复制到CUDA常量内存时，设备符号无效 [英] Invalid device symbol when copying to CUDA constant memory

问题描述

我有一个应用程序的几个文件在图像处理。因为图像的行和列数在做一些图像处理算法时不会改变，我试图将这些值放在常量内存中。我的应用程式如下：

I have several files for an app in image processing. As the number of rows and colums for an image does not change while doing some image processing algorithm I was trying to put those values in constant memory. My app looks like:

Imageproc.cuh

...
...
__constant__ int c_rows;
__constant__ int c_cols;

#ifdef __cplusplus
   extern "C"
   {
#endif
   ...
   ...
#ifdef __cplusplus
   }
#endif

Imageproc.cu

...
...

int algorithm(float *a, const int rows, const int cols){
   ...
   ...
   checkCudaError(cudaMemcpyToSymbol(&c_rows, &rows, sizeof(int)));
   checkCudaError(cudaMemcpyToSymbol(&c_cols, &cols, sizeof(int)));

   dim3 block(T, T);
   dim3 grid(cols/T+1, rows/T+1);

   kernel<<<grid, block>>>( ... );
   ...
   ...

}

它编译得很好，但是当试图运行程序时，我得到无效的设备符号cudaMemcpyToSymbol（& c_rows，& rows，sizeof（int））

It compiles well but when trying to run the program I get invalid device symbol cudaMemcpyToSymbol(&c_rows, &rows, sizeof(int))

无法将这些变量放在常量内存中或者我缺少什么？

Can't I put those variables in constant memory or what am I missing?

推荐答案

如果你的符号声明如下：

If your symbol is declared like this:

__constant__ int c_rows;

那么正确调用 cudaMemcpyToSymbol

int rows = 5;
cudaMemcpyToSymbol(c_rows, &rows, sizeof(int)));

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