On 7/29/21 1:14 AM, Peter Maydell wrote:
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.mayd...@linaro.org>
---
Not squashed into the previous patch because that one has already
been reviewed, so as this fixes a different edge case I thought
it clearer kept separate.
---
Reviewed-by: Richard Henderson <richard.hender...@linaro.org>
target/arm/mve_helper.c | 11 +++++++++--
1 file changed, 9 insertions(+), 2 deletions(-)
diff --git a/target/arm/mve_helper.c b/target/arm/mve_helper.c
index 5730b48f35e..1a4b2ef8075 100644
--- a/target/arm/mve_helper.c
+++ b/target/arm/mve_helper.c
@@ -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;
I'll note two things:
(1) The val == extval check could be sunk to the end of the function and shared with the
left shift,
(2) sat will never be unset, as #48 is encoded as sat=1 in the insn.
r~
