ios - 解除绑定(bind) Open GL ES 纹理的正确方法？

Question

我的绘图代码首先绘制没有纹理的背景，然后在其上叠加纹理。我目前正在使用 glBindTexture(GL_TEXTURE_2D, 0) 以便在执行第二批绘图后取消绑定(bind)纹理。

但是，当我在 Xcode 中分析 GPU 性能时，会生成以下警告:

当我删除取消绑定(bind)纹理的代码时，警告消失，但绘制不正确。我正在绘制的纹理没有启用 mipmapping，所以它不是来自那里。

绘图代码:

glClear(GL_COLOR_BUFFER_BIT);

{ // Drawing

    glViewport(0, 0, frameWidth, frameHeight);

    GLuint vertexCount = animBuffer.vertexCounts[animIndex];

    glBindBuffer(GL_ARRAY_BUFFER, animBuffer.vertexBuffers[animIndex]);
    glBindBuffer(GL_ELEMENT_ARRAY_BUFFER, animBuffer.indexBuffers[animIndex]);

    glVertexAttribPointer(positionSlot, 3, GL_FLOAT, GL_FALSE, sizeof(vertex), 0);
    glVertexAttribPointer(colorSlot, 4, GL_FLOAT, GL_FALSE, sizeof(vertex), (GLvoid*)(sizeof(float)*3));
    glVertexAttribPointer(textureCoordSlot, 2, GL_FLOAT, GL_FALSE, sizeof(vertex), (GLvoid*)(sizeof(float)*7));

    glDrawElements(GL_TRIANGLE_STRIP, vertexCount, GL_UNSIGNED_SHORT, 0);


    // glEnable(GL_TEXTURE_2D) (tried this, gives another warning)

    glActiveTexture(GL_TEXTURE0);
    glBindTexture(GL_TEXTURE_2D, texture);
    glUniform1i(textureUniform, 0);

    glBindBuffer(GL_ARRAY_BUFFER, textureColorBuffer); // Transparent color buffer to allow texture to be overlayed.
    glVertexAttribPointer(colorSlot, 4, GL_FLOAT, GL_FALSE, 0, (GLvoid*)(sizeof(float)*3));

    glDrawElements(GL_TRIANGLE_STRIP, vertexCount, GL_UNSIGNED_SHORT, 0);

    // Unbind texture.... this line is causing the error.
    glBindTexture(GL_TEXTURE_2D, 0);

    // glDisable(GL_TEXTURE_2D) (tried this, gives another warning)    
}

[context presentRenderbuffer:GL_RENDERBUFFER];

纹理加载代码:

CGImageRef textureImage = image.CGImage;

size_t width = CGImageGetWidth(textureImage);
size_t height = CGImageGetHeight(textureImage);

GLubyte* spriteData = malloc(width*height*4);

CGColorSpaceRef cs = CGImageGetColorSpace(textureImage);
CGContextRef c = CGBitmapContextCreate(spriteData, width, height, 8, width*4, cs, kCGImageAlphaPremultipliedLast | kCGBitmapByteOrder32Big);
CGColorSpaceRelease(cs);

CGContextScaleCTM(c, 1, -1);
CGContextTranslateCTM(c, 0, -CGContextGetClipBoundingBox(c).size.height);

CGContextDrawImage(c, (CGRect){CGPointZero, {width, height}}, textureImage);
CGContextRelease(c);

GLuint glTex;
glGenTextures(1, &glTex);
glBindTexture(GL_TEXTURE_2D, glTex);

glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MIN_FILTER, GL_LINEAR);
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_MAG_FILTER, GL_LINEAR);
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_S, GL_CLAMP_TO_EDGE);
glTexParameteri(GL_TEXTURE_2D, GL_TEXTURE_WRAP_T, GL_CLAMP_TO_EDGE);


glTexImage2D(GL_TEXTURE_2D, 0, GL_RGBA, (GLsizei)width, (GLsizei)height, 0, GL_RGBA, GL_UNSIGNED_BYTE, spriteData);

glBindTexture(GL_TEXTURE_2D, 0);

free(spriteData);

return glTex;

我应该如何解除绑定(bind)纹理以消除此警告？

编辑:

我已经尝试在纹理绘图之前和之后使用 glEnable(GL_TEXTURE_2D) & glDisable(GL_TEXTURE_2D)，尽管这现在给了我以下警告:

Answer 1

我认为避免此警告的最简单方法是保持纹理绑定(bind)，但只需向片段着色器发送一个 uniform 指示是否混合纹理。

所以，在绘制之前添加:

glUniform1i(drawTexInput, GL_FALSE); // Or GL_TRUE to draw.

在片段着色器中:

varying lowp vec4 destinationColor;

uniform bool drawTex;
varying lowp vec2 texCoordOut;
uniform sampler2D tex;

void main() {

    lowp vec4 result;
    if (drawTex) {
        lowp vec4 tex2D = texture2D(tex, texCoordOut);
        result = tex2D + vec4(1.0 - tex2D.a) * destinationColor; // Alpha blending
    } else {
        result = destinationColor;
    }

    gl_FragColor = result;
}