target/arm: Fix MVE 48-bit SQRSHRL for small right shifts

We got an edge case wrong in the 48-bit SQRSHRL implementation: if
the shift is to the right, although it always makes the result
smaller than the input value it might not be within the 48-bit range
the result is supposed to be if the input had some bits in [63..48]
set and the shift didn't bring all of those within the [47..0] range.

Handle this similarly to the way we already do for this case in
do_uqrshl48_d(): extend the calculated result from 48 bits,
and return that if not saturating or if it doesn't change the
result; otherwise fall through to return a saturated value.

Signed-off-by: Peter Maydell <peter.maydell@linaro.org>
Reviewed-by: Richard Henderson <richard.henderson@linaro.org>
This commit is contained in:
Peter Maydell 2021-08-13 17:11:48 +01:00
parent 95351aa76c
commit fdcf2269c4
1 changed files with 9 additions and 2 deletions

View File

@ -1563,6 +1563,8 @@ uint64_t HELPER(mve_uqrshll)(CPUARMState *env, uint64_t n, uint32_t shift)
static inline int64_t do_sqrshl48_d(int64_t src, int64_t shift,
bool round, uint32_t *sat)
{
int64_t val, extval;
if (shift <= -48) {
/* Rounding the sign bit always produces 0. */
if (round) {
@ -1572,9 +1574,14 @@ static inline int64_t do_sqrshl48_d(int64_t src, int64_t shift,
} else if (shift < 0) {
if (round) {
src >>= -shift - 1;
return (src >> 1) + (src & 1);
val = (src >> 1) + (src & 1);
} else {
val = src >> -shift;
}
extval = sextract64(val, 0, 48);
if (!sat || val == extval) {
return extval;
}
return src >> -shift;
} else if (shift < 48) {
int64_t extval = sextract64(src << shift, 0, 48);
if (!sat || src == (extval >> shift)) {