Swift 无法使用类型为“([Score]，Score)"的参数列表调用“find"，其中 Score 是一个结构

Swift Cannot invoke #39;find#39; with an argument list of type #39;([Score], Score)#39; where Score is a struct(Swift 无法使用类型为“([Score]，Score)的参数列表调用“find，其中 Score 是一个结构)

问题描述

虽然 find(["a", "b"], "c") 没有问题，但在尝试在结构数组中查找结构的索引时出现错误:

While find(["a", "b"], "c") works with no problems, I get an error when trying to find the index of a structure inside an array of structures:

struct Score
{
    //...
}

var scores: [Score] = //...
var score: Score = //...

find(self.scores, score) // Error: Cannot invoke 'find' with an argument list of type '([Score], Score)'

虽然默认情况下无法相互比较的结构可能是个问题.但是将 Score 的定义改为 class给我同样的错误.

I though it could be a problem with structures that can't be compared to each other by default. But changing Scores definition to class gives me the same error.

推荐答案

从 Swift 2.0 开始，现在有一个内置版本的 find 需要一个闭包，所以你不需要必须自己编写——而且，find 已重命名为 indexOf，现在是 CollectionType 的协议扩展，所以你可以这样称呼它一种方法:

as of Swift 2.0, there is now a built-in version of find that takes a closure so you don’t have to write your own – but also, find has been renamed indexOf, and is now a protocol extension of CollectionType, so you call it like a method:

// if you make `Score` conform to `Equatable:
if let idx = self.scores.indexOf(score) {

}

// or if you don't make it Equatable, you can just use a closure:
// (see original answer below for why you might prefer to do this)
if let idx = scores.indexOf({$0.scoreval == 3}) {

}

下面是 2.0 之前的原始答案

Original pre-2.0 answer below

虽然建议使您的类 Equatable 的答案可能会很好地工作，但我建议您在选择这样做之前要小心.原因是，正如文档所述，等式意味着可替代性，并且您的 == 运算符必须是自反的、对称的和可传递的.如果你不遵守这一点，在使用 equals、sort 等算法时可能会出现一些非常奇怪的行为.在实现 Equatable 在非最终类上.如果你确定你能满足要求，那就去做吧，find 就可以了.

func find<C: CollectionType>(source: C, match: C.Generator.Element -> Bool) -> C.Index {
    for idx in indices(source) {
        if match(source[idx]) { return idx }
    }
    return nil
}

let idx = find(scores) { $0 === $1 }