What appears on your screen will depend on the whole transformations you apply to it. I interpret your question in the way that you use an identity matrix both for the projection and the (model)view transformation. In that case, you are directly drawing in clip space. Technically, this will define a camera position at the origin, but the view frustum will be equivalent to an Ortho(-1.0, 1.0, -1.0, 1.0, 1.0, -1.0)
setup - so your view frustum will lie half behind the camera (which is perfectly fine for ortho projections, but might seem counterintuitive). For the camera orientation, that part is a bit more ambigous. Usually, GL had the convention that the camera looks into the -z direction, but in clip space, +z will point into the screen, and the classic Ortho
and Perspective
functions treat this by actually using z_near=-near
and z_far=-far
for the near
and far
parameters. (That is also is also the reason why identy transform is achieved through that Ortho
call with the last two signs flipped relative to the x and y ranges). So in that case, one could argue that the camera is actually looking along +z, or one could argue that the camera looks along -z, but the near and far planes are swapped by the projection. That is a matter of how you like to interpret these things. These conventions mostly matter to classic "fixed-function" GL, which uses the eye position for lighting and fog calculations (and maybe a view things else). And it also might assume the "camera looks along -z" convention for calculation of the specular lighting term if GL_LIGHT_MODEL_LOCAL_VIEWER
is not enabled (which is disabled by default, see glLightModel()
.
In modern GL, no such conventions exist besides the left-handedness of the clip space, and you can make your own conventions for an eye space (if you need one).