intel/compiler: fix brw_imm_w for negative 16-bit integers

16-bit immediates need to replicate the 16-bit immediate value
in both words of the 32-bit value. This needs to be careful
to avoid sign-extension, which the previous implementation was
not handling properly.

For example, with the previous implementation, storing the value
-3 would generate imm.d = 0xfffffffd due to signed integer sign
extension, which is not correct. Instead, we should cast to
uint16_t, which gives us the correct result: imm.ud = 0xfffdfffd.

We only had a couple of cases hitting this path in the driver
until now, one with value -1, which would work since all bits are
one in this case, and another with value -2 in brw_clip_tri(),
which would hit the aforementioned issue (this case only affects
gen4 although we are not aware of whether this was causing an
actual bug somewhere).

v2: Make explicit uint32_t casting for left shift (Jason Ekstrand)

Reviewed-by: Jason Ekstrand <jason@jlekstrand.net>
Cc: "18.0 18.1" <mesa-stable@lists.freedesktop.org>

diff --git a/src/intel/compiler/brw_reg.h b/src/intel/compiler/brw_reg.h
index dff9b970b2fd781514dbb839499fe0c966c3a879..ac12ab3d2dd7f3e6ebbbc96f51a5d44cdb8d55e8 100644 (file)
--- a/src/intel/compiler/brw_reg.h
+++ b/src/intel/compiler/brw_reg.h
@@ -705,7 +705,7 @@ static inline struct brw_reg
 brw_imm_w(int16_t w)
 {
    struct brw_reg imm = brw_imm_reg(BRW_REGISTER_TYPE_W);
-   imm.d = w | (w << 16);
+   imm.ud = (uint16_t)w | (uint32_t)(uint16_t)w << 16;
    return imm;
 }