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mempool: don't over-size the free block bitmap
Currently, the free block bitmap is roughly 4 times larger than it needs to, wasting memory. Let's assume maxsz = 128, minsz = 8 and n_max = 40. Z_MPOOL_LVLS(128, 8) returns 3. The block size for level #0 is 128, the block size for level #1 is 128/4 = 32, and the block size for level zephyrproject-rtos#2 is 32/4 = 8. Hence levels 0, 1, and 2 for a total of 3 levels. So far so good. Now let's look at Z_MPOOL_LBIT_WORDS(). We get: Z_MPOOL_LBIT_WORDS_UNCLAMPED(40, 0) = ((40 << 0) + 31) / 32 = 2 Z_MPOOL_LBIT_WORDS_UNCLAMPED(40, 1) = ((40 << 2) + 31) / 32 = 5 Z_MPOOL_LBIT_WORDS_UNCLAMPED(40, 2) = ((40 << 4) + 31) / 32 = 20 None of those are < 2 so Z_MPOOL_LBIT_WORDS() takes the results from Z_MPOOL_LBIT_WORDS_UNCLAMPED(). Finally, let's look at _MPOOL_BITS_SIZE(. It sums all possible levels with Z_MPOOL_LBIT_BYTES() which is: #define Z_MPOOL_LBIT_BYTES(maxsz, minsz, l, n_max) \ (Z_MPOOL_LVLS((maxsz), (minsz)) >= (l) ? \ 4 * Z_MPOOL_LBIT_WORDS((n_max), l) : 0) Or given what we already have: Z_MPOOL_LBIT_BYTES(128, 8, 0, 40) = (3 >= 0) ? 4 * 2 : 0 = 8 Z_MPOOL_LBIT_BYTES(128, 8, 1, 40) = (3 >= 1) ? 4 * 5 : 0 = 20 Z_MPOOL_LBIT_BYTES(128, 8, 2, 40) = (3 >= 2) ? 4 * 20 : 0 = 80 Z_MPOOL_LBIT_BYTES(128, 8, 3, 40) = (3 >= 3) ? 4 * ?? Wait... we're missing this one: Z_MPOOL_LBIT_WORDS_UNCLAMPED(40, 3) = ((40 << 6) + 31) / 32 = 80 then: Z_MPOOL_LBIT_BYTES(128, 8, 3, 40) = (3 >= 3) ? 4 * 80 : 0 = 320 Further levels yeld (3 >= 4), (3 >= 5), etc. so they're all false and produce 0. So this means that we're statically allocating 428 bytes to the bitmap when clearly only the first 3 Z_MPOOL_LBIT_BYTES() results for the corresponding 3 levels that we have should be summed e.g. only 108 bytes. Here the code logic gets confused between level numbers and the number levels, hence the extra allocation which happens to be exponential. Signed-off-by: Nicolas Pitre <[email protected]>
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