equivariant functions

[Denis Koroskin]

On Wed, 15 Oct 2008 00:59:44 +0400, Robert Fraser  
<fraserofthenight at gmail.com> wrote:

> Denis Koroskin wrote:
>> [snip]
>>  My concern is that it than 'inout' doesn't express that all the  
>> arguments marked as inout as bound to each other and actual type is  
>> deduced from them all. There is a relationship between their types.
>>  inout(A) min(inout(A1) a1, inout(A2) a2);
>>  A1 a1;
>> A2 a2;
>>  const(A1) ca1;
>> const(A2) ca2;
>>  invariant(A1) ia1;
>> invariant(A1) ia2;
>>  auto a = min(a1, ia2); // typeof(a) == const(A) even though neither a1  
>> nor ia2 of that type
>> auto a = min(a1, a2);  // typeof(a) == A
>>  You see, in the example above the return type is changed because type  
>> of 2nd argument is changed. Not only the return type but types of all  
>> the arguments, too.
>>  auto a = min(ia1, ia2);  // typeof(a) == invariant(A). Now return type  
>> is changed per 1st argument type change.
>
> Er.... isn't that the point. If I pass two mutable arguments to min(), I  
> want a mutable back. If I pass two invariant (immutable, hopefully!)  
> arguments to min() I want an invariant back. Otherwise, I want a const  
> back. That's why this syntax is needed at all.

Yes, it's all about this. My point is that 'inout' keyword is not very  
descriptive in this situation. The behaviour, however, is fine.