CODEWITHSUNDEEP
glsl
tay10r
tay10r

Reputation: 4337

GLSL Declarations

I'm reading the GLSL specification and noticed in chapter 9 in the grammar section that declarations can consist of only type qualifiers. Which leads me to believe that the specification allows declarations such as this:

const;
const a;
const a, b;

But this appears as a syntax error when I check it with glslangValidator:

test.vert
ERROR: 0:1: 'a' : identifier not previously declared 
ERROR: 1 compilation errors.  No code generated.

Are these kinds of declarations suppose to be valid? Was it suppose to be "type_specifier" instead of "type_qualifier" and this is just a mistake?

Here are the relevant bits from the specification:

invariant_qualifier :
  INVARIANT

interpolation_qualifier :
  SMOOTH
  FLAT
  NOPERSPECTIVE

layout_qualifier :
  LAYOUT LEFT_PAREN layout_qualifier_id_list RIGHT_PAREN

layout_qualifier_id_list :
  layout_qualifier_id
  layout_qualifier_id_list COMMA layout_qualifier_id

layout_qualifier_id :
  IDENTIFIER
  IDENTIFIER EQUAL constant_expression
  SHARED

storage_qualifier :
  CONST
  IN
  OUT
  INOUT
  CENTROID
  PATCH
  SAMPLE
  UNIFORM
  BUFFER
  SHARED
  COHERENT
  VOLATILE
  RESTRICT
  READONLY
  WRITEONLY
  SUBROUTINE
  SUBROUTINE LEFT_PAREN type_name_list RIGHT_PAREN

precise_qualifier :
  PRECISE

type_qualifier :
  single_type_qualifier
  type_qualifier single_type_qualifier

single_type_qualifier :
  storage_qualifier
  layout_qualifier
  precision_qualifier
  interpolation_qualifier
  invariant_qualifier
  precise_qualifier

declaration :
  function_prototype SEMICOLON
  init_declarator_list SEMICOLON
  PRECISION precision_qualifier type_specifier SEMICOLON
  type_qualifier IDENTIFIER LEFT_BRACE struct_declaration_list RIGHT_BRACE SEMICOLON
  type_qualifier IDENTIFIER LEFT_BRACE struct_declaration_list RIGHT_BRACE IDENTIFIER SEMICOLON
  type_qualifier IDENTIFIER LEFT_BRACE struct_declaration_list RIGHT_BRACE IDENTIFIER array_specifier SEMICOLON
  type_qualifier SEMICOLON
  type_qualifier IDENTIFIER SEMICOLON
  type_qualifier IDENTIFIER identifier_list SEMICOLON

Here is a link to the full specification.

Upvotes: 2

Views: 245

Answers (1)

Nicol Bolas
Nicol Bolas

Reputation: 473202

A language is more than just its grammatical structure. You can write a sentence that conforms to the rules of English grammar, yet it doesn't actually make any kind of sense.

The same goes here. It is grammatically legal to use just a type-qualifier. But that doesn't mean it is semantically valid to do so with every type-qualifier. Or even most of them.

So the declaration:

const;

Is not semantically meaningful, despite being grammatically legal.

Note that type-qualifier permits a sequence of qualifiers. And note that among the qualifiers are the storage-qualifiers in, out, uniform and buffer.

By allowing type-qualifier SEMICOLON, the standard permits this kind of grammar:

layout(std140) uniform;

This is valid both grammatically and in terms of meaning. It sets the default layout for all uniform blocks following this statement to std140.

Upvotes: 1

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