uint[3] not equivalent to void[12]?

[Ralph Doncaster]

On Friday, 9 February 2018 at 15:24:27 UTC, Mike Parker wrote:
> On Friday, 9 February 2018 at 15:05:33 UTC, Ralph Doncaster 
> wrote:
>> This seems odd to me.  Is there a way I can make a function 
>> that takes an array of any type but only of a specific size in 
>> bytes?
>>
>> void.d(8): Error: function void.foo (void[12] arr) is not 
>> callable using argument types (uint[3])
>> Failed: ["/usr/bin/dmd", "-v", "-o-", "void.d", "-I."]
>> void foo(void [12] arr)
>> {
>> }
>>
>> void main()
>> {
>>     uint[3] arr;
>>     foo(arr);
>> }
>
> void has no size, so what does it mean to have 12 of them?

according to the docs and my testing, the size of a void array 
element is 1, so the following code prints 12:
import std.stdio;

void foo(void [] arr)
{
     writeln("length: " arr.length);
}

void main()
{
     uint[3] arr;
     foo(arr);
}

I thought about using templates, but I was looking for a simple 
way of making a function that takes an array of 12 bytes, whether 
it is uint[3], ubyte[12], or ushort[6].