On Thu, May 12 2022 at 15:07, Matthew Wilcox wrote:
> On Thu, May 12, 2022 at 01:01:07PM +0000, David Laight wrote:
>> > +static inline int64_t sign_extend64(uint64_t value, int index)
>> > +{
>> > + int shift = 63 - index;
>> > + return (int64_t)(value << shift) >> shift;
>> > +}
>>
>> Shift of signed integers are UB.
>
> Citation needed.
I'll bite :)
C11/19: 6.5.7 Bitwise shift operators
4 The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated
bits are filled with zeros. If E1 has an unsigned type, the value of
the result is E1 × 2E2, reduced modulo one more than the maximum
value representable in the result type. If E1 has a signed type and
nonnegative value, and E1 × 2E2 is representable in the result type,
then that is the resulting value; otherwise, the behavior is
undefined.
This is irrelevant for the case above because the left shift is on an
unsigned integer. The interesting part is this:
5 The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1
has an unsigned type or if E1 has a signed type and a nonnegative
value, the value of the result is the integral part of the quotient
of E1/2E2. If E1 has a signed type and a negative value, the
resulting value is implementation-defined.
So it's not UB, it's implementation defined. The obvious choice is to
keep LSB set, i.e. arithmetic shift, what both GCC and clang do.
Thanks,
tglx
