glsl: do not set the 'smooth' qualifier by default on ES shaders

It leads to surprising states with integer inputs and outputs on
vertex processing stages (e.g. geometry stages). Instead, rely on the
driver to choose smooth interpolation by default.

We still allow varyings to match when one stage declares it as smooth
and the other declares it without interpolation qualifiers.

Reviewed-by: Marek Olšák <[email protected]>
Tested-by: Dieter Nützel <[email protected]>

1 files changed, 7 insertions, 1 deletions

diff --git a/src/mesa/main/shader_query.cpp b/src/mesa/main/shader_query.cpp
index 64e68b4a26d..6712bb45fb2 100644
--- a/src/mesa/main/shader_query.cpp
+++ b/src/mesa/main/shader_query.cpp
@@ -1634,7 +1634,13 @@ validate_io(struct gl_program *producer, struct gl_program *consumer)
        * Note that location mismatches are detected by the loops above that
        * find the producer variable that goes with the consumer variable.
        */
-      if (producer_var->interpolation != consumer_var->interpolation) {
+      unsigned producer_interpolation = producer_var->interpolation;
+      unsigned consumer_interpolation = consumer_var->interpolation;
+      if (producer_interpolation == INTERP_MODE_NONE)
+         producer_interpolation = INTERP_MODE_SMOOTH;
+      if (consumer_interpolation == INTERP_MODE_NONE)
+         consumer_interpolation = INTERP_MODE_SMOOTH;
+      if (producer_interpolation != consumer_interpolation) {
          valid = false;
          goto out;
       }