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>
fdcf2269c4