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GL_MAX_TEXTURE_SIZE barrier on OpenGL ES 2

8 posts in this topic

Good 'morrow, this is for Android game (possibly iOS later) : I'm currently hitting up against maximum texture size limits on some devices. I've decided to not bother with trying to support 1024x1024, but I'm aiming to support 2048x2048 as a lot of devices still have this limit (including my 2013 nexus 7).
Now for some games this isn't an issue, you just downsize the textures .. however I'm using a pixel perfect spritesheet for depth ordered billboards, and want to be able to draw everything with one drawcall, and I'm thinking I'm going to need at least 4096 x 2048. I can't just render with 2 drawcalls, as the sprites are interleaved randomly and have to be drawn in depth order.
So some options I have:
  1. Half size the spritesheet on load and upsize at rendering (ugly blocky)
  2. Half size the spritesheet and half the game resolution (very involved to support 2 version of the game, but would better support different screen sizes?)
  3. Use some shader jiggery pokery to get around the limit

I'm currently thinking along the lines of 3, something like :

  • Binding 2 halves of the spritesheet to active texture 0 and 1 (each 2048 x 2048)
  • Pass the uv coords in pixel space
  • In the fragment shader, if the u coord >= 2048, then -2048 and sample from texture 1 (and vice versa).
I could also maybe make the spritesheet generator avoid placing on the boundary, and decide which spritesheet in the vertex shader, if this would help.
Is this feasible? What kind of penalty should I be expecting for such an approach (with a conditional in the frag shader) versus the hardware supporting it natively?




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Posted (edited)

Option 3 with a 3D texture could work, seeing as you're also limited to 8 textures at once, then just select the layer based on your UV like you say.

Hm it's kinda funny considering back 5 years to when I was keeping fragment shaders uber simple, that's compared with today's crazy goings on with MSAA, lights galore, etc! But just like today I would say aim to optimize speed rather than size, perhaps even pass the texture layer as an attribute with each vertex.


edit: I was reading your post more than thinking; in terms of speed it'll just be your vertex shader passing a lowp int to the fragment shader. (or if adjusting the U element then an extra vertex instruction or two) [frag shader will be the same except for referencing a varying rather than a uniform]

Edited by jezham

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Yeah I guess I'll try passing the texture ID from the vertex shader and compare with just using the fragment shader. If there's not a lot of difference I might just go for the easier method so I don't have to adjust the spritesheet code.

If there is a performance hit I can have 2 sets of shaders, depending on whether the hardware supports > 2048, so the later hardware does not pay a price for the compatibility. :)

I've also had problems with devices not supporting frame buffers > 2048 for scrolling textures, but that's a whole other problem lol. Does make you wonder how the GPU guys decide what the limits should be... It must be intriguing what goes into GPU design, although I'm sure all top secret lol.


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Is this feasible? What kind of penalty should I be expecting for such an approach (with a conditional in the frag shader) versus the hardware supporting it natively?

Depends. If you do:

float4 val0 = texture( tex0, uv );
float4 val1 = texture( tex1, uv );

finalVal = uv.y > 0.5 ? val1 : val0;

There is no branch, and the cost is only slightly less performance, and higher power consumption (since you're consuming twice the bandwidth).

But if you do:

vec2 uvDdx = ddx( uv );
vec2 uvDdy = ddy( uv );

if( uv.y )
    val = texture( tex0, uv, uvDdx, uvDdy );
    val = texture( tex1, uv, uvDdx, uvDdy );

Then the performance impact could be quite serious (the cards you're targeting are very bad at hiding latency) but the power consumption should stay in check.

It boils down to whether the performance decrease you get puts you below the minimum you are targeting and decide to consume more power instead.


I've also had problems with devices not supporting frame buffers > 2048 for scrolling textures, but that's a whole other problem lol. Does make you wonder how the GPU guys decide what the limits should be... It must be intriguing what goes into GPU design, although I'm sure all top secret lol.

Smaller resolutions means less bits in the texture unit to perform addressing and filtering. Less bits means less transistors; which translates to lower costs, less power consumption therefore less heat and increased battery life.


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I'm not sure how feasible this is for your use case, but you could have a full mipmap chain for the texture, pack sprites into different miplevels, then do an explicit miplevel lookup (desktop GL has textureLod for this, I don't know GL ES but expect that it should have similar).

It won't give you the full extra texture space that a 4096x2048 texture would, but it will give you some extra.


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If you (miss-)use mipmaps you do not get very much out of it for making it more complicated. 33,3_% to be exact. And you can hardly make full use of the smaller mipmap levels.

My question would be if you really need more then a 2048x2048 texture for what you draw as ordered billboards? The only thing you normaly draw ordered without depth write are alpha transparent objects. And everything else (including stencil transparency) can be drawn before that with depth write enabled. So you can sort for texture affiliation and only need to bind each texture once per frame.


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Some good info there from Matias regarding the decision between branchless (more texture lookups) and branched shaders (have the same issue in some other shaders, I'll have to do some tests to compare performance .. it maybe the shader compiler decides it is quicker to do both the lookups even in the branched shader).

Using mipmaps is an interesting idea and something I hadn't thought of, but as osbios says this only gains a little extra texture space at quite a bit more complexity, so I'll probably go with the simpler approach.

My question would be if you really need more then a 2048x2048 texture for what you draw as ordered billboards? The only thing you normaly draw ordered without depth write are alpha transparent objects.

This is for that situation of alpha transparent depth sorted sprites (trees etc in the screenshot). :) 


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Posted (edited)

 I don't think it can be done much faster and smaller than just adjusting your UV coords in the vertex shader.

// vertex
uniform		lowp		int	u_baseTextureID;

attribute	mediump		vec2	a_uv; // normalized UV, .x also contains texture ID

varying		lowp		int	v_textureID;
varying		mediump		vec2	v_uv;

	v_textureID = a_uv.x < 1.0 ? u_baseTextureID : 1 + u_baseTextureID;
	v_uv.x = a_uv.x < 1.0 ? a_uv.x : a_uv.x - 1.0;
	v_uv.y = a_uv.y;

edit...Better still, plus allows more textures than GLES2 can shake a stick at:

    //ex. vec2(2.0, 0.0) is adjusted to vec2(0.0, 0.0) w/ texture id 2
    v_textureID = int(a_uv.x);
    v_uv = vec2(a_uv.x - float(v_textureID), a_uv.y);

I suggest your UVs are pointing at pixel centers vec2i(2047, 0) would be vec2(2047.5, 0.5) / textureDim


edit2: might as well go all the way; use GL_REPEAT then you only have to pass the textureID and leave the UVs as is (v_uv = a_uv). Note the advantage of doing this in the vertex shader rather than for every texel.

Edited by jezham

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