php小編香蕉今天要為大家介紹如何重寫 Pop() 方法。在程式設計中，Pop() 方法用於刪除並傳回數組的最後一個元素。然而，有時我們需要對Pop()方法進行自訂，以滿足特定需求。透過重寫Pop()方法，我們可以加入額外的邏輯或修改傳回的元素，從而更好地適應我們的程式碼。本文將詳細介紹如何重寫Pop()方法，並給予一些例子來幫助理解。讓我們開始吧！
在go的安裝下，他們在container/heap/example_pq_test.go中有一個優先權佇列的範例
我貼上整個文件的內容，以便我可以詢問 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
}如果我有以下的 pop() 方法（不建立原始切片的深層副本），可能會帶來什麼危害或是否存在謬誤
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，這一行為切片 old := *pq 建立一個深層副本（分配一個新的底層陣列）。這是真的嗎？
make函數建立的對象，這裡是map和slice，更像是指向資料位置的指針，而不是數據本身。
So old := *pq 的行為更像是別名，而不是資料複製。
以上是重寫 Pop() 方法的詳細內容。更多資訊請關注PHP中文網其他相關文章！
