On Wednesday, 15 June 2016 at 03:45:39 UTC, Walter Bright wrote:
On 6/14/2016 8:23 PM, tsbockman wrote:
This is specified fully in the template constraints:
if (isIntegral!N && isUnqual!N)
if ((isIntegral!N && !isUnqual!N) || isCheckedint!N)
The second overload simply forwards to the first, after
applying
BasicScalar!(Unqual!N).
Why would a checkedint be a base type for a checkedint?
Generic code (contrived, oversimplified example):
import checkedint.throws, checkedint.traits;
SmartInt!(typeof(A.init + B.init)) add(A, B)(const A a, const
B b)
if (isIntegral!A && isIntegral!B)
{
SmartInt!A ma = a;
SmartInt!B mb = b;
return ma + mb;
}
Of course I could force the user to write `BasicScalar`
everywhere - but why? The intent is just as clear this way, and
it's less verbose.
