Why do shift operators undergo integral promotion?

[Walter Bright]

On 8/9/2011 2:46 AM, Don wrote:
>  From a discussion on D.learn.
>
> If x and y are different integral types, then in an expression like
> x >> y
> the integral promotion rules are applied to x and y.
> This behaviour is obviously inherited from C, but why did C use such a
> counter-intuitive and bug-prone rule?
> Why isn't typeof(x >> y) simply typeof(x) ?
> What would break if it did?
>
> You might think the the rule is that typeof( x >> y) is typeof( x + y),
> but it isn't: the arithmetic conversions are NOT applied:
> typeof(int >> long) is int, not long, BUT
> typeof(short >> int) is int.
> And we have this death trap (bug 2809):
>
> void main()
> {
> short s = -1;
> ushort u = s;
> assert( u == s );
> assert ( (s >>> 1) == (u >>> 1) ); // FAILS
> }


That last is why we can't just change the behavior from C.