结构切片！=它实现的接口切片? [英] slice of struct != slice of interface it implements?

问题描述

我有一个接口Model，由struct Person实现.

I have an interface Model, which is implemented by struct Person.

为了获得模型实例，我有以下辅助函数:

To get a model instance, I have the following helper functions:

func newModel(c string) Model {
    switch c {
    case "person":
        return newPerson()
    }
    return nil
}

func newPerson() *Person {
    return &Person{}
}

上述方法允许我返回一个正确类型的 Person 实例(稍后可以使用相同的方法轻松添加新模型).

The above approach allows me to return a properly typed Person instance (can easily add new models later with same approach).

当我尝试为返回模型切片执行类似操作时，出现错误.代码:

When I attempted to do something similar for returning a slice of models, I get an error. Code:

func newModels(c string) []Model {
    switch c {
    case "person":
        return newPersons()
    }
    return nil
}

func newPersons() *[]Person {
    var models []Person
    return &models
}

Go 抱怨:cannot use newPersons() (type []Person) as type []Model in return argument

我的目标是返回请求的任何模型类型的切片(无论是[]Person、[]FutureModel、[]Terminator2000， 我们).我错过了什么，我该如何正确实施这样的解决方案?

My goal is to return a slice of whatever model type is requested (whether []Person, []FutureModel, []Terminator2000, w/e). What am I missing, and how can I properly implement such a solution?

推荐答案

这和我刚刚回答的一个问题非常相似:https://stackoverflow.com/a/12990540/727643

This is very similar to a question I just answered: https://stackoverflow.com/a/12990540/727643

简短的回答是你是对的.结构的切片不等于结构实现的接口的切片.

The short answer is that you are correct. A slice of structs is not equal to a slice of an interface the struct implements.

[]Person 和 []Model 具有不同的内存布局.这是因为它们切片的类型具有不同的内存布局.Model 是一个接口值，这意味着在内存中它的大小是两个字.一个词表示类型信息，另一个词表示数据.Person 是一个结构体，其大小取决于它包含的字段.为了从 []Person 转换为 []Model，您需要遍历数组并对每个元素进行类型转换.

A []Person and a []Model have different memory layouts. This is because the types they are slices of have different memory layouts. A Model is an interface value which means that in memory it is two words in size. One word for the type information, the other for the data. A Person is a struct whose size depends on the fields it contains. In order to convert from a []Person to a []Model, you will need to loop over the array and do a type conversion for each element.

由于这个转换是一个 O(n) 操作并且会导致一个新的切片被创建，Go 拒绝隐式这样做.您可以使用以下代码显式执行此操作.

Since this conversion is an O(n) operation and would result in a new slice being created, Go refuses to do it implicitly. You can do it explicitly with the following code.

models := make([]Model, len(persons))
for i, v := range persons {
    models[i] = Model(v)
}
return models

正如 dskinner 指出的，您很可能需要一个指针切片，而不是指向切片的指针.通常不需要指向切片的指针.

And as dskinner pointed out, you most likely want a slice of pointers and not a pointer to a slice. A pointer to a slice is not normally needed.

*[]Person        // pointer to slice
[]*Person        // slice of pointers

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