On Monday, 28 July 2014 at 09:46:34 UTC, bearophile wrote:
uint y = 3_000_000_000;
writeln(x, " ", y);
writeln(x * y);
D promotes int to uint, right? Which is a bad idea. It should
promote to long, right?
bool overflow = false;
immutable r1 = muls(x, y, overflow);
Why does muls accept this without an explicit cast to signed
operands long? If the result is signed, the operands should be
signed?
overflow = false;
immutable r2 = mulu(x, y, overflow);
Well, it overflows because you multiply 0xfffffffff with
3000000000 if it promotes int to uint, but the result would still
be incorrect without an overflow if the values were -1 and 1. You
would get 0xffffffff without an overflow/underflow.
(I might get it all wrong here, I didn't run the code.)
However, if you manually remove overflow checks, you probably
need to check that you still get vectorized output with
auto-vectorization. Maybe it is better that the backend deals
with it. Dunno.
