将为 GLSL 编写的片段 Shader 转换为 Metal Shader

Question

我想将以下为 glsl 编写的片段着色器转换为金属着色器。

const float PI = 3.14159265359;


mat2 rotate2d (float _angle) {
    return mat2 (cos (_angle), -sin (_angle),
                sin (_angle), cos (_angle));
}

void main (void) {

    vec2 st = (gl_FragCoord.xy * 2.0 --resolution) /min(resolution.x,resolution.y);
    float p = 0.0;
    
   st = rotate2d (sin (time) * PI) * st;
    
    vec2 c = max (abs (st) --0.2,0.0);
    p = length (c);
    p = ceil (p);
    
    vec3 color = vec3 (1.0-p);

    gl_FragColor = vec4 (color, 1.0);

}

当时理解为vec2等设置成float2等是没有问题的

应该怎么写？

Answer 1

如果没有关于当前渲染管道的任何信息，很难转换此着色器：

#include <metal_stdlib>

float2x2 rotate2d(float angle)
{
    return float2x2(float2(cos(angle), -sin(angle)),
                    float2(sin(angle),  cos(angle)));
}

fragment float4 fragmentShader(RasterizerData in [[stage_in]],
                               constant simd::float2 &resolution  [[ buffer(0) ]],
                               constant float &time  [[ buffer(1) ]])
{
    float2 st = (in.textureCoordinate.xy * 2.0 - resolution) / min(resolution.x, resolution.y);
    float p = 0.0;
    st = rotate2d(sin(time) * M_PI_F) *st;
    float2 c = max(abs(st) - 0.2, 0.0);
    p = length(c);
    p = ceil(p);
    float3 color = float3(1.0 - p);
    return float4(color, 1);
}

将为 GLSL 编写的片段 Shader 转换为 Metal Shader

convert the fragment Shader written for GLSL to Metal Shader

ios

metal