如何正确链接cuda头文件与设备功能？ [英] How to properly link cuda header file with device functions?

问题描述

我试图解耦我的代码，有点失败。编译错误：

I'm trying to decouple my code a bit and something fails. Compilation error:

error: calling a __host__ function("DecoupledCallGpu") from a __global__ function("kernel") is not allowed

代码摘录：

main.c （调用cuda主机函数）：

main.c (has a call to cuda host function):

#include "cuda_compuations.h"
...
ComputeSomething(&var1,&var2);
...

cuda_computations.cu 主机主函数，包括具有设备操作的标题）：

cuda_computations.cu (has kernel, host master functions and includes header which has device unctions):

#include "cuda_computations.h"
#include "decoupled_functions.cuh"
...
__global__ void kernel(){
...
DecoupledCallGpu(&var_kernel);
}

void ComputeSomething(int *var1, int *var2){
//allocate memory and etc..
...
kernel<<<20,512>>>();
//cleanup
...
}

decoupled_functions.cuh ：：

decoupled_functions.cuh:

#ifndef _DECOUPLEDFUNCTIONS_H_
#define _DECOUPLEDFUNCTIONS_H_

void DecoupledCallGpu(int *var);

#endif

decoupled_functions.cu：

#include "decoupled_functions.cuh"

__device__ void DecoupledCallGpu(int *var){
  *var=0;
}

#endif

汇编：

nvcc -g --ptxas-options = -v -arch = sm_30 -c cuda_computations.cu -o cuda_computations.o -lcudart

nvcc -g --ptxas-options=-v -arch=sm_30 -c cuda_computations.cu -o cuda_computations.o -lcudart

问题：为什么从主机函数调用 DecoupledCallGpu ，而不是内核

Question: why is it that the DecoupledCallGpu is called from host function and not a kernel as it was supposed to?

PS：如果你需要我可以共享它背后的实际代码。

P.S.: I can share the actual code behind it if you need me to.

推荐答案

将 __ device __ 装饰器添加到 decoupled_functions.cuh 中的原型。

Add the __device__ decorator to the prototype in decoupled_functions.cuh. That should take care of the error message you are seeing.

然后您需要使用您的模块之间单独的编译和链接。因此，与使用 -dc 编译 -c 不同。您的链接命令将需要修改。一个基本示例是此处。

Then you'll need to use separate compilation and linking amongst your modules. So instead of compiling with -c compile with -dc. And your link command will need to be modified. A basic example is here.

您的问题有点混乱：

问题：为什么DecoupledCallGpu从主机函数调用而不是内核，因为它是应该的？

Question: why is it that the DecoupledCallGpu is called from host function and not a kernel as it was supposed to?

我不知道你是否绊倒英语如果在这里有一个误会。实际的错误消息指出：

I can't tell if you're tripping over english or if there is a misunderstanding here. The actual error message states:

错误：调用 __ host __ 函数（DecoupledCallGpu

error: calling a __host__ function("DecoupledCallGpu") from a __global__ function("kernel") is not allowed

void DecoupledCallGpu(int *var);

此原型表示CUDA C中的未装饰函数， 等效于 __ host__ （仅）装饰函数：

This prototype indicates an undecorated function in CUDA C, and such functions are equivalent to __host__ (only) decorated functions:

__host__ void DecoupledCallGpu(int *var);

编译单元不知道decoupled_functions.cu中实际上是什么。

That compilation unit has no knowledge of what is actually in decoupled_functions.cu.

因此，当你有这样的内核代码：

Therefore, when you have kernel code like this:

__global__ void kernel(){       //<- __global__ function
...
DecoupledCallGpu(&var_kernel);  //<- appears as a __host__ function to compiler
}

编译器认为尝试从 __ global __ 函数调用 __ host __ 函数，这是非法的。

the compiler thinks you are trying to call a __host__ function from a __global__ function, which is illegal.

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