What to do about default function arguments

[Don Clugston]

On 26/04/12 11:21, Timon Gehr wrote:
> On 04/26/2012 09:54 AM, Walter Bright wrote:
>> On 4/26/2012 12:47 AM, Timon Gehr wrote:
>>> On 04/26/2012 05:44 AM, Walter Bright wrote:
>>>> A subtle but nasty problem - are default arguments part of the type, or
>>>> part of the declaration?
>>>>
>>>> See http://d.puremagic.com/issues/show_bug.cgi?id=3866
>>>>
>>>> Currently, they are both,
>>>
>>> That is how it should be.
>>>
>>>> which leads to the nasty behavior in the bug report.
>>>>
>>>
>>> It contributes, but it is not the main cause.
>>>
>>>> The problem centers around name mangling. If two types mangle the same,
>>>> then they are the same type.
>>>
>>> Then they are equal types.
>>
>> This is simply not tenable. What defines when they are "equal" types and
>> when they are "not equal"?
>
> This is a matter of terminology. For example, for 'equal' just exclude
> the default parameters from the comparison. For 'the same' include
> default parameters in the comparison. (therefore, 'the same' implies
> 'equal')

The language doesn't have the concepts of "same" and "equal" with 
respect to types. There is "equal" and "implicitly converts to", but 
that's not quite the same.

>>> The schizophrenic behavior occurs because the types cross-talk. Are
>>> mangled
>>> names kept unique in the compiler or what is the implementation issue
>>> exactly?
>>
>> It's a conceptual issue. When is one type the same as another, and when
>> is it not?
>>
>
> void function(int) is the same as void function(int) and both are equal
> void function(int=2) is not the same as void function(int=3), but both
> are equal.

The question was *when* are they same, not how you name them.