GLSL：无法从 FBO 读取纹理并使用片段着色器渲染到另一个 FBO

发布于 2025-01-04 13:56:34 字数 1366 浏览 0 评论 0原文

我正在尝试“读取”附加到第一个 FBO (fboA) 的纹理，修改它（使用片段着色器）并渲染到第二个 FBO (fboB)。

我无法弄清楚，我得到的只是黑色或白色的纹理。

这是代码：

glBindFramebufferEXT(GL_FRAMEBUFFER_EXT, fboB);
glEnable(GL_TEXTURE_RECTANGLE_ARB);
glActiveTexture(GL_TEXTURE0);
glBindTexture(GL_TEXTURE_RECTANGLE_ARB, textureFromFboA);

GLuint t1Location = glGetUniformLocation(shaderProgram, "texture");
glUniform1i(t1Location, 0);

glUseProgram(shaderProgram);

glBegin(GL_QUADS);

glTexCoord2f(0.0f, 0.0f);
glVertex2f (0.0f, 0.0f);

glTexCoord2f(imageRect.size.width, 0.0f); 
glVertex2f (imageRect.size.width, 0.0f);

glTexCoord2f(imageRect.size.width, imageRect.size.height);
glVertex2f (imageRect.size.width, imageRect.size.height);

glTexCoord2f(0.0f, imageRect.size.height);
glVertex2f (0.0f,imageRect.size.height);

glEnd();
glDisable(GL_TEXTURE_RECTANGLE_ARB);
glFlush();

glUseProgram(0);

这是片段着色器代码：

#version 120

uniform sampler2D texture;

void main(void)
{
    gl_FragColor = texture2D(texture, gl_TexCoord[0].st) * 0.8;
}

我希望在 fboB 中呈现较暗的纹理，但我只得到全黑纹理。如果我编写 gl_FragColor =texture2D(texture, gl_TexCoord[0].st);，也会发生这种情况。

相反，如果我写 gl_FragColor = vec2(1.0, 0.0, 0.0, 1.0); 我正确地得到了全红色纹理作为输出。

如果我注释掉 glUseProgram() 语句，则代码可以正常工作，并且 fboB 中的纹理是 fboA 中纹理的精确副本。

为什么会出现这种情况？我错过了什么吗？

原文

I'm trying to "read" the texture attached to a first FBO (fboA), modifying it (with fragment shader) and render to a second FBO (fboB).

I'm not able to figure it out, all I got is a black or white texture.

Here is the code:

glBindFramebufferEXT(GL_FRAMEBUFFER_EXT, fboB);
glEnable(GL_TEXTURE_RECTANGLE_ARB);
glActiveTexture(GL_TEXTURE0);
glBindTexture(GL_TEXTURE_RECTANGLE_ARB, textureFromFboA);

GLuint t1Location = glGetUniformLocation(shaderProgram, "texture");
glUniform1i(t1Location, 0);

glUseProgram(shaderProgram);

glBegin(GL_QUADS);

glTexCoord2f(0.0f, 0.0f);
glVertex2f (0.0f, 0.0f);

glTexCoord2f(imageRect.size.width, 0.0f); 
glVertex2f (imageRect.size.width, 0.0f);

glTexCoord2f(imageRect.size.width, imageRect.size.height);
glVertex2f (imageRect.size.width, imageRect.size.height);

glTexCoord2f(0.0f, imageRect.size.height);
glVertex2f (0.0f,imageRect.size.height);

glEnd();
glDisable(GL_TEXTURE_RECTANGLE_ARB);
glFlush();

glUseProgram(0);

This is the fragment shader code:

#version 120

uniform sampler2D texture;

void main(void)
{
    gl_FragColor = texture2D(texture, gl_TexCoord[0].st) * 0.8;
}

I would expect to have a darker texture rendered in fboB but I only get an all black texture. This happens also if I write gl_FragColor = texture2D(texture, gl_TexCoord[0].st);.

On the contrary, if I write gl_FragColor = vec2(1.0, 0.0, 0.0, 1.0); I correctly have an all red texture as output.

If I comment out the glUseProgram() statement, the code works fine and texture in fboB is an exact copy of texture in fboA.

Why this happens? Am I missing something?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

少钕鈤記 2025-01-11 13:56:34

uniform sampler2D texture;

矩形纹理不一样纹理类型作为 2D 纹理。是的，它们是二维的，但它们仍然保持独特的纹理类型。因此，无法通过 sampler2D 访问它们。

因此将其更改为 samplerRect。您还需要使用正确的纹理坐标，因为矩形纹理采用纹素空间坐标而不是标准化坐标。

或者，您可以只使用 2D 纹理。 NPOT 纹理已经存在了五年多了；您不必使用矩形纹理来获得非二次幂的渲染目标。

uniform sampler2D texture;

Rectangle textures are not the same texture type as 2D textures. Yes, they're two-dimensional, but they still maintain a distinct texture type. Therefore, they cannot be accessed via a sampler2D.

So change that to a samplerRect. You will also need to use proper texture coordinates, because rectangle textures take texel-space coordinates instead of normalized coordinates.

Alternatively, you can just use a 2D texture. NPOT textures have been around for well over half a decade; you don't have to use rectangle textures to have non-power-of-two render targets.

回复收藏 0 原文