Tango BitArray Initialization

[kris]

Bill Baxter wrote:
> Bill Baxter wrote:
> 
>> Sean Kelly wrote:
>>
>>> Bill Baxter wrote:
>>>
>>>> John Reimer wrote:
>>>>
>>>>> On Sun, 11 Feb 2007 19:41:48 -0800, Sean Kelly wrote:
>>>>>
>>>>
>>>>> Another simple alternative could employ a static opAssign.
>>>>>
>>>>> This would make things much simpler:
>>>>>
>>>>> BitArray bitbag = 0b11111000000;
>>>>>
>>>>> The value is limited to 64-bits, but at least it's clean and simple 
>>>>> for
>>>>> those situations where we don't have a long initialization value.
>>>>> (this would work for hexidecimal value also).  For any larger 
>>>>> values we
>>>>> can use an array literal assignment or something similar.
>>>>>
>>>>
>>>> Does opAssign work that way?  I think it has to be done in two lines:
>>>>
>>>>  BitArray bitbag;
>>>>  bitbag = 0b11111000000;
>>>>
>>>> Yes that does seem to be the case.  Otherwise you get, bizarrely, 
>>>> the error "no property 'opCall' for type 'Foo'".
>>>
>>>
>>> Oddly, if you make the opCall static, it works.
>>>
>>>
>>> Sean
>>
>>
>> Hmm.  Show me how.  This and several variations of this that I tried 
>> do not work:
>>
>> import std.stdio;
>> struct Struct
>> {
>>     static void opAssign(int i) {
>>         val = i;
>>     }
>>     int val = 0;
>> }
>>
>> void main()
>> {
>>     Struct s = 2;
>>     writefln("S.val=%s", s.val);
>> }
> 
> 
> Oh, wait.  "if you make the *opCall* static".  Weird.  This does work:
> 
> import std.stdio;
> 
> struct Struct
> {
>     static Struct opCall(int i) {
>         Struct s;
>         s.val = i;
>         return s;
>     }
>     void opAssign(int i) {
>         val = i;
>     }
>     int val = 0;
> }
> 
> void main()
> {
>     Struct s = 2;
>     writefln("S.val=%s", s.val);
> }
> 
> Neat!
> --bb


No opCall needed. Just return the struct from the static opAssign() instead