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Reputation: 7190

GLSL, literal constant Input Layout Qualifiers

I wonder if I may have something like this:

layout (location = attr.POSITION) in vec3 position;

Where for example Attr is a constant structure

const struct Attr
{
    int POSITION;
} attr = Attr(0);

I already tried, but it complains

Shader status invalid: 0(34) : error C0000: syntax error, unexpected integer constant, expecting identifier or template identifier or type identifier at token ""

Or if there is no way with structs, may I use something else to declare a literal input qualifier such as attr.POSITION?

Upvotes: 1

Views: 1300

Answers (1)

Nicol Bolas
Nicol Bolas

Reputation: 473352

GLSL has no such thing as a const struct declaration. It does however have compile time constant values:

const int position_loc = 0;

The rules for constant expressions say that a const-qualified variable which is initialized with a constant expression is itself a constant expression.

And there ain't no rule that says that the type of such a const-qualified variable must be a basic type:

struct Attr
{
  int position;
};
const Attr attr = {1};

Since attr is initialized with an initialization list containing constant expressions, attr is itself a constant expression. Which means that attr.position is an constant expression too, one of integral type.

And such a compile-time integral constant expression can be used in layout qualifiers, but only if you're using GLSL 4.40 or ARB_ehanced_layouts:

layout(location = attr.position) in vec3 position;

Before that version, you'd be required to use an actual literal. Which means the best you could do would be a #define:

#define position_loc 1
layout(location = position_loc) in vec3 position;

Now personally, I would never rely on such integral-constant-expression-within-struct gymnastics. Few people rely on them, so driver code rarely gets tested in this fashion. So the likelihood of encountering a driver bug is fairly large. The #define method is far more likely to work in practice.

Upvotes: 2

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