检查字符串是否以十进制数字开头？

It looks the following works, is it a good approach?

var thestr = "192.168.0.1"
if (thestr[0]>= '0' && thestr[0] <= '9'){
    //...
}

Since you are comparing by character and no characters are between 1 and 9, I think your solution is ok, but it does not account for the other numbers following.

For example, if thestr was "192.something.invalid" it's no longer an IP.

I'd recommend using a regular expression to check the IP.

something like

\b(?:\d{1,3}\.){3}\d{1,3}\b

Please do not use regexps for that simple task :)

What I would change in this case:

• add check for empty string before checking for the first rune
• I would rephrase it as "starts with a digit" as the number semantic is too broad. .5e-45 is a number, but probably it is not what you want. 0's semantic is also not simple: https://math.stackexchange.com/questions/238737/why-do-some-people-state-that-zero-is-not-a-number

Your solution is completely fine.

But note that strings in Go are stored as a read-only byte slice where the bytes are the UTF-8 encoded byte sequence, and indexing a string indexes its bytes, not its runes (characters). But since a decimal digit ('0'..'9') has exactly one byte, it is ok in this case to test the first byte, but first you should test if len(s) > 0 or s != "".

Here are some other alternatives, try all on the Go Playground:

1) Testing the byte range:

This is your solution, probably the fastest one:

s := "12asdf"
fmt.Println(s[0] >= '0' && s[0] <= '9')

2) Using fmt.Sscanf():

Note: this also accepts if the string starts with a negative number, decide if it is a problem for you or not (e.g. accepts "-12asf").

i := 0
n, err := fmt.Sscanf(s, "%d", &i)
fmt.Println(n > 0, err == nil) // Both n and err can be used to test

3) Using unicode.IsDigit():

fmt.Println(unicode.IsDigit(rune(s[0])))

4) Using regexp:

I would probably never use this as this is by far the slowest, but here it is:

r := regexp.MustCompile(`^\d`)
fmt.Println(r.FindString(s) != "")

Or:

r := regexp.MustCompile(`^\d.*`)
fmt.Println(r.MatchString(s))