Christopher Haster
2fe2078f50
It doesn't make sense to test more complex logic, such as t2_btree.toml, when the logic it is built on, t1_rbyd.toml, does not past testing. The test runner already guarantees a consistent lexicographic order, so all we need to do is renamed these from test_* -> tn_*. Note, if we every have more than 10 tests, we will need to bump up the number of digits for all tests, so t1_rbyd.toml -> t01_rbyd.toml. This is the main downside of lexicographic ordering. But we'll cross that bridge when we get to it.
112 lines
3.0 KiB
TOML
112 lines
3.0 KiB
TOML
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# maximize lookahead buffer, we don't actually gc so we only get one pass
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# of the disk for these tests
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defines.LOOKAHEAD_SIZE = 'BLOCK_COUNT / 8'
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[cases.b2_btree_lookup]
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defines.N = [8, 16, 32, 64, 128, 256, 1024]
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# 0 = in-order
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# 1 = reversed-order
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# 2 = random-order
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defines.ORDER = [0, 1, 2]
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defines.SEED = 42
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in = 'lfs.c'
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code = '''
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lfs_t lfs;
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lfs_init(&lfs, cfg) => 0;
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// create free lookahead
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memset(lfs.lookahead.buffer, 0, lfs.cfg->lookahead_size);
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lfs.lookahead.start = 0;
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lfs.lookahead.size = lfs_min(8*lfs.cfg->lookahead_size,
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lfs.cfg->block_count);
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lfs.lookahead.next = 0;
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lfs_alloc_ack(&lfs);
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uint32_t prng = SEED;
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// create a tree with N elements
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lfsr_btree_t btree = LFSR_BTREE_NULL;
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const char *alphas = "abcdefghijklmnopqrstuvwxyz";
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for (lfs_size_t i = 0; i < N; i++) {
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lfs_off_t i_
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= (ORDER == 0) ? i
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: (ORDER == 1) ? 0
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: BENCH_PRNG(&prng) % (lfsr_btree_weight(&btree)+1);
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lfsr_btree_push(&lfs, &btree, i_, LFSR_TAG_INLINED, 1,
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LFSR_DATA_BUF(&alphas[i % 26], 1)) => 0;
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}
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// assume an unfetched btree
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btree.root.off = 0;
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// bench lookup
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BENCH_START();
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lfs_size_t i = BENCH_PRNG(&prng) % N;
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uint8_t buffer[4];
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lfsr_tag_t tag_;
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lfs_size_t weight_;
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lfsr_btree_get(&lfs, &btree, i,
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&tag_, &weight_, buffer, 4) => 1;
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assert(tag_ == LFSR_TAG_INLINED);
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assert(weight_ == 1);
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BENCH_STOP();
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'''
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[cases.b2_btree_commit]
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defines.N = [8, 16, 32, 64, 128, 256, 1024]
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# 0 = in-order
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# 1 = reversed-order
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# 2 = random-order
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defines.ORDER = [0, 1, 2]
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defines.SEED = 42
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defines.AMORTIZED = false
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in = 'lfs.c'
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code = '''
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lfs_t lfs;
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lfs_init(&lfs, cfg) => 0;
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// create free lookahead
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memset(lfs.lookahead.buffer, 0, lfs.cfg->lookahead_size);
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lfs.lookahead.start = 0;
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lfs.lookahead.size = lfs_min(8*lfs.cfg->lookahead_size,
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lfs.cfg->block_count);
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lfs.lookahead.next = 0;
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lfs_alloc_ack(&lfs);
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uint32_t prng = SEED;
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// create a tree with N elements
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if (AMORTIZED) {
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BENCH_START();
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}
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lfsr_btree_t btree = LFSR_BTREE_NULL;
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const char *alphas = "abcdefghijklmnopqrstuvwxyz";
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for (lfs_size_t i = 0; i < N; i++) {
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lfs_off_t i_
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= (ORDER == 0) ? i
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: (ORDER == 1) ? 0
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: BENCH_PRNG(&prng) % (lfsr_btree_weight(&btree)+1);
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lfsr_btree_push(&lfs, &btree, i_, LFSR_TAG_INLINED, 1,
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LFSR_DATA_BUF(&alphas[i % 26], 1)) => 0;
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}
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// bench appending a new id
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if (!AMORTIZED) {
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BENCH_START();
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}
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lfs_size_t i = BENCH_PRNG(&prng) % N;
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lfsr_btree_push(&lfs, &btree, i, LFSR_TAG_INLINED, 1,
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LFSR_DATA_BUF(&alphas[i % 26], 1)) => 0;
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BENCH_STOP();
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uint8_t buffer[4];
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lfsr_tag_t tag_;
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lfs_size_t weight_;
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lfsr_btree_get(&lfs, &btree, i,
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&tag_, &weight_, buffer, 4) => 1;
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assert(tag_ == LFSR_TAG_INLINED);
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assert(weight_ == 1);
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'''
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