On Wed, 2024-04-10 at 18:07 +0800, LIU Hao wrote: > 在 2024-04-10 17:52, Xi Ruoyao via Gcc-help 写道: > > No, the OP is still thinking it's a wrong-code. > > Would you read the Subject please? > > > u32 x = a * b; > u64 r = x; > return r; > > This is same as > > u32 x = (int) a * (int) b; > u64 r = x; > return r; > > and > > return (u64)(u32) ((int) a * (int) b); > > > The code requests an `int` be zero-extended to a `u64` (if the result is written to EAX then this is > no-op), but GCC performs sign extension anyway. Do you still consider it not a bug? It is a missed-optimization, but not wrong code. The code does not requests an *arbitrary* int to be zero-extended to a u64. It requests a value which can be proven as non-negative to be zero-extended. Thus doing a sign-extension may be sub-optimal (depending on the context and the target feature etc), but not wrong. u32 x = (int) a * (int) b; Here a and b are u16, so both (int) a and (int) b are non-negative. And since signed-overflow is UB, x is non-negative too. And for a non- negative int x, (u64)x is just same as (u64)(u32)x. -- Xi Ruoyao <xry111@xxxxxxxxxxx> School of Aerospace Science and Technology, Xidian University
