泛型 – 使用Swift语言的自定义对象的泛型类型

我想知道有没有办法在下面的泛型函数中比较两个泛型类型实例与==运算符：

func compare<T>(T a,T b) -> Bool {
    if a == b{
       // do something
       return true; 
    }else{
       // do another thing
       return false;
    }   
 }

这是我的自定义对象：

class MyObj{
  var id = 3
  var name: String?
}

来自Apple Developer Resources,

Not every type in Swift can be compared with the equal to operator (==). If you create your own class or structure to represent a complex data model,for example,then the meaning of “equal to” for that class or structure is not something that Swift can guess for you. Because of this,it is not possible to guarantee that this code will work for every possible type T,and an appropriate error is reported when you try to compile the code.

All is not lost,however. The Swift standard library defines a protocol called
Equatable,which requires any conforming type to implement the equal to operator (==)
and the not equal to operator (!=) to compare any two values of that type. All of
Swift’s standard types automatically support the Equatable protocol.

Any type that is Equatable can be used safely with the findindex function,because it is guaranteed to support the equal to operator. To express this fact,you write a type constraint of Equatable as part of the type parameter’s deFinition when you define the function:

func findindex<T: Equatable>(array: T[],valuetoFind: T) -> Int? {
    for (index,value) in enumerate(array) {
        if value == valuetoFind {
            return index
        }
    }
    return nil
}

以下是他们的文档中的示例,说明如何覆盖==

struct MyStruct: Equatable {
    var name = "Untitled"
}
func == (lhs: MyStruct,rhs: MyStruct) -> Bool {
    return lhs.name == rhs.name
}

let value1 = MyStruct()
var value2 = MyStruct()
let firstCheck = value1 == value2
// firstCheck is true

value2.name = "A New Name"
let secondCheck = value1 == value2
// secondCheck is false

在你的情况下你会这样做,

class MyObj{
  var id = 3
  var name: String?
}

func == (lhs: MyObj,rhs: MyObj) -> Bool {
    return lhs.id == rhs.id
}