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

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

# 罗勇刚(Yonggang Luo) (10 years ago)

For example:

function addWithType(a = int64(0), b = int64(0)) { return a + b; }

This function is input add two int64 and return a int64

# Benjamin Gruenaum (10 years ago)

What about non-default parameters?

# Jordan Harband (10 years ago)

What if I have an "int64" function already defined? Will this skip over my int64 function? What about a = something(0) - will that call my "something" function, whereas "int64" is special and skipped over?

# Edwin Reynoso (10 years ago)

Don't you just mean:

function addWithType(a=0, b=0) {
  return int64(a) + int64(b);
}
# Emanuel Allen (10 years ago)

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

# Caitlin Potter (10 years ago)

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.

# Gary Guo (10 years ago)

If strong type is needed, why not use TypeScript?

# 罗勇刚(Yonggang Luo) (10 years ago)

Well, native support means better performance for browsers.

2015-05-20 17:02 GMT+08:00 Gary Guo <nbdd0121 at hotmail.com>: