OpenGL ES2颜色和纹理着色器

Question

I want to write one vertex shader to work with colors and textures I used this code

precision mediump float;
varying vec4 v_Color;
uniform sampler2D u_TextureUnit;
varying vec2 v_TextureCoordinates;

void main() {
    gl_FragColor= (v_Color*texture2D(u_TextureUnit, v_TextureCoordinates));
}

it works when the model has a texture map but if the model has only color it loads black, I want to check the sampler2D u_TextureUnit before putting gl_FragColor

Answer 1

You have written a shader that reads from a texture unit; you should generally avoid conditionality in shaders as it potentially splits the execution path. Usually a compiler can figure it out if the thing you're acting conditionally upon is a uniform but you may end up paying for a recompile every time you set the uniform. ES 2 has no means of introspecting the current properties of a sampling unit so the prima facie solution might be something like:

uniform bool u_ApplyTexture;
...
if(u_ApplyTexture)
    gl_FragColor = {what you already have};
else
    gl_FragColor = v_Color;

A non-conditional alternative possibly to profile as an alternative might be:

uniform float u_TextureWeight;
...
gl_FragColor = v_Color*
                   mix(vec4(1.0), 
                       texture2D(u_TextureUnit, v_TextureCoordinates), 
                       u_TextureWeight);

As that would evaluate to v_Color*vec4(1.0) when u_TextureWeight is 0.0 and v_Color*texture2D(u_TextureUnit, v_TextureCoordinates) when u_TextureWeight is 1.0 .

If you were really looking to be hacky, you might just upload your textures as negative images and negative them again in the shader:

gl_FragColor = v_Color*(vec4(1.0) - texture2D(u_TextureUnit, v_TextureCoordinates));

As then obviously if you're getting vec4(0.0) from the sampling unit because no texture is attached, that ends up multiplying v_Color by 1.0 .