how to calculate the number of M9k that nios onchip memory occupied?

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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

hi, thanks for replying.

i check the fitter report:

http://www.alteraforum.com/forum/attachment.php?attachmentid=11266&stc=1

i also think that may be my problem, but the fitter report shows that the 163840 bit occupy 32M9ks

by the way, when i change the on chip memory size from 20480 bytes to 16384 bytes, the fitter report told me it takes only 16 M9ks

can you give me more advise? thanks!

Something that caught my eye: "Note: Fitter may spread logical memories into multiple blocks to improve timing. The actual required RAM blocks can be found in the Fitter Resource Usage section."

You can try changing SEED, system frequency.

What version are you using before 15.0?