Why Doesn\'t My Go Regex Match?

Regex Not Working in Go: Understanding the Problem

Despite the apparent simplicity of a regular expression pattern, why does it fail to match in Go? Let's delve into the issue and discover the solution.

Original Code and Output

Consider the following Go code:

package main

import (
    "fmt"
    "regexp"
)

func main() {
    a := "parameter=0xFF"
    regex := "^.+=\b0x[A-F][A-F]\b$"
    result, err := regexp.MatchString(regex, a)
    fmt.Println(result, err)
}

Surprisingly, the output is false , indicating a failed match. This contradicts the expectation of a match based on the provided input and pattern.

Raw Strings to the Rescue
The key to resolving this issue lies in using a raw string literal in defining the regular expression pattern. A raw string literal is a sequence of characters enclosed in back quotes (`) instead of double quotes (").

By using a raw string literal, you prevent Go from interpreting backslashes as escape characters. This is crucial because the pattern contains escape sequences (b and x) that define specific matching conditions.

Updated Code

With the raw string literal, the code becomes:

var regex string = `^.+=\b0x[A-F][A-F]\b$`

Explanation of the Pattern

The updated pattern matches strings that meet the following criteria:

• Starts with any number of characters (. )
• Contains an equals sign (=)
• Followed by the prefix "0x" (b0x)
• Contains two hexadecimal characters ([A-F][A-F])
• Ends with a word boundary (b)

Conclusion

By using a raw string literal, we ensure that the regular expression pattern is interpreted correctly in Go. This allows the pattern to accurately match the desired input strings, as intended.

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