Question

我正在使用Thrust库的thrust :: copy_if函数，加上计算迭代器以获取数组中非零元素的索引。我还需要获取复制元素的数量。

我正在使用'counting_iterator.cu'示例中的代码，除了在我的应用程序中我需要重用预先分配的数组，所以我用thrust :: device_ptr包装它们然后将它们传递给thrust :: copy_if功能。这是代码：

using namespace thrust;

int output[5];
thrust::device_ptr<int> tp_output = device_pointer_cast(output);

float stencil[5];
stencil[0] = 0;
stencil[1] = 0;
stencil[2] = 1;
stencil[3] = 0;
stencil[4] = 1;
device_ptr<float> tp_stencil = device_pointer_cast(stencil);

device_vector<int>::iterator output_end = copy_if(make_counting_iterator<int>(0), 
     make_counting_iterator<int>(5), 
     tp_stencil, 
     tp_output, 
     _1 == 1);

int number_of_ones = output_end - tp_output;

如果我注释最后一行代码，该函数会正确填充输出数组。但是，当我取消注释它时，我得到以下编译错误：

1＆gt; C：\ Program Files \ NVIDIA GPU Computing Toolkit \ CUDA \ v5.5 \ include \ thrust / iterator / iterator_adaptor.h（223）：error：no operator“ - ”匹配这些操作数

1＆GT;操作数类型是：int * const - const thrust :: device_ptr

1＆GT;检测期间： 1 GT;实例化“thrust :: iterator_adaptor :: difference_type thrust :: iterator_adaptor :: distance_to（const thrust :: iterator_adaptor＆amp;）const [with Derived = thrust :: detail :: normal_iterator＆gt ;,Base = thrust :: device_ptr，Value = thrust :: use_default，System = thrust :: use_default，Traversal = thrust :: use_default，Reference = thrust :: use_default，Difference = thrust :: use_default，OtherDerived = thrust :: device_ptr，OtherIterator = int *，V = signed int，S = thrust :: device_system_tag，T = thrust :: random_access_traversal_tag，R = thrust :: device_reference，D = ptrdiff_t]“ 1 GT; C：\ Program Files \ NVIDIA GPU Computing Toolkit \ CUDA \ v5.5 \ include \ thrust / iterator / iterator_facade.h（181）：here 1 GT;实例化“Facade1 :: difference_type thrust :: iterator_core_access :: distance_from（const Facade1＆amp;，const Facade2＆amp;，thrust :: detail :: true_type）[与Facade1 = thrust :: detail :: normal_iterator＆gt;，Facade2 =推力：：device_ptr]” 1 GT; C：\ Program Files \ NVIDIA GPU Computing Toolkit \ CUDA \ v5.5 \ include \ thrust / iterator / iterator_facade.h（202）：here 1 GT;实例化“thrust :: detail :: distance_from_result :: type thrust :: iterator_core_access :: distance_from（const Facade1＆amp;，const Facade2＆amp;）[与Facade1 = thrust :: detail :: normal_iterator＆gt ;, Facade2 = thrust :: device_ptr ]” 1 GT; C：\ Program Files \ NVIDIA GPU Computing Toolkit \ CUDA \ v5.5 \ include \ thrust / iterator / iterator_facade.h（506）：here 1 GT;实例化“thrust :: detail :: distance_from_result，thrust :: iterator_facade＆gt; :: type thrust :: operator-（const thrust :: iterator_facade＆amp;，const thrust :: iterator_facade＆amp;）[with Derived1 = thrust :: detail：：normal_iterator＆gt;，Value1 = signed int，System1 = thrust :: device_system_tag，Traversal1 = thrust :: random_access_traversal_tag，Reference1 = thrust :: device_reference，Difference1 = signed int，Derived2 = thrust :: device_ptr，Value2 = signed int，System2 = thrust :: device_system_tag，Traversal2 = thrust :: random_access_traversal_tag，Reference2 = thrust :: device_reference，Difference2 = signed int]“ 1 GT; C：/ ProgramData / NVIDIA Corporation / CUDA Samples / v5.5 / 7_CUDALibraries / nsgaIIparallelo_23ott / rank_cuda.cu（70）：here

如果我使用thrust :: device_vector代替输出数组，那么一切都还可以：

using namespace thrust;

thrust::device_vector<int> output(5);

float stencil[5];
stencil[0] = 0;
stencil[1] = 0;
stencil[2] = 1;
stencil[3] = 0;
stencil[4] = 1;
device_ptr<float> tp_stencil = device_pointer_cast(stencil);

device_vector<int>::iterator output_end = copy_if(make_counting_iterator<int>(0), 
     make_counting_iterator<int>(5), 
     tp_stencil, 
     output.begin(), 
     _1 == 1);

int number_of_ones = output_end - output.begin();

你能建议解决这个问题吗？谢谢。

Answer 1

尝试在copy_if调用中使用device_ptr而不是device_vector :: iterator：

thrust::device_ptr<int> output_end = copy_if(make_counting_iterator<int>(0),
 make_counting_iterator<int>(5),
 tp_stencil,
 tp_output,
 _1 == 1);

Thrust：使用device_ptr时如何获取copy_if函数复制的元素数量

1 个答案: