数组扩展以按值删除对象

问题描述

extension Array {
    func removeObject<T where T : Equatable>(object: T) {
        var index = find(self, object)
        self.removeAtIndex(index)
    }
}

但是，我在 var index = find(self, object)

'T' 不能转换为 'T'

我也尝试过这个方法签名:func removeObject(object: AnyObject)，但是，我得到了同样的错误:

I also tried with this method signature: func removeObject(object: AnyObject), however, I get the same error:

'AnyObject' 不能转换为 'T'

这样做的正确方法是什么?

What is the proper way to do this?

推荐答案

从 Swift 2 开始，这可以通过协议扩展方法来实现.removeObject() 被定义为所有符合类型的方法到 RangeReplaceableCollectionType(特别是在 Array 上)如果集合的元素是 Equatable:

As of Swift 2, this can be achieved with a protocol extension method.removeObject() is defined as a method on all types conformingto RangeReplaceableCollectionType (in particular on Array) ifthe elements of the collection are Equatable:

extension RangeReplaceableCollectionType where Generator.Element : Equatable {

    // Remove first collection element that is equal to the given `object`:
    mutating func removeObject(object : Generator.Element) {
        if let index = self.indexOf(object) {
            self.removeAtIndex(index)
        }
    }
}

示例:

var ar = [1, 2, 3, 2]
ar.removeObject(2)
print(ar) // [1, 3, 2]

Swift 2/Xcode 7 beta 2 的更新: 正如 Airspeed Velocity 所注意到的在评论中，现在实际上可以在对模板有更多限制的泛型类型上编写方法，因此该方法现在实际上可以定义为 Array 的扩展:

Update for Swift 2 / Xcode 7 beta 2: As Airspeed Velocity noticedin the comments, it is now actually possible to write a method on a generic type that is more restrictive on the template, so the methodcould now actually be defined as an extension of Array:

extension Array where Element : Equatable {

    // ... same method as above ...
}

协议扩展仍然具有适用于更大的一组类型.

The protocol extension still has the advantage of being applicable toa larger set of types.

Swift 3 的更新:

extension Array where Element: Equatable {

    // Remove first collection element that is equal to the given `object`:
    mutating func remove(object: Element) {
        if let index = index(of: object) {
            remove(at: index)
        }
    }
}

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