我可以不从渲染器的着色器访问预乘颜色吗

Question

我正在使用 pixi.js 版本 4.8.2。我希望在 pixi.js 应用程序中访问渲染器的着色器而不是预乘颜色。

我设置的透明是'notMultiplied'，但是我可以olny访问premultiplied rgb颜色...

有没有办法访问不相乘的颜色？

我想为 GPGPU 使用纹理，所以我想访问纹理的原始 RGBA。

我把代码和结果放在这里。

// init with not multiply mode
    var app = new PIXI.Application(800, 600, {
      backgroundColor : 0xcccccc,
      transparent: 'notMultiplied'
    });
    document.body.appendChild(app.view);

    // draw circle graphics with red and alpha 0.5 ( drawn at display left )
    var graphic = new PIXI.Graphics();
    graphic.alpha = 0.5;
    graphic.beginFill(0xff0000);
    graphic.drawCircle(100,100,100);
    graphic.endFill();
    app.stage.addChild(graphic);

    // use graphics as a texture ( drawn at display right )
    var mesh = new PIXI.mesh.Mesh( graphic.generateCanvasTexture() );
    mesh.position.set(300,100);
    app.stage.addChild(mesh);

    // replace MeshRenderer shader for test premultiply effect
    app.renderer.plugins.mesh.shader = new PIXI.Shader(
      app.renderer.gl,
      // vertex shader is same as original MeshRender's one
      ` 
          attribute vec2 aVertexPosition;
          attribute vec2 aTextureCoord;

          uniform mat3 projectionMatrix;
          uniform mat3 translationMatrix;
          uniform mat3 uTransform;

          varying vec2 vTextureCoord;

          void main(void) {
              gl_Position = vec4((projectionMatrix * translationMatrix * vec3(aVertexPosition, 1.0)).xy, 0.0, 1.0);
              vTextureCoord = (uTransform * vec3(aTextureCoord, 1.0)).xy;
          }
      `,

      // I changed change fragment shader for test
      `
          varying vec2 vTextureCoord;
          uniform vec4 uColor;
          uniform sampler2D uSampler;

          void main(void) {
              //gl_FragColor = texture2D(uSampler, vTextureCoord) * uColor; <- remove
              gl_FragColor = vec4(texture2D(uSampler, vTextureCoord).rgb, 1.0); 
          }
      `
    );

    // render graphics and mesh.
    app.render();

<html>
    <head>

    </head>
    <body>
        <script src="https://cdnjs.cloudflare.com/ajax/libs/pixi.js/4.8.2/pixi.js"></script>
    </body>
 </html>

执行结果

理想的结果是这样的

Answer 1

您的问题并非如您所想的那样是由预乘 alpha 引起的。该问题是由您设置的不透明度属性 .alpha 引起的。

graphic.alpha = 0.5;

此属性应用于（相乘）对象的所有颜色通道和 Alpha 通道。

如果您不希望此属性影响红绿蓝颜色通道，则必须在片段着色器中将 RGB 通道除以不透明度 .alpha：

varying vec2 vTextureCoord;

uniform vec4      uColor;
uniform float     uAlpha;
uniform sampler2D uSampler;

void main(void) {
    vec4 texColor = texture2D(uSampler, vTextureCoord);
    gl_FragColor  = vec4(texColor.rgb / uAlpha, 1.0); 
}

看例子：

var app = new PIXI.Application(800, 600, {
  backgroundColor : 0xcccccc,
  transparent: 'notMultiplied',
});
document.body.appendChild(app.view);

var graphic = new PIXI.Graphics();
graphic.alpha = 0.5;
graphic.beginFill(0xff0000);
graphic.drawCircle(100,100,100);
graphic.endFill();
app.stage.addChild(graphic);

var mesh = new PIXI.mesh.Mesh( graphic.generateCanvasTexture() );
mesh.position.set(300,100);
app.stage.addChild(mesh);
app.renderer.plugins.mesh.shader = new PIXI.Shader(
  app.renderer.gl,
  ` 
    attribute vec2 aVertexPosition;
    attribute vec2 aTextureCoord;

    uniform mat3 projectionMatrix;
    uniform mat3 translationMatrix;
    uniform mat3 uTransform;

    varying vec2 vTextureCoord;

    void main(void) {
        gl_Position   = vec4((projectionMatrix * translationMatrix * vec3(aVertexPosition, 1.0)).xy, 0.0, 1.0);
        vTextureCoord = (uTransform * vec3(aTextureCoord, 1.0)).xy;
    }
  `,
  `
    varying vec2 vTextureCoord;
    
    uniform vec4      uColor;
    uniform float     uAlpha;
    uniform sampler2D uSampler;

    void main(void) {
        vec4 texColor = texture2D(uSampler, vTextureCoord);
        gl_FragColor  = vec4(texColor.rgb / uAlpha, 1.0); 
    }
  `
);
app.render();

<script src="https://cdnjs.cloudflare.com/ajax/libs/pixi.js/4.8.2/pixi.min.js"></script>

我可以不从渲染器的着色器访问预乘颜色吗

Can I access not premultiply color from renderer's shader

javascript

glsl

pixi.js