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The new genericized LRU cache no longer needs to have the IDs separately allocated: name old time/op new time/op delta Add-8 494ns ± 2% 388ns ± 2% -21.46% (p=0.000 n=10+9) name old alloc/op new alloc/op delta Add-8 176B ± 0% 152B ± 0% -13.64% (p=0.000 n=10+10) name old allocs/op new allocs/op delta Add-8 5.00 ± 0% 3.00 ± 0% -40.00% (p=0.000 n=10+10)
80 lines
1.6 KiB
Go
80 lines
1.6 KiB
Go
package bloblru
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import (
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"math/rand"
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"testing"
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"github.com/restic/restic/internal/restic"
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rtest "github.com/restic/restic/internal/test"
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)
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func TestCache(t *testing.T) {
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var id1, id2, id3 restic.ID
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id1[0] = 1
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id2[0] = 2
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id3[0] = 3
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const (
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kiB = 1 << 10
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cacheSize = 64*kiB + 3*overhead
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)
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c := New(cacheSize)
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addAndCheck := func(id restic.ID, exp []byte) {
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c.Add(id, exp)
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blob, ok := c.Get(id)
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rtest.Assert(t, ok, "blob %v added but not found in cache", id)
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rtest.Equals(t, &exp[0], &blob[0])
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rtest.Equals(t, exp, blob)
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}
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// Our blobs have len 1 but larger cap. The cache should check the cap,
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// since it more reliably indicates the amount of memory kept alive.
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addAndCheck(id1, make([]byte, 1, 32*kiB))
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addAndCheck(id2, make([]byte, 1, 30*kiB))
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addAndCheck(id3, make([]byte, 1, 10*kiB))
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_, ok := c.Get(id2)
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rtest.Assert(t, ok, "blob %v not present", id2)
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_, ok = c.Get(id1)
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rtest.Assert(t, !ok, "blob %v present, but should have been evicted", id1)
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c.Add(id1, make([]byte, 1+c.size))
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_, ok = c.Get(id1)
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rtest.Assert(t, !ok, "blob %v too large but still added to cache")
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c.c.Remove(id1)
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c.c.Remove(id3)
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c.c.Remove(id2)
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rtest.Equals(t, cacheSize, c.size)
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rtest.Equals(t, cacheSize, c.free)
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}
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func BenchmarkAdd(b *testing.B) {
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const (
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MiB = 1 << 20
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nblobs = 64
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)
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c := New(64 * MiB)
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buf := make([]byte, 8*MiB)
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ids := make([]restic.ID, nblobs)
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sizes := make([]int, nblobs)
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r := rand.New(rand.NewSource(100))
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for i := range ids {
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r.Read(ids[i][:])
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sizes[i] = r.Intn(8 * MiB)
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}
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b.ResetTimer()
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b.ReportAllocs()
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for i := 0; i < b.N; i++ {
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c.Add(ids[i%nblobs], buf[:sizes[i%nblobs]])
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}
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}
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