Google's Go

[Steven Schveighoffer]

On Mon, 25 Jan 2010 09:05:11 -0500, Lars T. Kyllingstad  
<public at kyllingen.nospamnet> wrote:

> Steven Schveighoffer wrote:
>> On Sun, 24 Jan 2010 04:25:21 -0500, grauzone <none at example.net> wrote:
>>
>>> Or how inout(T) isn't just a shortcut to avoid writing  
>>> const/immutable-aware code 3 times or putting it into a template?
>>  The benefits are:
>>  - Single implementation where all that is different is the type  
>> qualifier.  (also allows the function to be virtual)
>> - No propogation of contract through an accessor.  In other words,  
>> using an inout accessor on an object or struct does not alter the  
>> caller's contract with the data itself.
>>  The latter function is almost essential for properties, for without  
>> such a mechanism, you are forced to write your property definitions in  
>> triplicate.
>>  i.e.
>>  class C {}
>>  struct S
>> {
>>    C c;
>> }
>>  immutable s1 = S(new C);
>> S s2 = S(new C);
>>  immutable c1 = s1.c;
>> C c2 = s2.c;
>>  Now, change S.c into a property.
>>  The first line of thinking is, "well, accessing c doesn't change the  
>> object itself, so it should be const."  But that means you must return  
>> a const(C), so it breaks defining c1 and c2 (can't assign immutable or  
>> mutable from const).
>>  So, you say, "I'll just define it without const," but then you can't  
>> call the property unless S is a mutable type, so that only works in  
>> c2's case
>>  Maybe you think you can get away with just mutable and immutable, but  
>> again, it doesn't work if the whole object is const, since you can't  
>> call either function from there.
>>  Templates won't work here, you cannot template the 'this' pointer.  So  
>> you end up with 3 identical implementations, and *no* const guarantee  
>> on the mutable one:
>>  @property C c() { return _c; }
>> @property const(C) c() const { return _c; }
>> @property immutable(C) c() immutable { return _c; }
>
>
> Out of curiosity:  How does inout(T) fix this?  I thought inout was all  
> about transporting the const-ness of the input type to the return type,  
> and in this example there are no input parameters.


inout is applied to the hidden input parameter -- this:

@property inout(C) inout {return _c; }

-Steve