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why in project A it only occupy 20 M9Ks?
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This is fine and makes perfect sense. 1024 bytes per M9K block. 20 x 1024 bytes = 20 M9K blocks.
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why in project B it occupy 32 M9Ks?
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It shouldn't. I can only presume you are not (quite) doing what you think you are in copying the project.
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i am a little confused,
in binary system, the 20480 bytes should occupy the same space with 2048*16 bytes, that is 2048*16*8=83,886,080bits=28.4 M9Ks, that is 29M9Ks
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I dare say you are 'a little confused'. 20480 bytes fits into 20 x M9K blocks, as project A confirms. 28.4, 29 or 32 are not necessary.
I suggest you do a little more digging.
Cheers,
Alex
Honored Contributor
