M.V. Hutz
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## Description Currently, the name of `bucket` is a bit confusing, because it is considered a 'table' in literature (as well as the whole hash table). A `bucket` is better described as a 'subtable', which is used by the total hash table to perform cuckoo hashing. In addition, the constructors `NewTable`, `NewTableBy`, and `NewCustomTable` were given shorter names, because the package name `cuckoo` already implies that `New*` would create a hash table with cuckoo hashing. This package has one use-case, and so it unambiguous what constructors produce. ## Changes - `NewTable` -> `New` - `NewTableBy` -> `NewBy` - `NewCustomTable` -> `NewCustom` - `bucket` -> `subtable` ### Design Decisions - I would have renamed `Table` and `subtable` to map equivalents, but 'submap' implies that a certain subsection of the map is contained within it, which isn't quite right. - I chose not to go with `Map` and `table`, because of the split naming convention. ## Checklist - [x] Tests pass - [x] Docs updated Reviewed-on: #22 Co-authored-by: M.V. Hutz <git@maximhutz.me> Co-committed-by: M.V. Hutz <git@maximhutz.me>
49 lines
1.3 KiB
Go
49 lines
1.3 KiB
Go
package cuckoo_test
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import (
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"fmt"
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"git.maximhutz.com/tools/go-cuckoo"
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)
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// This example demonstrates what happens when EqualFunc and Hash disagree on
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// equality. Although 'isEqual' only compares user IDs, but the hashes use the
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// entire 'User' object. So, two objects with the same ID but different names
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// hash to different slots, so the table cannot find them.
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func ExampleEqualFunc_badEqualFunc() {
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type User struct{ ID, Name string }
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makeHash := func(seed uint64) cuckoo.Hash[User] {
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return func(u User) uint64 {
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digest := seed
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for _, c := range u.ID + u.Name {
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digest ^= uint64(c)
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}
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return digest
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}
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}
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// Two users with the same ID are equal.
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isEqual := func(a, b User) bool { return a.ID == b.ID }
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userbase := cuckoo.NewCustom[User, bool](makeHash(1), makeHash(2), isEqual)
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(userbase.Put(User{"1", "Robert Doe"}, true))
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fmt.Println("Has Robert?", userbase.Has(User{"1", "Robert Doe"}))
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fmt.Println("Has Johanna?", userbase.Has(User{"2", "Johanna Smith"}))
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// The hashes are different, so even though the equal function returns true,
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// the table does not recognize it.
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fmt.Println("Equal?", isEqual(User{"1", "Rob Doe"}, User{"1", "Robert Doe"}))
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fmt.Println("Has Rob?", userbase.Has(User{"1", "Rob Doe"}))
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// Output:
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// Has Robert? true
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// Has Johanna? false
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// Equal? true
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// Has Rob? false
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}
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