extension?Record{
????public?static?func?<(lhs:?Self,?rhs:?Self)?->?Bool{
????????return?lhs.winningPercent()?<?rhs.winningPercent()
????}

????public?static?func?==(lhs:?Self,?rhs:?Self)?->?Bool{
????????return?lhs.winningPercent()?==?rhs.winningPercent()
????}
}

用上一視頻使用的例子，創(chuàng)建了三個(gè)遵守Record這個(gè)協(xié)議的結(jié)構(gòu)體(BaseballRec, BasketballRec, FootballRec)

其三者的實(shí)例(baseTeamRec,basketTeamRec,footTeamRec)是可以兩兩互相比較的，也可以調(diào)用isPrizable()方法，但當(dāng)帶入函數(shù)topPrizable時(shí)

func topPrizable<T: Record & Prizable>(list: [T]) -> T?

topPrizable(list: [baseTeamRec, basketTeamRec, footTeamRec] )

會(huì)提示類型不匹配的錯(cuò)誤，要怎麼修正呢？

Cannot convert value of type '[Any]' to expected argument type '[_]'

且要將這些實(shí)例存成Record的Array也有錯(cuò)誤，提示Record Protocol 要成爲(wèi)類型得要用泛型約束，gameRecords的類型要如何定義呢？

let gameRecords: [Record] =?[baseTeamRec, basketTeamRec, footTeamRec]

Protocol 'Record' can only be used as a generic constraint because it has Self or associated type requirements