening and how to solve it. Thank.

+3
source share
1 answer

CUDA , . , , cudaThreadSynchonize, .

, , cudaThreadSynchronize .

, malloc. CUDA, , @Harrism, ??? printf- , . , - .

  • printf , cudaMalloc'ed

    A()
{
    d = NULL;
    cutilSafeCall(cudaMalloc((void**)&d, sizeof(S)*3));
    printf("D: %p\n", d);
}

  • , cudaMalloc cudaMemcopy ( ).

    void A::Init()
{
    for (int i=0;i<3;i++)
    {
        h[i].a = 0;
        h[i].b = 1;
        h[i].c = 2;
        h[i].d = 3;
    }
    cutilSafeCall(cudaMalloc((void**)&d, sizeof(S)*3)); // here!..
    cutilSafeCall(cudaMemcpy(d, h, 3*sizeof(S), cudaMemcpyHostToDevice));
}

.

+3

More articles:


All Articles