On 2013-06-11 07:35, Adam D. Ruppe wrote:
On Tuesday, 11 June 2013 at 10:12:27 UTC, Temtaime wrote:
ubyte k = 10;
ubyte c = k + 1;
This code fails to compile because of: Error: cannot implicitly
convert expression (cast(int)k + 1) of type int to ubyte
The reason is arithmetic operations transform the operands into ints,
that's why the error says cast(int)k. Then it thinks int is too big
for ubyte. It really isn't about overflow, it is about truncation.
That's why uint + 1 is fine. The result there is still 32 bits so
assigning it to a 32 bit number is no problem, even if it does
overflow. But k + 1 is promoted to int first, so it is a 32 bit
number and now the compiler complains that you are trying to shove it
into an 8 bit variable. Unless it can prove the result still fits in
8 bits, it complains, and it doesn't look outside the immediate line
of code to try to prove it. So it thinks k can be 255, and 255 + 1 =
256, which doesn't fit in 8 bits.
The promotion to int is something D inherited from C and probably
isn't going anywhere.
i think part of the problem is that '1' is an int. so the calculation must be
promoted to integer.
if we had byte and ubyte integer literals (suffix b/B and ub/UB?), then if all
RHS arguments are (unsigned) bytes the compiler could infer that we are serious
with sticking to bytes...
ubyte k = 10; // or optional 10ub
ubyte c = k + 1ub;
