glsl: stop considering unnamed and named structures equal

Previously, if an unnamed and a named struct contained the same fields,
they were considered the same type during linking of globals.

The discussion around commit e018ea81bf58 ("glsl: Structures must have
same name to be considered same type.") doesn't seem to have considered
this thoroughly, and I see no evidence that an unnamed struct should
ever be considered to be the same type as a named struct.

Reviewed-by: Timothy Arceri <tarceri@itsqueeze.com>

diff --git a/src/compiler/glsl_types.cpp b/src/compiler/glsl_types.cpp
index 063d3f1edc2eda1233b6f37bce06e4115fbd8ff8..188b72f345ac411cc2435488d1874059a5dfd2dd 100644 (file)
--- a/src/compiler/glsl_types.cpp
+++ b/src/compiler/glsl_types.cpp
@@ -927,13 +927,9 @@ glsl_type::record_compare(const glsl_type *b, bool match_locations) const
     *     type definitions, and field names to be considered the same type."
     *
     * GLSL ES behaves the same (Ver 1.00 Sec 4.2.4, Ver 3.00 Sec 4.2.5).
-    *
-    * Note that we cannot force type name check when comparing unnamed
-    * structure types, these have a unique name assigned during parsing.
     */
-   if (!this->is_anonymous() && !b->is_anonymous())
-      if (strcmp(this->name, b->name) != 0)
-         return false;
+   if (strcmp(this->name, b->name) != 0)
+      return false;
 
    for (unsigned i = 0; i < this->length; i++) {
       if (this->fields.structure[i].type != b->fields.structure[i].type)