[PULL 07/44] target/arm: Fix MVE 48-bit SQRSHRL for small right shifts

[Peter Maydell]

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>
---
 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;
     } else if (shift < 48) {
         int64_t extval = sextract64(src << shift, 0, 48);
         if (!sat || src == (extval >> shift)) {
-- 
2.20.1