为什么定义类头，没有CUDA __device__属性工程？ （C ++） [英] Why defining class headers without CUDA __device__ attribute works? (C++)

问题描述

我有一个带有以下声明的.h文件：

I have a .h file with the following declarations:

class Foo{
public:
    inline int getInt();
};

且我的.cu档定义如下：

and my .cu file defines the following:

__device__ int Foo::getInt(){
   return 42;
}

这真的很棒，因为虽然我不能调用 getInt 从主机，我可以包括.h文件在.cpp文件，所以我有类型声明可见的主机。但对我来说，它似乎不工作，所以为什么我不需要把 __ device __ 属性在.h文件？

This is pretty awesome, because althought I cannot actually call getInt from host, I can include the .h file in .cpp files so I have the type declaration visible for the host. But for me it doesn't seem it should work, so why I dont need to put the __device__ attribute on the .h file?

推荐答案

如果它工作，它不应该。这可能是CUDA编译器中的一个错误，它可能会在将来得到修复 - 所以不要依赖它。

If it works, it should not. It is probably a bug in a CUDA compiler and it might get fixed in the future - so do not rely on it.

但是，如果你想让类可见对于主机（和非cuda编译器），但你有一些 __ device __ 功能，你不需要在主机上，你总是可以封装这些函数与 #ifdef __CUDACC __ - #endif 。 __ CUDACC __ 在使用nvcc进行编译时是预定义的，否则不是。所以你可以写在你的头像：

However, if you want the class to be visible for the host (and non-cuda compiler), but you have some __device__ functionality which you don't need on the host, you can always encapsulate those functions with the #ifdef __CUDACC__ -- #endif. The __CUDACC__ is predefined when compiling with nvcc, otherwise it is not. So you can write in your header something like:

class Foo{
public:
#ifdef __CUDACC__
    inline __device__ int getInt();
#endif
};

如果你害怕有太多的预处理器ifdefs，你也可以做一个小技巧： / p>

If you are afraid of having too many preprocessor ifdefs, you can also do a trick as follows:

#ifdef __CUDACC__
#define HOST __host__
#define DEVICE __device__
#else
#define HOST
#define DEVICE
#endif

...

class Foo{
public:
    inline HOST DEVICE int getInt();
};

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