ios - 如何使用MTLSamplerState而不是在片段着色器代码中声明采样器？

我在下面的着色器中定义了采样器(constexpr sampler textureSampler (mag_filter::linear,min_filter::linear);)。

    using namespace metal;

    struct ProjectedVertex {
      'float4 position [[position]];
      'float2 textureCoord;
    };

    fragment float4 fragmentShader(const ProjectedVertex in [[stage_in]],
                                   const texture2d<float> colorTexture [[texture(0)]],
                                   constant float4 &opacity [[buffer(1)]]){

      constexpr sampler textureSampler (mag_filter::linear,min_filter::linear);
      const float4 colorSample = colorTexture.sample(textureSampler, in.textureCoord);
      return colorSample*opacity[0];

    }

现在，我想避免在着色器代码中几乎不定义此采样器。我找到了MTLSamplerState但我不知道如何使用它

最佳答案

要创建采样器，请首先创建 MTLSamplerDescriptor 对象，然后配置描述符的属性。然后在将使用此采样器的 MTLDevice 对象上调用newSamplerStateWithDescriptor:方法。创建采样器后，可以释放描述符或重新配置其属性以创建其他采样器。

// Create default sampler state
MTLSamplerDescriptor *samplerDesc = [MTLSamplerDescriptor new];
samplerDesc.rAddressMode = MTLSamplerAddressModeRepeat;
samplerDesc.sAddressMode = MTLSamplerAddressModeRepeat;
samplerDesc.tAddressMode = MTLSamplerAddressModeRepeat;
samplerDesc.minFilter = MTLSamplerMinMagFilterLinear;
samplerDesc.magFilter = MTLSamplerMinMagFilterLinear;
samplerDesc.mipFilter = MTLSamplerMipFilterNotMipmapped;
id<MTLSamplerState> ss = [device newSamplerStateWithDescriptor:samplerDesc];

为片段函数设置采样器状态:

    id<MTLRenderCommandEncoder> encoder = [commandBuffer renderCommandEncoderWithDescriptor: passDescriptor];
     ...
    [encoder setFragmentSamplerState: ss atIndex:0];

从着色器访问:

fragment float4 albedoMainFragment(ImageColor in [[stage_in]],
                                texture2d<float> diffuseTexture [[texture(0)]],
                                sampler smp [[sampler(0)]]) {

    float4 color = diffuseTexture.sample(smp, in.texCoord);
    return color;
}

关于ios - 如何使用MTLSamplerState而不是在片段着色器代码中声明采样器？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/59002795/