什么是接口断言？ [英] What is an interface assertion?

问题描述

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type Logger interface {
    Debug(msg string, keyvals ...interface{}) error
    Info(msg string, keyvals ...interface{}) error
    Error(msg string, keyvals ...interface{}) error
}

type tmLogger struct {
    srcLogger kitlog.Logger
}

// Interface assertions
var _ Logger = (*tmLogger)(nil) // What is this?

// ... interface definition ...

这是什么接口断言？

推荐答案

它将具体类型的nil指针指定给接口类型的变量。这是一种常见的做法，用于证明具体类型满足接口 - 如果不满足，则该行不会编译，给出错误，即不能将具体类型分配给接口类型的变量以及原因。

It assigns a nil pointer to a concrete type to a variable of the interface type. This is a common practice to prove that the concrete type fulfills the interface - if it doesn't, that line won't compile, giving an error that the concrete type can't be assigned to a variable of the interface type and why.

正如@JimB所说，界面断言是由作者组成的术语。 Go没有这样的术语。具体来说，这是一个类型转换，转换 nil 指向 tmLogger 的指针，然后将键入的nil指针指定给空白标识符变量 Logger 。如果 * tmLogger 不满足 Logger ，则分配将无法编译;但是，在运行时，这不占用任何内存，因为它使用的是nil值。

As @JimB noted, "interface assertion" is a term made up by the author. Go has no such term. This is, specifically, a type conversion, converting nil to a pointer to tmLogger, then assigning the typed nil pointer to a blank identifier variable of the interface type Logger. If *tmLogger does not satisfy Logger, the assignment won't compile; but, at runtime, this takes no memory because it's using a nil value.

据推测，作者使用这个术语的单位测试意义上的断言比type assertionsense - 该行代码断言该类型实现了接口，如果没有，则该行将失败。

Presumably the author uses this term more in the unit-testing sense of "assertion" than the "type assertion" sense - that line of code asserts that the type implements the interface, and if it doesn't, the line will fail.

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