On 6/3/21 10:26 AM, Thomas Monjalon wrote: > 03/06/2021 09:06, Andrew Rybchenko: >> On 6/2/21 11:35 PM, Thomas Monjalon wrote: >>> + * Allocate a chunk of memory on the GPU. >> >> Looking a below function it is required to clarify here if >> the memory is visible or invisible to GPU (or both allowed). > > This function allocates on the GPU so it is visible by the GPU. > I feel I misunderstand your question.
Below function says rte_gpu_malloc_visible() and its description highlights that allocated memory is visible to GPU. My problem that I don't understand what's the difference between these two functions. >>> + */ >>> +__rte_experimental >>> +int rte_gpu_malloc(uint16_t gpu_id, size_t size, void **ptr); >> >> May be *malloc() should return a pointer and "negative" >> values used to report various errnos? > > I don't understand what you mean by negative values if it is a pointer. > We could return a pointer and use rte_errno. I was talking about something like (void *)(-ENOMEM), but it is a bad idea. NULL + rte_errno is much better. However, may be I'd kept callback as is set rte_error in lib if negative value is returned by the callback. This way we'll be safe against lost rte_errno update in drivers.
