2008/2/29 Brendan Eich <[EMAIL PROTECTED]>:
> On Feb 28, 2008, at 7:29 AM, Lars Hansen wrote:
>
> > In fact the RI doesn't work as one would like it to, as the third test
> > below shows:
> >
> >>> function h({x:x, y:y} = { x:10, y:20 }) [x,y]
> >>> h({x:1,y:2})
> > 1,2
> >>> h()
> > 10,20
> >>> h({x:1})
> > 1,
> >
> > What appears to happen here is what one would expect: if the entire
> > structure argument is omitted then the default value is used, but
> > if the
> > structure is partly omitted then the default structure is not
> > consulted
> > at all, and I think that's what Steven would like to see happen.
>
> Confusion: Steven, do you want to see ES4 do what the RI is doing? I
> thought you were looking for "member-wise" defaulting, so that the
> last call, to h({x:1}), would return [1, 20].
My initial concern was 'native' support for optional named arguments.
In the absence of that, I'd indeed like 'member-wise' defaulting.
However, I'm not sure how this affects the language.
Firstly, I can see 2 interpretations of how default values would work
from the end-programmer's perspective:
Step 1. SET arg = default value
Step 2. SET arg = specified value
or:
Step 1. IF no specified value, THEN SET arg = default value
Secondly, destructuring currently works as follows:
>> ({x:a,y:b} = {x:1,y:2}); [a,b]
1,2
>> ({x:a,y:b} = {x:1}); [a,b]
1,
>> typeof(b)
undefined
>>
Suppose we took the first interpretation of default values and object
destructuring were changed so that omitted (or undefined) members
weren't assigned at all. Only then would a result in which calling
h({x:1}) returns [1, 20] be consistent with the rest of the language.
A third interpretation would involve treating destructuring defaults
differently to other defaults, but I don't like that kind of special
casing (from an end-programmer's perspective).
As mentioned, an alternative is an operator for defaults in
destructuring. Lars, I assume you meant by this the following (in
shorthand):
function h({x ?= 10, y ?= 20}) [x,y]
or in longhand in a statement:
({x:x ?= 10, y:y ?= 20} = {x: 1})
Of the alternatives, I'd probably be happiest with this kind of operator.
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