why ushort alias casted to int?

[Patrick Schluter]

On Friday, 22 December 2017 at 10:14:48 UTC, crimaniak wrote:
> My code:
>
> alias MemSize = ushort;
>
> struct MemRegion
> {
> 	MemSize start;
> 	MemSize length;
> 	@property MemSize end() const { return start+length; }
> }
>
> Error: cannot implicitly convert expression 
> `cast(int)this.start + cast(int)this.length` of type `int` to 
> `ushort`
>
> Both operands are the same type, so as I understand casting to 
> longest type is not needed at all, and longest type here is 
> ushort in any case. What am I doing wrong?

@property MemSize end() const { return cast(int)(start+length); }

The rule of int promotion of smaller types comes from C as they 
said. There are 2 reason to do it that way. int is supposed in C 
to be the natural arithmetic type of the CPU it runs on, i.e. the 
default size the processor has the least difficulties to handle. 
The second reason is that it allows to detect easily without much 
hassle if the result of the operation is in range or not. When 
doing arithmetic with small integer types, it is easy that the 
result overflows. It is not that easy to define portably this 
overflow behaviour. On some cpus it would require extra 
instructions. D has inherited this behaviour so that copied 
arithmetic code coming from C behaves in the same way.

@property MemSize end() const
{
   MemSize result = start+length;
   assert(result <= MemSize.max);
   return cast(int)result;
}

with overflow arithmetic this code is not possible (as is if 
MemSize was uint or ulong).