切片struct！=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.

要获取模型实例，我具有以下帮助函数:

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{}
}

以上方法使我可以返回类型正确的Person实例(以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).

当我尝试执行类似的操作以返回模型切片时，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

我的目标是返回任何要求的模型类型的切片([]Person，[]FutureModel，[]Terminator2000，w/e).我缺少什么，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

这与我刚刚回答的问题非常相似: https: //stackoverflow.com/a/12990540/727643

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

简短的回答是您是正确的.一片结构不等于该结构实现的接口的一部分.

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

A []Person和[]Model具有不同的内存布局.这是因为它们是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为两个字.一个单词代表类型信息，另一个单词代表数据. Person是一个结构，其大小取决于它包含的字段.为了从[]Person转换为[]Model，您将需要遍历数组并为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.

由于此转换是O(n)操作，并且会导致创建新的切片，所以Go拒绝隐式地执行此操作.您可以使用以下代码明确地做到这一点.

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

并且正如 dskinner指出的，您很可能需要切片指针而不是切片指针.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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