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glsl: don't set uniform block as used when its not

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The spec requires unused uniform block to be set as active in the
program resource list. To support this we tell opt dead code not to
remove them. However we can mark them as unused internally and
avoid unnecessarily state changes.

This change is also required for the folowing clean-up patch.

Reviewed-by: Tapani Pälli <tapani.palli@intel.com>
This commit is contained in:
Timothy Arceri 2019-12-04 00:14:03 +11:00
parent 50dc4b77f6
commit f0cb0fe1c0
2 changed files with 10 additions and 2 deletions
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3
src/compiler/glsl/link_uniforms.cpp
View file

@ -1059,7 +1059,8 @@ private:
this->uniforms[id].opaque[shader_type].index = ~0;
this->uniforms[id].opaque[shader_type].active = false;
this->uniforms[id].active_shader_mask |= 1 << shader_type;
if (current_var->data.used || base_type->is_subroutine())
this->uniforms[id].active_shader_mask |= 1 << shader_type;
/* This assigns uniform indices to sampler and image uniforms. */
handle_samplers(base_type, &this->uniforms[id], name);

9
src/compiler/glsl/opt_dead_code.cpp
View file

@ -144,8 +144,15 @@ do_dead_code(exec_list *instructions, bool uniform_locations_assigned)
*/
if (entry->var->is_in_buffer_block()) {
if (entry->var->get_interface_type_packing() !=
GLSL_INTERFACE_PACKING_PACKED)
GLSL_INTERFACE_PACKING_PACKED) {
/* Set used to false so it doesn't get set as referenced by
* the shader in the program resource list. This will also
* help avoid the state being unnecessarily flushed for the
* shader stage.
*/
entry->var->data.used = false;
continue;
}
}
if (entry->var->type->is_subroutine())