(gdb) p sizeof(key_map)
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$1 = 48
|
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(gdb) p sizeof(ulonglong)
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$2 = 8
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So, 64-bit bitmap takes 8 bytes, 128-bit bitmap would take 16 bytes, 256 bit bitmap would take 32 bytes, and that
1. would still be less than current size of 48 bytes.
2. would not have a risk of not calling Bitmap::~Bitmap somewhere, and leaking memory
On the other hand, solution that uses Bitmap<n> is already coded, while we would still need to code the in-place bitmaps.
gadnium ok