CODEWITHSUNDEEP
cuda
Yasser
Yasser

Reputation: 1785

Using std::vector in CUDA device code

The question is that: is there a way to use the class "vector" in Cuda kernels? When I try I get the following error:

error : calling a host function("std::vector<int, std::allocator<int> > ::push_back") from a __device__/__global__ function not allowed

So there a way to use a vector in global section? I recently tried the following:

  1. create a new Cuda project
  2. go to properties of the project
  3. open Cuda C/C++
  4. go to Device
  5. change the value in "Code Generation" to be set to this value: compute_20,sm_20

........ after that I was able to use the printf standard library function in my Cuda kernel.

is there a way to use the standard library class vector in the way printf is supported in kernel code? This is an example of using printf in kernel code:

// this code only to count the 3s in an array using Cuda
//private_count is an array to hold every thread's result separately 

__global__ void countKernel(int *a, int length, int* private_count) 
{
    printf("%d\n",threadIdx.x);  //it's print the thread id and it's working

    // vector<int> y;
    //y.push_back(0); is there a possibility to do this?

    unsigned int offset  = threadIdx.x * length;
    int i = offset;
    for( ; i < offset + length; i++)
    {
        if(a[i] == 3)
        {
            private_count[threadIdx.x]++;
            printf("%d ",a[i]);
        }
    }   
}

Upvotes: 56

Views: 53807

Answers (5)

einpoklum
einpoklum

Reputation: 131547

You can't use std::vector in device-side code. Why?

It's not marked to allow this

The "formal" reason is that, to use code in your device-side function or kernel, that code itself has to be in a __device__ function; and the code in the standard library, including, std::vector is not. (There's an exception for constexpr code; and in C++20, std::vector does have constexpr methods, but CUDA does not support C++20 at the moment, plus, that constexprness is effectively limited.)

You probably don't really want to

The std::vector class uses allocators to obtain more memory when it needs to grow the storage for the vectors you create or add into. By default (i.e. if you use std::vector<T> for some T) - that allocation is on the heap. While this could be adapted to the GPU - it would be quite slow, and incredibly slow if each "CUDA thread" would dynamically allocate its own memory.

Now, you could say "But I don't want to allocate memory, I just want to read from the vector!" - well, in that case, you don't need a vector per se. Just copy the data to some on-device buffer, and either pass a pointer and a size, or use a CUDA-capable span, like in cuda-api-wrappers or cuda-kat.

Another option, though a bit "heavier", is to use the NVIDIA thrust library's device_vector class. Under the hood, it's quite different from the standard library vector though.

Upvotes: 2

Zhong Hu
Zhong Hu

Reputation: 272

In the CUDA library Thrust, you can use thrust::device_vector<classT> to define a vector on the device, and the data transfer between host STL vector and device_vector is very straightforward. You can refer to this useful link to find some useful examples.

Note however, that device_vector itself can not be used in device code either. Only its pointers/iterators can be used there.

Upvotes: 14

Robin Lew
Robin Lew

Reputation: 483

I think you can implement a device vector by youself, because CUDA supports dynamic memory alloction in device codes. Operator new/delete are also supported. Here is an extremely simple prototype of device vector in CUDA, but it does work. It hasn't been tested sufficiently.

template<typename T>
class LocalVector
{
private:
    T* m_begin;
    T* m_end;

    size_t capacity;
    size_t length;
    __device__ void expand() {
        capacity *= 2;
        size_t tempLength = (m_end - m_begin);
        T* tempBegin = new T[capacity];

        memcpy(tempBegin, m_begin, tempLength * sizeof(T));
        delete[] m_begin;
        m_begin = tempBegin;
        m_end = m_begin + tempLength;
        length = static_cast<size_t>(m_end - m_begin);
    }
public:
    __device__  explicit LocalVector() : length(0), capacity(16) {
        m_begin = new T[capacity];
        m_end = m_begin;
    }
    __device__ T& operator[] (unsigned int index) {
        return *(m_begin + index);//*(begin+index)
    }
    __device__ T* begin() {
        return m_begin;
    }
    __device__ T* end() {
        return m_end;
    }
    __device__ ~LocalVector()
    {
        delete[] m_begin;
        m_begin = nullptr;
    }

    __device__ void add(T t) {

        if ((m_end - m_begin) >= capacity) {
            expand();
        }

        new (m_end) T(t);
        m_end++;
        length++;
    }
    __device__ T pop() {
        T endElement = (*m_end);
        delete m_end;
        m_end--;
        return endElement;
    }

    __device__ size_t getSize() {
        return length;
    }
};

Upvotes: 6

yyfn
yyfn

Reputation: 767

you can't use std::vector in device code, you should use array instead.

Upvotes: 7

Edvard Pedersen
Edvard Pedersen

Reputation: 773

You can't use the STL in CUDA, but you may be able to use the Thrust library to do what you want. Otherwise just copy the contents of the vector to the device and operate on it normally.

Upvotes: 26

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