glsl: Allow elimination of uniform block members

glGetActiveUniform is not supposed to report block members that are not
active even if they are included in the layout of the block.  The block
layout is determined from the GLSL_TYPE_INTERFACE that defines the
block, so eliminating the ir_variables that correspond to the individual
fields is safe.

Fixes gles3conform test
uniform_buffer_object_getuniformindices_for_for_nonexistent_or_not_active_uniform_names.

This also fixes the assertion failures (added in the previous commit) in
gles3conform uniform_buffer_object_index_of_not_active_block,
uniform_buffer_object_inherit_and_override_layouts, and
uniform_buffer_object_repeat_global_scope_layouts.

Signed-off-by: Ian Romanick <ian.d.romanick@intel.com>
Reviewed-by: Jordan Justen <jordan.l.justen@intel.com>

src/glsl/opt_dead_code.cpp

@@ -98,15 +98,10 @@ do_dead_code(exec_list *instructions, bool uniform_locations_assigned)
 	 /* uniform initializers are precious, and could get used by another
 	  * stage.  Also, once uniform locations have been assigned, the
 	  * declaration cannot be deleted.
-	  *
-	  * Also, GL_ARB_uniform_buffer_object says that std140
-	  * uniforms will not be eliminated.  Since we always do
-	  * std140, just don't eliminate uniforms in UBOs.
 	  */
 	 if (entry->var->mode == ir_var_uniform &&
 	     (uniform_locations_assigned ||
-	      entry->var->constant_value ||
+	      entry->var->constant_value))
-	      entry->var->is_in_uniform_block()))
 	    continue;
 
 	 entry->var->remove();