从推力::设备向量到原始指针并返回?

2022-01-10 00:00:00 cuda c++ thrust

我了解如何从向量转到原始指针，但我跳过了关于如何倒退的节拍.

I understand how to go from a vector to a raw pointer but im skipping a beat on how to go backwards.

// our host vector
thrust::host_vector<dbl2> hVec;

// pretend we put data in it here

// get a device_vector
thrust::device_vector<dbl2> dVec = hVec;

// get the device ptr
thrust::device_ptr devPtr = &d_vec[0];

// now how do i get back to device_vector?
thrust::device_vector<dbl2> dVec2 = devPtr; // gives error
thrust::device_vector<dbl2> dVec2(devPtr); // gives error

有人可以解释/给我举个例子吗?

Can someone explain/point me to an example?

推荐答案

您像标准容器一样初始化和填充推力向量，即通过迭代器:

You initialize and populate thrust vectors just like standard containers, i.e. via iterators:

#include <thrust/device_vector.h>
#include <thrust/device_ptr.h>

int main()
{
  thrust::device_vector<double> v1(10);                    // create a vector of size 10
  thrust::device_ptr<double> dp = v1.data();               // or &v1[0]

  thrust::device_vector<double> v2(v1);                    // from copy
  thrust::device_vector<double> v3(dp, dp + 10);           // from iterator range
  thrust::device_vector<double> v4(v1.begin(), v1.end());  // from iterator range
}

在您的简单示例中，无需通过指针绕道，因为您可以直接复制另一个容器.一般来说，如果你有一个指向数组开头的指针，如果你提供数组大小，你可以使用 v3 的版本.

In your simple example there's no need to go the detour via pointers, as you can just copy the other container directly. In general, if you have a pointer to the beginning of an array, you can use the version for v3 if you supply the array size.

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