问题描述
在这种情况下cudaMemcpy函数将如何工作？
How will the cudaMemcpy function work in this case?
我已经声明了这样的矩阵
I have declared a matrix like this
float imagen[par->N][par->M];
，我想将其复制到cuda设备，所以我做到了
and I want to copy it to the cuda device so I did this
float *imagen_cuda;
int tam_cuda=par->M*par->N*sizeof(float);
cudaMalloc((void**) &imagen_cuda,tam_cuda);
cudaMemcpy(imagen_cuda,imagen,tam_cuda,cudaMemcpyHostToDevice);
将2d数组复制到1d数组中可以吗？
Will this copy the 2d array into a 1d array fine?
我如何复制到另一个2d数组？
And how can I copy to another 2d array? can I change this and will it work?
float **imagen_cuda;
推荐答案
处理加倍下标的C数组并非易事在主机和设备之间复制数据时。在大多数情况下， cudaMemcpy （包括 cudaMemcpy2D ）期望用于源和目标的普通指针，而不是指向
It's not trivial to handle a doubly-subscripted C array when copying data between host and device. For the most part, cudaMemcpy (including cudaMemcpy2D) expect an ordinary pointer for source and destination, not a pointer-to-pointer.
最简单的方法（我认为）是压平主机和设备上的2D数组，并使用索引算法来模拟2D坐标：
The simplest approach (I think) is to "flatten" the 2D arrays, both on host and device, and use index arithmetic to simulate 2D coordinates:
float imagen[par->N][par->M];
float *myimagen = &(imagen[0][0]);
float myval = myimagen[(rowsize*row) + col];
然后您可以使用普通的cudaMemcpy操作来处理转移（使用 myimagen 指针）：
You can then use ordinary cudaMemcpy operations to handle the transfers (using the myimagen pointer):
float *d_myimagen;
cudaMalloc((void **)&d_myimagen, (par->N * par->M)*sizeof(float));
cudaMemcpy(d_myimagen, myimagen, (par->N * par->M)*sizeof(float), cudaMemcpyHostToDevice);
如果您真的想处理动态大小（即编译时未知）的双下标数组，您可以查看。
If you really want to handle dynamically sized (i.e. not known at compile time) doubly-subscripted arrays, you can review this question/answer.
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