重写 Pop() 方法

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！

问题内容

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？

解决方法

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。

So old := *pq 的行为更像是别名，而不是数据复制。

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