Home >Backend Development >Golang >Why Does fmt.Println Print 'bad error' Instead of 5 When an Integer Implements the Error Interface in Go?

Why Does fmt.Println Print 'bad error' Instead of 5 When an Integer Implements the Error Interface in Go?

Barbara Streisand
Barbara StreisandOriginal
2024-12-04 02:47:09489browse

Why Does fmt.Println Print

Golang Interface: Understanding the Transformation of an Integer to "Bad Error"

When working with interfaces in Golang, it's essential to comprehend the behavior of types that implement the Error() method. As exemplified in the given code snippet:

type T int

func (t T) Error() string {
    return "bad error"
}

func main() {
    var v interface{} = T(5)
    fmt.Println(v) //output: bad error, not 5
}

The intent is to define the Error() method for type T, expecting to print the value 5. However, the output surprisingly displays "bad error." To clarify this phenomenon, we must delve into the documentation of the fmt package:

"If an operand implements the error interface, the Error method will be invoked to convert the object to a string, which will then be formatted as required by the verb (if any)."

Additionally, "For each Printf-like function, there is also a Print function that takes no format and is equivalent to saying %v for every operand. Another variant Println inserts blanks between operands and appends a newline."

In this case, Println() is used without specifying a format, which is equivalent to using %v. Since T implements the error interface, the Error() method is invoked, converting the value to "bad error," which is subsequently printed.

To print the integer value, it's necessary to utilize a specific format specifier, such as:

fmt.Printf("%d", v) // Will print the integer value

The above is the detailed content of Why Does fmt.Println Print 'bad error' Instead of 5 When an Integer Implements the Error Interface in Go?. For more information, please follow other related articles on the PHP Chinese website!

Statement:
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