Split string along regex, but keep matches

Question

I want to split a string on a regular expresion, but preserve the matches.

I have tried splitting the string on a regex, but it throws away the matches. I have also tried using this, but I am not very good at translating code from language to language, let alone C#.

re := regexp.MustCompile(`\d`)
array := re.Split("ab1cd2ef3", -1)

I need the value of array to be ["ab", "1", "cd", "2", "ef", "3"], but the value of array is ["ab", "cd", "ef"]. No errors.

Answer 1

The kind of regex support in the link you have pointed out is NOT available in Go regex package. You can read the related discussion.

What you want to achieve (as per the sample given) can be done using regex to match digits or non-digits.

package main

import (
    "fmt"
    "regexp"
)

func main() {
    str := "ab1cd2ef3"
    r := regexp.MustCompile(`(\d|[^\d]+)`)
    fmt.Println(r.FindAllStringSubmatch(str, -1))
}

Playground: https://play.golang.org/p/L-ElvkDky53

Output:

[[ab ab] [1 1] [cd cd] [2 2] [ef ef] [3 3]]