CUDA 5.5 cudaMemcpyToSymbol，__constant__和超出范围的错误 [英] CUDA 5.5 cudaMemcpyToSymbol, __constant__ and out of scope error

问题描述

我正在尝试编译一个CUDA示例，它具有：

cuda.cu：

  __ constant__ unsigned VERTICES; 
 __constant__ unsigned TRIANGLES; 
  

和main.cpp中的相应代码;

  cudaMemcpyToSymbol（VERTICES，& verticesNo，sizeof（int））; 
 cudaMemcpyToSymbol（TRIANGLES，& trianglesNo，sizeof（int））; 
 

如何避免编译main.cpp时出现VERTICES未在此范围内声明错误？ p>



TIA。

干杯，

解决方案

CUDA __ constant __ 变量具有文件范围链接。这意味着 cudaMemcpyToSymbol 必须在定义 __ constant __ 变量的相同.cu文件中。

您可以向.cu文件添加包装器函数，并从.cpp文件中调用此函数。

cuda.cu：

  __ constant__ unsigned VERTICES; 
 __constant__ unsigned TRIANGLES; 
 
 void wrapper_fn（unsigned * verticesNo，unsigned * trianglesNo）
 {
 cudaMemcpyToSymbol（VERTICES，verticesNo，sizeof（unsigned））; 
 cudaMemcpyToSymbol（TRIANGLES，trianglesNo，sizeof（unsigned））; 
} 
  

然后只调用 wrapper_fn 在您的main.cpp。

I'm trying to compile a CUDA example which has;

cuda.cu:

__constant__ unsigned VERTICES;
__constant__ unsigned TRIANGLES;

and the corresponding code in main.cpp;

cudaMemcpyToSymbol(VERTICES, &verticesNo, sizeof(int));
cudaMemcpyToSymbol(TRIANGLES, &trianglesNo, sizeof(int));

How to avoid "VERTICES not declared in this scope" error when compiling the main.cpp?

TIA.

cheers,

解决方案

CUDA __constant__ variables have a file scope linkage. That means that the cudaMemcpyToSymbol have to be in the same .cu file where the __constant__ variable is defined.

You can add a wrapper function to the .cu file and call this one from your .cpp file.

sample for cuda.cu:

__constant__ unsigned VERTICES;
__constant__ unsigned TRIANGLES;

void wrapper_fn(unsigned *verticesNo, unsigned *trianglesNo)
{
  cudaMemcpyToSymbol(VERTICES, verticesNo, sizeof(unsigned));
  cudaMemcpyToSymbol(TRIANGLES, trianglesNo, sizeof(unsigned));
}

Then only call wrapper_fn in your main.cpp.

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