2D array allocation and copy ?

Accelerated Computing CUDA CUDA Programming and Performance
1

hi everyone !

i’ve created my 2D host array : h_ucPicture[8192][16] ( BIG ARRAY :) )

and my 2D device array : *d_ucPicture=0;

Now, I’d like to allocate this device array. I want to use the cudaMallocPitch function :

this is what i’ve already done :

unsigned char	h_ucPicture[HEIGHT][WIDTH];

unsigned char*	d_ucPicture=0;

size_t pitch=sizeof(unsigned char)*HEIGHT*WIDTH;

cutilSafeCall( cudaMallocPitch((void**) &d_ucPicture, &pitch, WIDTH, HEIGHT));

Is this right ? because i dont really understand how this function works…

After, i decide to copy all datas from h_ucPicture to d_ucPicture :

I remember that WIDTH = 8192 and WIDTH = 16

It compiles but after excecuting i have a runtime error …

thanks for help! ! :unsure: :unsure:

2

cudaMallocPitch((void**) &d_ucPicture, &pitch, WIDTH, HEIGHT)), the third parameter should be WIDTH*sizeof(datatype).

You can check the detailed definition in http://developer.download.nvidia.com/compu…6e49be4a0b6d8db

And it isn’t need to assign the value to pitch because it will be filled after calling cudaMallocPitch.