GLSL break command - glsl

Question

Currently I am learning how to create shaders in GLSL for a game engine I am working on, and I have a question regarding the language which puzzles me. I have learned that in shader versions lower than 3.0 you cannot use uniform variables in the condition of a loop. For example the following code would not work in shader versions older than 3.0.
for (int i = 0; i < uNumLights; i++)
{
...............
}
But isn't it possible to replace this with a loop with a fixed amount of iterations, but containing a conditional statement which would break the loop if i, in this case, is greater than uNumLights?. Ex :
for (int i = 0; i < MAX_LIGHTS; i++)
{
if(i >= uNumLights)
break;
..............
}
Aren't these equivalent? Should the latter work in older versions GLSL? And if so, isn't this more efficient and easy to implement than other techniques that I have read about, like using a different version of the shader for different number of lights?
I know this might be a silly question, but I am a beginner and I cannot find a reason why this shouldn't work.

Answer 1

GLSL can be confusing insofar as for() suggests to you that there must be conditional branching, even when there isn't because the hardware is unable to do it at all (which applies to if() in the same way).
What really happens on pre-SM3 hardware is that the HAL inside your OpenGL implementation will completely unroll your loop, so there is actually no jump any more. And, this explains why it has difficulties doing so with non-constants.
While technically possible to do it with non-constants anyway, the implementation would have to recompile the shader every time you change that uniform, and it might run against the maximum instruction count if you're just allowed to supply any haphazard number.
That is a problem because... what then? That's a bad situation.
If you supply a too big constant, it will give you a "too many instructions" compiler error when you build the shader. Now, if you supply a silly number in an uniform, and the HAL thus has to produce new code and runs against this limit, what can OpenGL do?
You most probably validated your program after compiling and linking, and you most probably queried the shader info log, and OpenGL kept telling you that everything was fine. This is, in some way, a binding promise, it cannot just decide otherwise all of a sudden. Therefore, it must make sure that this situation cannot arise, and the only workable solution is to not allow uniforms in conditions on hardware generations that don't support dynamic branching.
Otherwise, there would need to be some form of validation inside glUniform that rejects bad values. However, since this depends on successful (or unsuccessful) shader recompilation, this would mean that it would have to run synchronously, which makes it a "no go" approach. Also, consider that GL_ARB_uniform_buffer_object is exposed on some SM2 hardware (for example GeForce FX), which means you could throw a buffer object with unpredictable content at OpenGL and still expect it to work somehow! The implementation would have to scan the buffer's memory for invalid values after you unmap it, which is insane.
Similar to a loop, an if() statement does not branch on SM2 hardware, even though it looks like it. Instead, it will calculate both branches and do a conditional move.

Answer 2

(I'm assuming you are talking about pixel shaders).
Second variant is going to work only on gpu which supports shader model >= 3. Because dynamic branching (such as putting variable uNumLights into IF condition) is not supported on gpu shader model < 3 either.
Here you can compare what is and isn't supported between different shader models.

Answer 3

There is a fun work around I just figured out. Seems stupid and I can't promise you that it's a healthy choice, but it appears to work for me right now:
Set your for loop to the maximum you allow. Put a condition inside the loop to skip over the heavy routines, if the count goes beyond your uniform value.
uniform int iterations;
for(int i=0; i<10; i++){
if(i<iterations){
//do your thing...
}
}