Consider javascript already support for default parameters, so maybe we can use the default parameter to specify the strong type.

[Caitlin Potter]

How do you distinguish between `function f(a = defaultComesFromEvaluatingSomeFunction(true))` from `function f(a = StrongType(true))`?

It’s ambiguous which one is meant, and that doesn’t really work.

> On May 18, 2015, at 2:35 PM, Emanuel Allen <emanuelallen at hotmail.com> wrote:
> 
> I like to use what the language already have so if we can denote the type by constructor:
> 
> Function foo(a=Number(64),b=String()){
>  return b+a;
> }
> 
> Sent from my iPhone
> 
> On May 18, 2015, at 2:23 PM, Edwin Reynoso <eorroe at gmail.com <mailto:eorroe at gmail.com>> wrote:
> 
>> Don't you just mean:
>> 
>> ```
>> function addWithType(a=0, b=0) {
>>   return int64(a) + int64(b);
>> }
>> ```
>> 
>> On Mon, May 18, 2015 at 2:16 PM, Benjamin Gruenaum <benjamingr at gmail.com <mailto:benjamingr at gmail.com>> wrote:
>> What about non-default parameters?
>> 
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