glsl-lang-quote offers proc-macros to quote GLSL syntax in Rust, using the [glsl_lang] crate.
```rust use glsllangquote::glsl;
// Parse a translation unit at compile time let ast = glsl! { void main() { gl_FragColor = vec4(1., 0., 0., 1.); } };
// There is exactly one external declaration assert_eq!(ast.0.len(), 1); ```
This crate offers a set of features quote-expr, and quote-statement. Enabling those
features enable the respective parser-expr, and parser-statement in [glsl_lang], which
creates dedicated parsers for those types of GLSL fragments of the language grammar.
This is the most efficient option for parsing lots of expressions and statements at
compile-time, however this will slow down the initial compilation of glsl-lang-quote since
the generated parser file in [glsl_lang] will be much larger.
This is why by default, this crate enables the quote-parsable feature, which uses
[glsllang]'s [glsllang::parse::Parsable] trait, whose limitations apply. Whichever method you
chose, the following code will work:
```rust use glsllangquote::glsl_expr;
// Parse an expression let ast = glsl_expr! { a = vec4(1., 0., 0., 1.) }; ```
Since [glsl_lang]'s parser has a stateful lexer (to handle the fact that GLSL's grammar is not
context-free), declaring a type (a struct for example) and using it must happen in the same
macro invocation, otherwise the parser will forget about the previously declared types. The
best is to only parse whole translation units ([glsl!] macro), or parse unambiguous fragments
(such as expressions, but not statements).
```rust use glsllangquote::{glsl, glsl_statement};
// This is ok: let ast = glsl! { struct PointLight { vec3 pos; vec3 color; }
PointLight p;
};
// This will not compile, PointLight can't be parsed as a type name without extra state let statement = glsl_expr! { PointLight p; };
Vincent Tavernier vince.tavernier@gmail.com
BSD-3-Clause