Golang地址运算符和（* int）（Type）语法

Question

Starting to play around with golang and was looking at a custom json.Unmarshal. In a blog post the had the following:

type FlexInt int

func (fi *FlexInt) UnmarshalJSON(data []byte) error {
    if data[0] != '"' {
        return json.Unmarshal(data, (*int)(fi))
    }
    var s string
    if err := json.Unmarshal(data, &s); err != nil {
        return err
    }
    i, err := strconv.Atoi(s)
    if err != nil {
        return err
    }
    *fi = FlexInt(i)
    return nil
}

And I understand what it is doing - but I dont understand (*int)(fi) part. Looking at the value of the fi pointer it is the same as (*int)(fi) - (*int)(fi) == fi. Yet when I change that line to simply fi it then does an infinite loop

Answer 1

The expression converts fi to an *int. The result contains the same address, but has a different type.

If a *FlexInt is passed to json.Unmarshal, then json.Unmarshal will call the *FlexInt.UnmarshalJSON method which calls json.Unmarshal and so on.