Integer promotion issue

[Steven Schveighoffer]

On 1/25/20 7:07 PM, Evan wrote:
> On Saturday, 25 January 2020 at 23:58:00 UTC, Steven Schveighoffer wrote:
>> On 1/25/20 6:46 PM, Evan wrote:
>>> [...]
>>
>> Not really. You can cast yourself, write your own type that does it, 
>> or use masking to have the compiler accept the result (i.e. (a + b) & 
>> 0xff).
>>
>> I'd recommend your own type, as it will look a lot cleaner.
>>
> 
> If I make my own type then I have to write more casts from my type to 
> built in types. I want to create less work for myself not more.

I'm not sure what you mean. You should be able to alias-this the type to 
reduce the casting. Whipped up in a couple minutes:

struct U8
{
     ubyte _val;
     alias _val this;
     this(ubyte x)
     {
         _val = x;
     }

     U8 opBinary(string op, T)(T other)
             if ((is(T == U8) || is(T == ubyte)) && (op == "+" || op == 
"-" /* || op == ... */))
     {
         return mixin("U8(cast(ubyte)(_val " ~ op ~ " cast(ubyte)other))");
     }
     // opBinaryRight etc...
}

void main()
{
     import std.stdio;

     U8 a = 0x01;
     U8 b = 0x01;
     ubyte c = a + b; //Error: cannot implicitly convert expression 
cast(int)a + cast(int)b of type int to ubyte
     writeln(c); // 2
     ubyte d = b - a;
     writeln(d); // 0
     writeln(a); // 1 (uses alias this)
}

-Steve