Why Don\'t Type Assertions Work for Named Types in Go?

Patricia Arquette
Release: 2024-11-05 04:41:01
Original
701 people have browsed it

Why Don't Type Assertions Work for Named Types in Go?

Type Assertions vs. Conversions in Go

In Go, named type assertions are used to test whether a value implements a given interface. However, this approach doesn't directly apply to named types that redefine pre-defined types with a new name.

Consider the following code:

<code class="go">type Answer string

func acceptMe(str string) {
    fmt.Println(str)
}

func main() {
    type Answer string
    var ans Answer = "hello"

    // Type assertion fails:
    acceptMe(ans.(string))

    // Conversion to underlying type succeeds:
    acceptMe(string(ans))
}</code>
Copy after login

The type assertion ans.(string) fails because it attempts to check if the Answer type implements the string interface, which it doesn't. Instead, Answer is a custom type that redefines string.

On the other hand, the conversion string(ans) succeeds because it directly converts the Answer value to its underlying string type. Since Answer is a named type for string, this conversion is always valid.

Why doesn't the type assertion work for named types?

Type assertions are designed for interfaces. An interface can have multiple implementations (known as concrete types), and type assertions are used to check whether a value implements that interface. However, a named type has a fixed underlying type, which is known at compile time. Therefore, type assertions are unnecessary and cannot be performed for named types.

When to use conversion instead of type assertion?

Use conversion when you want to convert a named type to its underlying type. This is useful when you need to pass a value to a function or interface that expects a specific type but can accept its underlying type.

The above is the detailed content of Why Don\'t Type Assertions Work for Named Types in Go?. For more information, please follow other related articles on the PHP Chinese website!

source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Latest Articles by Author
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template
About us Disclaimer Sitemap
php.cn:Public welfare online PHP training,Help PHP learners grow quickly!