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

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.