Derek Parnell wrote:
On Fri, 12 Jun 2009 02:08:14 +0200, Don wrote:
Walter Bright wrote:
davidl wrote:
It seems that comparing two different operands with different size
makes no sense. The compiler should issue an error against that.
Consider:
byte b;
if (b == 1)
here you're comparing two different sizes, a byte and an int.
Disallowing such (in its various incarnations) is a heavy burden, as the
user will have to insert lots of ugly casts.
There really isn't any escaping from the underlying representation of
2's complement arithmetic with its overflows, wrap-arounds, sign
extensions, etc.
The problem is a lot more specific than that.
The unexpected behaviour comes from the method used to promote two types
to a common type, when both are smaller than int, but of different
signedness. Intuitively, you expect the common type of {byte, ubyte} to
be ubyte, by analogy to {int, uint}->uint, and {long, ulong}->ulong. But
instead, the common type is int!
I think that the common type for byte and ubyte is short. Byte and ubyte
have overlapping ranges of values (-127 to 127) and (0 to 255) so a common
type would have to be able to hold both these ranges at least, and short
(16-bit signed integer) does that.
But then you still have the problem that the high half of the short was
extended from the low half in two different ways, once by sign-extend,
once by zero-extend. Mixing sign-extend and zero-extend in the same
expression is asking for trouble.
