I think that 6x bigger and 5x slower pretty much makes my point.
Holy crap, now I'm curious. Only 5x slower is faster than what I would have guessed. But 6x larger is crazy. Anyone know how much boxing Perl does? I would have thought that the rough estimate for the size of a scalar number would be say 16 bytes (4 bytes for a pointer, 4 bytes for type/reference-count information, 8 bytes for a double precision floating point number). And the overhead for an array at maybe 16 bytes (8 bytes for a length field, 4 bytes for type/reference-count info, 4 bytes for a pointer to the array of pointers). For the tree structure above (an array composed of one scalar and two arrays) that would be 16 (array overhead)+3*4(three elements in first array, 4 byte pointers (32 bit machine))+16(the scalar)+2*16(the left and right branches) = 76 bytes. I guess that's starting to add up, but it is still shy of the 112 bytes measured above. Perl must preallocate space for each array to make growing it faster (maybe 12 elements initially?). Does that sound about right? Any way to get a more slimed down data structure in pure Perl?

In reply to Re^5: Perl memory usage by Anonymous Monk
in thread Performance, Abstraction and HOP by pg

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