On Tue, 31 Jan 2023, Richard Biener wrote:
> On Tue, 31 Jan 2023, Xianmiao Qu wrote:
>
> > In the architecture where sign defaults to unsigned, the 'f' will be zero
> > extended to int type in the expression 'd = ~(f & ~2880764155)', then the
> > 'd' will become -1 wich cause the case to fail.
>
On Tue, 31 Jan 2023, Xianmiao Qu wrote:
> In the architecture where sign defaults to unsigned, the 'f' will be zero
> extended to int type in the expression 'd = ~(f & ~2880764155)', then the
> 'd' will become -1 wich cause the case to fail.
> So it's ok for the architectures where sign defaults
In the architecture where sign defaults to unsigned, the 'f' will be zero
extended to int type in the expression 'd = ~(f & ~2880764155)', then the
'd' will become -1 wich cause the case to fail.
So it's ok for the architectures where sign defaults to signed like x86,
but failed for the
