What is the correct behavior when both a 1D and a 2D texture are bound in OpenGL?

Question

Say you have something like this:

glBindTexture(GL_TEXTURE_2D, my2dTex);
glBindTexture(GL_TEXTURE_1D, my1dTex);
glBegin...

What is the correct OpenGL behavior? To draw the 1d texture, the 2d or both? For each active texture are there actually multiple textures that can be bound to it at the same time (i.e. a 1d, 2d, 3d cube map, etc.)?

Answer 1

The GL state for the bindings is one texture name per target (i.e. 1D/2D/3D/cube). So when calling

glBindTexture(GL_TEXTURE_2D, my2dTex)
glBindTexture(GL_TEXTURE_1D, my1dTex)

the GL will remember both settings.

Now, the answer of which one GL will use depends on whether you have a shader on.

If a shader is on, the GL will use whatever the shader says to use. (based on sampler1d/sampler2d...).

If no shader is on, then it first depends on which glEnable call has been made.

glEnable(GL_TEXTURE_2D)
glEnable(GL_TEXTURE_1D)

If both are enabled, there is a static priority rule in the spec (3.8.15 Texture Application in the GL 1.5 spec).

Cube > 3D > 2D > 1D

So in your case, if both your texture targets are enabled, the 2D one will be used.

As a side note, notice how a shader does not care whether or not the texture target is Enabled...

Edit to add:

And for the people who really want to get to the gritty details, you always have a texture bound for each target * each unit. The name 0 (the default for the binding state) corresponds to a series of texture objects, one per target. glBindTexture(GL_TEXTURE_2D, 0) and glBindTexture(GL_TEXTURE_1D, 0) both bind a texture, but not the same one...

This is historical, specified to match the behavior of GL 1.0, where texture objects did not exist yet. I am not sure what the deprecation in GL3.0 did with this, though.