工作一个bytes.Buffer而无需使用字符串，strvconv等

Question

I would simply like to do this without resorting to strconv & strings, but I am not proficient working in bytes only:

func rangeSeq(b *bytes.Buffer) ([][]byte, bool) {
    q := bytes.Split(b.Bytes(), []byte{SEQ_RANGE})
    if len(q) == 2 {
        var ret [][]byte
        il, lt := string(q[0]), string(q[1])
        initial, err := strconv.ParseInt(il, 10, 64)
        last, err := strconv.ParseInt(lt, 10, 64)
        if err == nil {
            if initial < last {
                for i := initial; i <= last; i++ {
                    out := strconv.AppendInt([]byte{}, i, 10)
                    ret = append(ret, out)
                }
            }
            return ret, true
        }
    }
    return nil, false
}

suggestions?

Answer 1

There is no []byte equivalent to the strconv.Parse* functions (see issue 2632). Using strconv is currently the easiest way to handle this.

You are ignoring the first error however, which is a bug. You can also shorten a couple things, and use the more common idiom of returning early instead of increasing indentation. I would also return an error, instead of a bool for more contextual information.

func rangeSeq(b *bytes.Buffer) ([][]byte, error) {
    q := bytes.Split(b.Bytes(), sep)
    if len(q) != 2 {
        return nil, fmt.Errorf("invalid value: %s", b)
    }

    var ret [][]byte

    initial, err := strconv.Atoi(string(q[0]))
    if err != nil {
        return nil, err
    }

    last, err := strconv.Atoi(string(q[1]))
    if err != nil {
        return nil, err
    }

    if initial < last {
        for i := initial; i <= last; i++ {
            ret = append(ret, strconv.AppendInt(nil, i, 10))
        }
    }
    return ret, nil
}