在Swift中将新元素插入数组 [英] Insert a new element into an array in Swift

问题描述

代码:

let oldNums: [Int] = [1, 2, 3, 4, 5 ,6 , 7, 8, 9, 10]
var newArray = oldNums[1..<4]
newArray.insert(99, atIndex: 0) // <-- crash here
newArray.insert(99, atIndex: 1) // <-- work very well

我感谢newArray是一个新的可变变量.所以我感到困惑. 为什么?我无法在"newArray"中插入新元素

I thank the newArray is a new mutable variable. So I get confuse. Why? I can't insert a new element into "newArray"

推荐答案

oldNums[1..<4]不是数组，而是

oldNums[1..<4] is not an array, but an ArraySlice:

类似Array的类型，代表任何Array，ContiguousArray或其他ArraySlice的子序列.

An Array-like type that represents a sub-sequence of any Array, ContiguousArray, or other ArraySlice.

数组切片的索引不是从零开始的，而是对应的 到原始数组的索引.这是一个改变 随Swift 2一起提供，并记录在

The indices of array slices are not zero-based, but correspond to the indices of the original array. This is a change that came with Swift 2 and is documented in the Xcode 7.0 Release notes:

为了保持一致性并更好地编写通用代码，ArraySlice 索引不再总是从零开始，而是直接映射到 他们正在切片的集合的索引并维护该映射 即使发生了突变.

For consistency and better composition of generic code, ArraySlice indices are no longer always zero-based but map directly onto the indices of the collection they are slicing and maintain that mapping even after mutations.

在您的情况下:

let oldNums: [Int] = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
var newArray = oldNums[1..<4]
print(newArray.indices)
// 1..<4

所以0是insert()的无效索引，这就是为什么

So 0 is an invalid index for insert(), and that's why

newArray.insert(99, atIndex: 0)

崩溃.要在切片的开头插入元素，您可以 可以使用

crashes. To insert an element at the beginning of the slice, you can use

newArray.insert(99, atIndex: newArray.startIndex)

要创建真实"数组而不是切片，请使用Array() 构造函数:

To create a "real" array instead of a slice, use the Array() constructor:

let oldNums: [Int] = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
var newArray = Array(oldNums[1..<4])
print(newArray.indices)
// 0..<3

newArray.insert(99, atIndex:0)
print(newArray)
// [99, 2, 3, 4]

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