4a366fcfee
We do that by re-render visited pages that is not already in the stack. This may potentially do some double work, but that small penalty should be well worth it. Fixes #5281
97 lines
2.6 KiB
Go
97 lines
2.6 KiB
Go
// Copyright 2017-present The Hugo Authors. All rights reserved.
|
|
//
|
|
// Licensed under the Apache License, Version 2.0 (the "License");
|
|
// you may not use this file except in compliance with the License.
|
|
// You may obtain a copy of the License at
|
|
// http://www.apache.org/licenses/LICENSE-2.0
|
|
//
|
|
// Unless required by applicable law or agreed to in writing, software
|
|
// distributed under the License is distributed on an "AS IS" BASIS,
|
|
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
|
|
// See the License for the specific language governing permissions and
|
|
// limitations under the License.
|
|
|
|
// Package types contains types shared between packages in Hugo.
|
|
package types
|
|
|
|
import (
|
|
"sync"
|
|
)
|
|
|
|
// EvictingStringQueue is a queue which automatically evicts elements from the head of
|
|
// the queue when attempting to add new elements onto the queue and it is full.
|
|
// This queue orders elements LIFO (last-in-first-out). It throws away duplicates.
|
|
// Note: This queue currently does not contain any remove (poll etc.) methods.
|
|
type EvictingStringQueue struct {
|
|
size int
|
|
vals []string
|
|
set map[string]bool
|
|
mu sync.Mutex
|
|
}
|
|
|
|
// NewEvictingStringQueue creates a new queue with the given size.
|
|
func NewEvictingStringQueue(size int) *EvictingStringQueue {
|
|
return &EvictingStringQueue{size: size, set: make(map[string]bool)}
|
|
}
|
|
|
|
// Add adds a new string to the tail of the queue if it's not already there.
|
|
func (q *EvictingStringQueue) Add(v string) {
|
|
q.mu.Lock()
|
|
if q.set[v] {
|
|
q.mu.Unlock()
|
|
return
|
|
}
|
|
|
|
if len(q.set) == q.size {
|
|
// Full
|
|
delete(q.set, q.vals[0])
|
|
q.vals = append(q.vals[:0], q.vals[1:]...)
|
|
}
|
|
q.set[v] = true
|
|
q.vals = append(q.vals, v)
|
|
q.mu.Unlock()
|
|
}
|
|
|
|
// Contains returns whether the queue contains v.
|
|
func (q *EvictingStringQueue) Contains(v string) bool {
|
|
q.mu.Lock()
|
|
defer q.mu.Unlock()
|
|
return q.set[v]
|
|
}
|
|
|
|
// Peek looks at the last element added to the queue.
|
|
func (q *EvictingStringQueue) Peek() string {
|
|
q.mu.Lock()
|
|
l := len(q.vals)
|
|
if l == 0 {
|
|
q.mu.Unlock()
|
|
return ""
|
|
}
|
|
elem := q.vals[l-1]
|
|
q.mu.Unlock()
|
|
return elem
|
|
}
|
|
|
|
// PeekAll looks at all the elements in the queue, with the newest first.
|
|
func (q *EvictingStringQueue) PeekAll() []string {
|
|
q.mu.Lock()
|
|
vals := make([]string, len(q.vals))
|
|
copy(vals, q.vals)
|
|
q.mu.Unlock()
|
|
for i, j := 0, len(vals)-1; i < j; i, j = i+1, j-1 {
|
|
vals[i], vals[j] = vals[j], vals[i]
|
|
}
|
|
return vals
|
|
}
|
|
|
|
// PeekAllSet returns PeekAll as a set.
|
|
func (q *EvictingStringQueue) PeekAllSet() map[string]bool {
|
|
all := q.PeekAll()
|
|
set := make(map[string]bool)
|
|
for _, v := range all {
|
|
set[v] = true
|
|
}
|
|
|
|
return set
|
|
}
|