Hi Jesse
Thanks for the response.
Got it.
On Sun, May 17, 2020 at 7:52 AM Jesse McNelis <jes...@jessta.id.au> wrote:
> 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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>> .
>>
>
>
> --
> =====================
> http://jessta.id.au
>
>
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