恐慌:接口转换:*不是接口X:缺少方法xxx

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英文:

Panic: interface conversion: * is not inteface X: missing method xxx

问题

我正在为一个项目工作,该项目从Kafka消费消息,使用MessagePack或JSON格式进行解包,构建SQL并将其插入TDengine数据库。

接口定义如下:

type TaosEncoder interface {
	TaosDatabase() string
	TaosSTable() string
	TaosTable() string
	TaosTags() []interface{}
	TaosCols() []string
	TaosValues() []interface{}
}

对于一个名为Record的对象:

type Record struct {
	DeviceID  string        `json:"deviceID"` // 用于表名
	Timestamp time.Time     `json:"timestamp"`
	Cols      []interface{} `json:"cols"`
}

实现该接口:

func (r *Record) TaosCols() []string {
	var tags []string
	return tags
}
// 其他方法也要实现

一个方法将使用接口方法:

func ToTaosBatchInsertSql(l []interface{}) (string, error) {
	records := make([]string, len(l))
	for i, t := range l {
		// type TaosEncoderT = *TaosEncoder
		record, err := ToTaosRecordSql(t.(TaosEncoder))
		if err != nil {
			return "", err
		}
		records[i] = record
	}
	return fmt.Sprintf("insert into %s", strings.Join(records, " ")), nil
}

使用方法如下:

records := make([]interface{}, size)
for i := 0; i < size; i++ {
	item := <-q.queue // 一个Record类型的通道,定义为:queue chan interface{}
	records[i] = item
}
sqlStr, err := utils.ToTaosBatchInsertSql(records)

它可以编译通过,但在运行时失败,如下所示:

panic: interface conversion: record.Record is not utils.TaosEncoder: missing method TaosCols

goroutine 71 [running]:
xxx/pkg/utils.ToTaosBatchInsertSql(0xc000130c80, 0xc8, 0xc8, 0xc8, 0x0, 0x0, 0x0)

但是,当我更改Record的实现 - 移除*时:

func (r Record) TaosCols() []string {
	var tags []string
	return tags
}

然后它就可以正常工作了。

我被告知在接口实现中使用*,所以问题是:

  1. 使用func (r Record) xxx实现接口是否可以?
  2. 如果不行,我该怎么办?
英文:

I'm working on a project that consume messages from Kafka, unmushal using messagepack or json format, build sql of it and insert them into TDengine database.

The inteface defined as:

type TaosEncoder interface {
	TaosDatabase() string
	TaosSTable() string
	TaosTable() string
	TaosTags() []interface{}
	TaosCols() []string
	TaosValues() []interface{}
}

For an object called Record:

type Record struct {
	DeviceID  string        `json:&quot;deviceID&quot;` // for table name
	Timestamp time.Time     `json:&quot;timestamp&quot;`
	Cols      []interface{} `json:&quot;cols&quot;`
}

Implement the interface:

func (r *Record) TaosCols() []string {
	var tags []string
	return tags
}
// other methods are implemented too

A method will use the interface methods:

func ToTaosBatchInsertSql(l []interface{}) (string, error) {
	records := make([]string, len(l))
	for i, t := range l {
		// type TaosEncoderT = *TaosEncoder
		record, err := ToTaosRecordSql(t.(TaosEncoder))
		if err != nil {
			return &quot;&quot;, err
		}
		records[i] = record
	}
	return fmt.Sprintf(&quot;insert into %s&quot;, strings.Join(records, &quot; &quot;)), nil
}

Use it as this:

records := make([]interface{}, size)
for i := 0; i &lt; size; i++ {
	item := &lt;-q.queue # an channel of Record, defined as: queue chan interface{}
	records[i] = item
}
sqlStr, err := utils.ToTaosBatchInsertSql(records)

It compiles ok, but failed while running like this:

panic: interface conversion: record.Record is not utils.TaosEncoder: missing method TaosCols

goroutine 71 [running]:
xxx/pkg/utils.ToTaosBatchInsertSql(0xc000130c80, 0xc8, 0xc8, 0xc8, 0x0, 0x0, 0x0)

But when I change the implementations of Record - remove *

func (r Record) TaosCols() []string {
	var tags []string
	return tags
}

Then it just works.

I've told to use * in interface implements, so the question is:

  1. Is it ok to implement interface with func (r Record) xxx?
  2. If not, what can I do?

答案1

得分: 2

这个错误的原因与类型的方法集有关。

假设有一个类型 T,它有一些接收者类型为 T 的方法,还有一些接收者类型为 *T 的方法。

那么,如果一个变量的类型是 T,它可以使用所有接收者类型为 T 的方法,但不能使用接收者类型为 *T 的方法

如果一个变量的类型是 *T,它可以使用两组方法,一组是接收者类型为 T 的方法,另一组是接收者类型为 *T 的方法。

在你的情况下,类型 Record 没有 TaosCols() 方法,因此它没有完全实现接口 TaosEncoder

如果将类型为 Record 的变量转换为 *Record 类型的变量,它就可以使用所有的方法,并且可以实现该接口。

Method Sets in: Frequently Asked Questions (FAQ) - The Go Programming Language

Method Sets in: The Go Programming Language Specification - The Go Programming Language

英文:

The cause of this error is related to the method sets of a type.

Assume a type T that has methods with a receiver type of T and some more methods with a receiver type of *T.

Then, if a variable is of type T, it has all methods available that have a pointer receiver of T but none of the methods with receiver type *T.

If a variable is of type *T, it has both sets of methods available, those with receiver type T, and those with receiver type *T.

In your case, type Record has no TaosCols() method, and so it does not implement interface TaosEncoder completely.

If you turn the variable of type Record into a *Record variable, it has all methods available and then would implement the interface.

Method Sets in: Frequently Asked Questions (FAQ) - The Go Programming Language

Method Sets in: The Go Programming Language Specification - The Go Programming Language

答案2

得分: 0

一个 Record 类型和 *Record 类型不同。如果你使用指针接收器来定义实现接口的方法,那么只有 *Record 类型才实现了 TaosEncoder 接口。

根据这个代码片段:

item := <-q.queue // 一个 Record 类型的通道

看起来你需要发送一个 *Record 类型的值到 q.queue,而不是一个 Record 类型的值。

英文:

A Record is not the same type as a *Record. If you use a pointer receiver for defining the methods that implement the interface, then only *Record implements TaosEncoder.

Based on this:

item := &lt;-q.queue # an channel of Record

it looks like you will need to send a value of *Record on q.queue, rather than a Record.

huangapple
  • 本文由 发表于 2021年7月16日 18:08:29
  • 转载请务必保留本文链接:https://go.coder-hub.com/68407216.html
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