如何(简洁地)从 Go 中的切片中删除第一个元素? [英] How do I (succinctly) remove the first element from a slice in Go?

问题描述

我在 Go 中构建了一个简单的队列.它使用内部切片来跟踪其元素.元素通过附加到切片被推送到队列中.我想通过删除 elements 中的第一个元素来实现 .Pop().

I've built a simple queue in Go. It uses an internal slice to keep track of its elements. Elements are pushed onto the queue by appending to the slice. I'd like to implement .Pop() by removing the first element in elements.

在许多其他语言中，弹出"列表的第一个元素是单行的，这让我相信我下面的实现是草率和冗长的.有没有更好的办法?

In many other languages, "popping" the first element of a list is a one-liner, which leads me to believe my implementation below is sloppy and verbose. Is there a better way?

type Queue struct {
    elements []interface{}
}

func (queue *Queue) Push(element interface{}) {
    queue.elements = append(queue.elements, element)
}

func (queue *Queue) Pop() interface{} {
    element := queue.elements[0]
    if len(queue.elements) > 1 {
        queue.elements = queue.elements[1:]
    } else {
        queue.elements = make([]interface{}, 0)
    }
    return element
}

请注意，如果 len(queue.elements) == 0，我希望 Queue 恐慌.我不检查边界并不是疏忽.

Please note that I wish for the Queue to panic if len(queue.elements) == 0. It's not an oversight that I don't check the bounds.

推荐答案

这些你试过了吗?

从队列中弹出

x, a = a[0], a[1:]

从栈中弹出

x, a = a[len(a)-1], a[:len(a)-1]

推

a = append(a, x)

来自:https://code.google.com/p/go-wiki/wiki/SliceTricks

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