EAL malloc layer assumed all free elements content
is filled with zeros ("clean"), as opposed to uninitialized ("dirty").
This assumption was ensured in two ways:
1. EAL memalloc layer always returned clean memory.
2. Freed memory was cleared before returning into the heap.
Clearing the memory can be as slow as around 14 GiB/s.
To save doing so, memalloc layer is allowed to return dirty memory.
Such segments being marked with RTE_MEMSEG_FLAG_DIRTY.
The allocator tracks elements that contain dirty memory
using the new flag in the element header.
When clean memory is requested via rte_zmalloc*()
and the suitable element is dirty, it is cleared on allocation.
When memory is deallocated, the freed element is joined
with adjacent free elements, and the dirty flag is updated:
a) If the joint element contains dirty parts, it is dirty:
dirty + freed + dirty = dirty => no need to clean
freed + dirty = dirty the freed memory
Dirty parts may be large (e.g. initial allocation),
so clearing them could create unpredictable slowdown.
b) If the only dirty part of the joint element
is the freed memory, the joint element can be made clean:
clean + freed + clean = clean => freed memory
clean + freed = clean must be cleared
freed + clean = clean
freed = clean
This logic naturally reproduces the old behavior
and always applies in modes when EAL memalloc layer
returns only clean segments.
As a result, memory is either cleared on free, as before,
or it will be cleared on allocation if need be, but never twice.
Signed-off-by: Dmitry Kozlyuk <dkozlyuk@nvidia.com>
Reviewed-by: Anatoly Burakov <anatoly.burakov@intel.com>
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