php小编香蕉今天为大家介绍如何重写 Pop() 方法。在编程中,Pop() 方法用于删除并返回数组的最后一个元素。然而,有时我们需要对Pop()方法进行自定义,以满足特定需求。通过重写Pop()方法,我们可以添加额外的逻辑或修改返回的元素,从而更好地适应我们的代码。本文将详细介绍如何重写Pop()方法,并给出一些实例来帮助理解。让我们开始吧!
在go的安装下,他们在container/heap/example_pq_test.go
中有一个优先级队列的示例
我粘贴整个文件的内容,以便我可以询问 pop() 方法。
// copyright 2012 the go authors. all rights reserved. // use of this source code is governed by a bsd-style // license that can be found in the license file. // this example demonstrates a priority queue built using the heap interface. package heap_test import ( "container/heap" "fmt" ) // an item is something we manage in a priority queue. type item struct { value string // the value of the item; arbitrary. priority int // the priority of the item in the queue. // the index is needed by update and is maintained by the heap.interface methods. index int // the index of the item in the heap. } // a priorityqueue implements heap.interface and holds items. type priorityqueue []*item func (pq priorityqueue) len() int { return len(pq) } func (pq priorityqueue) less(i, j int) bool { // we want pop to give us the highest, not lowest, priority so we use greater than here. return pq[i].priority > pq[j].priority } func (pq priorityqueue) swap(i, j int) { pq[i], pq[j] = pq[j], pq[i] pq[i].index = i pq[j].index = j } func (pq *priorityqueue) push(x any) { n := len(*pq) item := x.(*item) item.index = n *pq = append(*pq, item) } func (pq *priorityqueue) pop() any { old := *pq n := len(old) item := old[n-1] old[n-1] = nil // avoid memory leak item.index = -1 // for safety *pq = old[0 : n-1] return item } // update modifies the priority and value of an item in the queue. func (pq *priorityqueue) update(item *item, value string, priority int) { item.value = value item.priority = priority heap.fix(pq, item.index) } // this example creates a priorityqueue with some items, adds and manipulates an item, // and then removes the items in priority order. func example_priorityqueue() { // some items and their priorities. items := map[string]int{ "banana": 3, "apple": 2, "pear": 4, } // create a priority queue, put the items in it, and // establish the priority queue (heap) invariants. pq := make(priorityqueue, len(items)) i := 0 for value, priority := range items { pq[i] = &item{ value: value, priority: priority, index: i, } i++ } heap.init(&pq) // insert a new item and then modify its priority. item := &item{ value: "orange", priority: 1, } heap.push(&pq, item) pq.update(item, item.value, 5) // take the items out; they arrive in decreasing priority order. for pq.len() > 0 { item := heap.pop(&pq).(*item) fmt.printf("%.2d:%s ", item.priority, item.value) } // output: // 05:orange 04:pear 03:banana 02:apple }
如果我有如下的 pop() 方法(不创建原始切片的深层副本),可能会带来什么危害或者是否存在谬误
func (pq *PriorityQueue) Pop2() any { n := len(*pq) item := (*pq)[n-1] (*pq)[n-1] = nil // avoid memory leak item.index = -1 // for safety *pq = (*pq)[: n-1] return item }
我相信原始的 pop()
方法,这一行为切片 old := *pq
创建一个深层副本(分配一个新的底层数组)。这是真的吗?pop()
方法,这一行为切片 old := *pq
创建一个深层副本(分配一个新的底层数组)。这是真的吗?
make
函数创建的对象,这里是map
和slice
So old := *pq
make
函数创建的对象,这里是map
和slice
,更像是指向数据位置的指针,而不是数据本身。
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