Swift init（count：，repeatedValue :)是否有效？

Tested this from the reference: https://developer.apple.com/documentation/swift

从参考资料中测试过：https：//developer.apple.com/documentation/swift

var string = String(count: 5, repeatedValue: "a")
// string is "aaaaa"

I got this error:

我收到了这个错误：

Playground execution failed: error: :5:14: error: could not find an overload for 'init' that accepts the supplied arguments var string = String(count: 5, repeatedValue: "a")

Playground执行失败：错误:: 5：14：错误：无法找到接受提供的参数的'init'的重载var string = String（count：5，repeatedValue：“a”）

Does this actually work?

这实际上有用吗？

6 个解决方案

#1

It seems that you have to explicitly pass in a Character type to it to function. This works for me.

看来你必须明确传入一个Character类型才能运行。这对我有用。

let char = Character("a")
let string = String(count: 5, repeatedValue: char)

Although, there may be bug mixed in with all this as well. I believe the way you were doing this should have worked on its own. And I can't seem to get code completion on this initializer at all.

虽然，所有这些都可能有混杂的错误。我相信你这样做的方式应该是独立的。而且我似乎无法在此初始化程序上完成代码完成。

Edit: I'm going with bug. The following compiles just fine.

编辑：我要跟bug一起去。以下编译就好了。

let array = Array(count: 5, repeatedValue: "a")

#2

For the benefit of future searchers: as of Swift 3, use init(repeating:count:).

为了未来的搜索者的利益：从Swift 3开始，使用init（重复：计数:)。

#3

This works just fine :

这很好用：

var str9 = String(count: 5,repeatedValue: Character("c"))

#4

For anyone in swift 3.x its now something like this this will work like a charm.

对于swift 3.x中的任何人来说，现在这样的东西就像魅力一样。

var string = String(repeating: "a", count: 5)

#5

I know this is an old question and already has an answer. However I think I know why String(count: 5, repeatedValue: "a") does not work.

我知道这是一个老问题，已经有了答案。但是我想我知道为什么String（计数：5，repeatedValue：“a”）不起作用。

The thing is String has two similar looking initialisers:

问题是String有两个类似的初始化器：

init(count: Int, repeatedValue: Character)
init(count: Int, repeatedValue: UnicodeScalar)

So in this case compiler can't tell whether a literal is a Character or UnicodeScalar, hence compile time error if you don't pass explicit Character. To confirm that "a" can be interpreted as UnicodeScalar you can check that this line compiles:

因此，在这种情况下，编译器无法判断文字是字符还是UnicodeScalar，因此如果不传递显式字符，则编译时错误。要确认“a”可以解释为UnicodeScalar，您可以检查此行是否编译：

let a: UnicodeScalar = "a"

#6

Swift 3:

斯威夫特3：

var array = Array(repeating: 0, count: 5)

Output: [0, 0, 0, 0, 0]

输出：[0,0,0,0,0]

#1