@disable

[Pelle Månsson]

On 01/16/2010 01:46 AM, Leandro Lucarella wrote:
> Ali Çehreli, el 15 de enero a las 16:01 me escribiste:
>>> http://www.digitalmars.com/d/2.0/declaration.html#AutoDeclaration
>>
>> It is news to me that the following works without 'auto':
>>
>> struct S
>> {
>>      int i;
>>
>>      this(int i)
>>      {
>>          this.i = i;
>>      }
>> }
>>
>> void main()
>> {
>>      const c = S(42);   //<-- no 'auto' needed
>> }
>>
>> Now I realize that 'auto' is for when we want type inference for
>> mutable variables because this doesn't work:
>>
>>      c = S(42);   //<-- compiler ERROR
>>
>> So we have to use a keyword:
>>
>>      auto c = S(42);
>>      ++c.i;
>>
>> If I understand it correctly, 'auto' serves as the nonexistent
>> 'mutable' keyword in this case.
>>
>> I think to be consistent, I will continue using 'auto' even for when
>> a storage class is specified:
>>
>>      const auto c = S(42);  // works too
>>
>> For me, that gives 'auto' a single meaning: "the inferred type".
>>
>> Do I get it right? :)
>
> I don't think so. auto means in D the same that in C/C++, the difference
> is that D do type inference when a *storage class* is given. const,
> static, immutable, shared are other storage classs, so when you used
> them, you can infer the type too (if no type is given).
>
> You can do const auto c = 1; (I think), but I can't do static auto c = 1;
> (I think too). You can omit auto when declaring automatic variables if you
> specify the type (seen the other way :), because it defaults to auto. And
> you can omit the type if you use a storage class, because it defaults
> to the infered type.
>
Makes sense, but static auto totally works.

I think auto just means inferred type.