@@ -85,18 +85,31 @@ elements in contiguous memory locations. Once again to avoid memory allocation i
8585code these should be pre-allocated and passed as arguments or as bound objects.
8686
8787When passing slices of objects such as ``bytearray `` instances, Python creates
88- a copy which involves allocation. This can be avoided using a ``memoryview ``
89- object:
88+ a copy which involves allocation of the size proportional to the size of slice.
89+ This can be alleviated using a ``memoryview `` object. ``memoryview `` itself
90+ is allocated on heap, but is a small, fixed-size object, regardless of the size
91+ of slice it points too.
9092
9193.. code :: python
9294
93- ba = bytearray (100 )
94- func(ba[3 : 10 ]) # a copy is passed
95- mv = memoryview (ba)
96- func(mv[3 : 10 ]) # a pointer to memory is passed
95+ ba = bytearray (10000 ) # big array
96+ func(ba[30 : 2000 ]) # a copy is passed, ~2K new allocation
97+ mv = memoryview (ba) # small object is allocated
98+ func(mv[30 : 2000 ]) # a pointer to memory is passed
9799
98100 A ``memoryview `` can only be applied to objects supporting the buffer protocol - this
99- includes arrays but not lists.
101+ includes arrays but not lists. Small caveat is that while memoryview object is live,
102+ it also keeps alive the original buffer object. So, memoryviews isn't universal
103+ panacea. For instance, in the example above, if you are done with 10K buffer and
104+ just need those bytes 30:2000 from it, it may be better to make a slice, and let
105+ the 10K buffer go (be ready for garbage collection), instead of making a
106+ long-living memoryview and keeping 10K blocked for GC.
107+
108+ Nonetheless, ``memoryview `` is indispensable for advanced preallocated buffer
109+ management. ``.readinto() `` method discussed above puts data at the beginning
110+ of buffer and fills in entire buffer. What if you need to put data in the
111+ middle of existing buffer? Just create a memoryview into the needed section
112+ of buffer and pass it to ``.readinto() ``.
100113
101114Identifying the slowest section of code
102115---------------------------------------
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