glslang/Test/hlsl.type.identifier.frag
John Kessenich 0320d090e2 HLSL: Recognize types declared as identifiers as identifiers.
E.g., in

    float float;
    (float) * float;

The "(float)" is not a type cast, it is an expression.
2017-06-13 22:22:52 -06:00

32 lines
587 B
GLSL

struct foo_t {
float float;
};
float fn(float float) { return float; }
float4 main() : SV_Target0
{
float float = 7;
bool bool[2] = { float, float };
int int = bool[1];
uint uint = float + int;
min16float min16float = uint;
min10float min10float = min16float;
half half = 0.5;
{
foo_t float;
float.float = 42;
}
bool[0] = bool[1];
float = float + int + uint + min16float + min10float + (bool[0] ? int : float) + fn(float);
half2x3 half2x3;
half2x3._11 = (float) * float;
return float + half2x3._11;
}