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Results 1 - 3 of 3 for NewLRU (0.04 sec)

  1. tests/lru_test.go

    func TestLRU_Add_ExistingKey_UpdatesValueAndExpiresAt(t *testing.T) {
    	lru := lru.NewLRU[string, int](10, nil, time.Hour)
    	lru.Add("key1", 1)
    	lru.Add("key1", 2)
    
    	if value, ok := lru.Get("key1"); !ok || value != 2 {
    		t.Errorf("Expected value to be updated to 2, got %v", value)
    	}
    }
    
    func TestLRU_Add_NewKey_AddsEntry(t *testing.T) {
    	lru := lru.NewLRU[string, int](10, nil, time.Hour)
    	lru.Add("key1", 1)
    
    Registered: Sun Sep 07 09:35:13 UTC 2025
    - Last Modified: Sun May 25 07:40:40 UTC 2025
    - 10.4K bytes
    - Viewed (0)
  2. internal/lru/lru.go

    const numBuckets = 100
    
    // NewLRU returns a new thread-safe cache with expirable entries.
    //
    // Size parameter set to 0 makes cache of unlimited size, e.g. turns LRU mechanism off.
    //
    // Providing 0 TTL turns expiring off.
    //
    // Delete expired entries every 1/100th of ttl value. Goroutine which deletes expired entries runs indefinitely.
    Registered: Sun Sep 07 09:35:13 UTC 2025
    - Last Modified: Fri Apr 25 08:22:26 UTC 2025
    - 12.5K bytes
    - Viewed (0)
  3. internal/stmt_store/stmt_store.go

    		size = defaultMaxSize
    	}
    
    	if ttl <= 0 {
    		ttl = defaultTTL
    	}
    
    	onEvicted := func(k string, v *Stmt) {
    		if v != nil {
    			go v.Close()
    		}
    	}
    	return &lruStore{lru: lru.NewLRU[string, *Stmt](size, onEvicted, ttl)}
    }
    
    type lruStore struct {
    	lru *lru.LRU[string, *Stmt]
    }
    
    func (s *lruStore) Keys() []string {
    	return s.lru.Keys()
    }
    
    Registered: Sun Sep 07 09:35:13 UTC 2025
    - Last Modified: Sun Apr 27 06:05:16 UTC 2025
    - 6K bytes
    - Viewed (0)
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