Hi Clément,
Le 22/02/2017 à 09:44, Clément David a écrit :
.../...
At runtime in release mode, Scilab 5 `who()` was able to return the size of the
allocated data but this is not implemented yet as this is not clear to us if we
should return the size of the allocated raw data or the size of the Scilab
value. For exemple :
a = zeros(10,10);
1. `a` is 10x10 matrix of double, and thus with size: 10*10*8 bytes.
2. `a` is an ArrayOf<double> with size : sizeof(types::Double) + 10*10*8 bytes
Both computation are valuable, to determine either the data weight in memory or
the memory allocated by scilab to use them.
Up to now, the memory used by the container was included in who() results.
Why not allowing both results, and implementing an option to let the
user choosing the counting mode?
IMHO, returning the data memory without the container could become the
default. The relative difference between both gets very small as the
sizes of data increase.
For sparse encodings, the indices locating each non-zero or true value
could be accounted with the values (so not being considered as part of
the container), because the memory used to store indices is proportional
to the number of non-zero values. The container being just the
"structural" remainder.
By the way, it would be much clearer to return the number of bytes
instead of the equivalent number of decimal numbers. This unit (8-bytes
sometimes called a "word" in the help) looks screwy. It is
straightforward to understand it only for decimal numbers (and only with
real values), while bytes are datatype independent.
If back-compatibility is of concern, this could be returned with a new
unit="bytes" option. Otherwise, it could become the default.
HTH
Samuel
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