How can I map bytes to a matrix of structures?

[Steven Schveighoffer]

On Fri, 09 Sep 2011 11:43:04 -0400, Timon Gehr <timon.gehr at gmx.ch> wrote:

> On 09/09/2011 05:19 PM, teo wrote:
>> Here is an example of what I am after:
>>
>> struct DATA
>> {
>>    ubyte D1;
>>    ubyte D2;
>>    ubyte D3;
>>    ubyte D4;
>> }
>>
>> void main()
>> {
>>    ubyte[16] a = [ 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, 0x08, 0x01,
>> 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, 0x08 ];
>>    auto b = (cast(DATA*)a.ptr)[0 .. 4];
>>    auto c = (cast(DATA[]*)b.ptr)[0 .. 2][0 .. 2];
>> }
>>
>> I need to have a DATA[2][2]. That code compiles but gives me a
>> segmentation fault.
>
> If you actually want a dynamic DATA[2][] array of length 2, this works:
> auto b=(*(cast(DATA[2][2]*)a.ptr))[];
>
> Otherwise:
>
> A simple reinterpret cast should do:
> auto b=*(cast(DATA[2][2]*)a.ptr);
>
> but note that this copies the data, because static arrays have value  
> semantics.
>
> If you want to have refer the new array to the same location, you can  
> use a union.
>
> void main(){
>      union Myunion{
> 	ubyte[16] a = [ 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, 0x08, 0x01,
> 			0x02, 0x03, 0x04, 0x05, 0x06, 0x07, 0x08 ];
> 	DATA[2][2] b;
>      }
>      Myunion myunion;
>      assert(*(cast(DATA[2][2]*)myunion.a.ptr)==myunion.b);
> }

You can also use ref, but you have to use a function, as it's impossible  
to declare a ref local variable except as a function parameter.

void main()
{
    ubyte[16] a = ...;
    void foo(ref DATA[2][2] b)
    {
       ...
    }
    foo(*(cast(DATA[2][2]*)a.ptr));
}

-Steve