On Thu, 19 Nov 2015 15:08:32 -0800
Andrew Morton wrote:
> I don't think we should apply this unless someone can runtime test it.
> Presumably the current code works OK, but we just don't know what
> nasties the fixed version might expose.
I fully agree. But who can test it?
> The best I can
On Thu, 19 Nov 2015 15:08:32 -0800
Andrew Morton wrote:
> I don't think we should apply this unless someone can runtime test it.
> Presumably the current code works OK, but we just don't know what
> nasties the fixed version might expose.
I fully agree. But who can
On Thu, 19 Nov 2015 21:13:19 +0100 Michael B__sch wrote:
> The expression (~0 >> x) will always yield all-ones, because the right
> shift is an arithmetic right shift that will always shift ones in.
> Hence the old fault code bits will not be cleared before being ORed
> with the new fault code.
The expression (~0 >> x) will always yield all-ones, because the right
shift is an arithmetic right shift that will always shift ones in.
Hence the old fault code bits will not be cleared before being ORed
with the new fault code.
Fix this by forcing a logical right shift instead of an arithmetic
The expression (~0 >> x) will always yield all-ones, because the right
shift is an arithmetic right shift that will always shift ones in.
Hence the old fault code bits will not be cleared before being ORed
with the new fault code.
Fix this by forcing a logical right shift instead of an arithmetic
The expression (~0 >> x) will always yield all-ones, because the right
shift is an arithmetic right shift that will always shift ones in.
Hence the old fault code bits will not be cleared before being ORed
with the new fault code.
Fix this by forcing a logical right shift instead of an arithmetic
The expression (~0 >> x) will always yield all-ones, because the right
shift is an arithmetic right shift that will always shift ones in.
Hence the old fault code bits will not be cleared before being ORed
with the new fault code.
Fix this by forcing a logical right shift instead of an arithmetic
On Thu, 19 Nov 2015 21:13:19 +0100 Michael B__sch wrote:
> The expression (~0 >> x) will always yield all-ones, because the right
> shift is an arithmetic right shift that will always shift ones in.
> Hence the old fault code bits will not be cleared before being ORed
> with the
The expression (~0 >> x) will always yield all-ones, because the right
shift is an arithmetic right shift that will always shift ones in.
Hence the old fault code bits will not be cleared before being ORed
with the new fault code.
Fix this by forcing a logical right shift instead of an arithmetic
The expression (~0 x) will always yield all-ones, because the right
shift is an arithmetic right shift that will always shift ones in.
Hence the old fault code bits will not be cleared before being ORed
with the new fault code.
Fix this by forcing a logical right shift instead of an arithmetic
10 matches
Mail list logo