Split string using regular expression in Go

I made a regex-split function based on the behavior of regex split function in java, c#, php.... It returns only an array of strings, without the index information.

func RegSplit(text string, delimeter string) []string {
    reg := regexp.MustCompile(delimeter)
    indexes := reg.FindAllStringIndex(text, -1)
    laststart := 0
    result := make([]string, len(indexes) + 1)
    for i, element := range indexes {
            result[i] = text[laststart:element[0]]
            laststart = element[1]
    }
    result[len(indexes)] = text[laststart:len(text)]
    return result
}

example:

fmt.Println(RegSplit("a1b22c333d", "[0-9]+"))

result:

[a b c d]

You can use regexp.Split to split a string into a slice of strings with the regex pattern as the delimiter.

package main

import (
    "fmt"
    "regexp"
)

func main() {
    re := regexp.MustCompile("[0-9]+")
    txt := "Have9834a908123great10891819081day!"

    split := re.Split(txt, -1)
    set := []string{}

    for i := range split {
        set = append(set, split[i])
    }

    fmt.Println(set) // ["Have", "a", "great", "day!"]
}