On Tue, 11 Jun 2013 07:46:11 -0400, Temtaime <temta...@gmail.com> wrote:
No. I means, that
uint a = uint.max;
uint b = a + 1;
writeln(b);
Works OK.
Why? Compiler doesn't know if a + b fits in uint, right?
Then why overflow with ints are accepted?
CPU performs math at int level. So even if you add two ubytes, they are
added at integer level.
For instance, this is valid, and does not produce a truncated uint:
ubyte x = ubyte.max;
uint y = x + x;
assert(y == 255 + 255);
Essentially the compiler is making sure you wanted to throw away that
extra precision that it had to create anyways because that's what the CPU
supports.
It can take some getting used to.
Note that += does not have this problem:
ubyte c = k;
c += 1; // ok
This I find extremely inconsistent...
-Steve
