Hi,

As i understand from the description, it looks like the above memcpy statement takes 40ms on target for 2534400 bytes. But i could not the second statement, Do you mean if the gl buffer is allocated using malloc, it takes only 5ms? Typically malloc buffer is cached, so can you check if gl buffer is cached? This can be one reason why it takes some much time in copying..

Regards,

Brijesh

How do I check if the buffer is cached?

I am not sure how GL allocates the buffer. Can you please go inside the implementation and check?

I'm also not sure how GL allocates buffers. And then if we assume that it does have no cache, is there a way to fix it? Or is there a better plan for me?

Why dont you directly use it for the display? In this case, there will not be need for the memcpy.

If there is someway to get the physical address of this buffer, we can directly use it for the display..

Rgds,

Brijesh

Unsigned char * ptr = (unsigned char *) glMapBufferRange (GL_PIXEL_PACK_BUFFER, 0, static_cast < glSizeIptr > (iSize), GL_MAP_READ_BIT);
"ptr" is two different addresses (using double buffering). TDA4's display interface is created with an address specified as input, so I can't directly use it. Or is there a good way I can use the pointer returned by the display interface as GL's framebuffer?

Hi,

Are you using OpenVX interface for accessing display? There are some OpenVX APIs using which you could provide this buffer pointer to the display node. 

How do you get the dPtr in below statement?

memcpy （d st ，ptr ，iSize ）;

Regards,

Brijesh

Hi,

I can get "dPtr", I have a new idea about how to use "dPtr" to create an EGLImageKHR, do you have relevant information to introduce to me?

Sorry, did not get your question. what is the new idea?

I was asking, do you use OpenVX framework and do you get dPtr from vx_image? 

Regards,

Brijesh