Hi,
You can simply do:
func P(args ...interface{}) {
print("\n", args...)
}
The problem you encountered was that you were passing the args as a single
argument to print()
The 'var args' syntax collects the arguments and puts them in a slice. So
args in P() is a []interface{}.
You then pass that whole []interface{} as an argument to print(), for args
in print() gets a []interface{} where the first item in the slice is an
interface{} that contains a []interface{}, which isn't what you wanted.
By using the spread operator when passing the args slice to print(), you
get the pass the args slices multiple arguments to print().
On Sun, May 17, 2020 at 12:06 PM Amarjeet Anand <
amarjeetanandsi...@gmail.com> wrote:
> Hi
>
> Why does the type of a parameter passed to another function changes?
> For example:
>
> func P(args ...interface{}) {
> print("\n", args)
> }
>
> func print(seperator string, args ...interface{}) {
> for _, arg := range args {
> fmt.Println("arg type: ", reflect.TypeOf(arg))
>
> str := ""
> switch v := arg.(type) {
> case byte:
> str = string(v)
> case rune:
> str = string(v)
> case int:
> str = strconv.Itoa(v)
> case int64:
> str = strconv.FormatInt(v, 10)
> case string:
> str = v
> }
> out.WriteString(str + seperator)
> }
> }
>
>
> Calling *P("one"[2]) *gives type as *[]interface {}*
> where as directly calling *print("one"[2]) *gives type as *byte*
>
> *Why is it like this? *
> *And any way i can achieve this without putting switch in P() function?*
>
>
>
>
>
>
>
>
>
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>
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