OpenGL ES 2.0 NVIDIA TEGRA 2展開

Question

我的一個新着色器出現問題。

當我在不同的設備上嘗試着色器時，着色器編譯並順利運行。

當我嘗試它的的NVIDIA Tegra 2的GPU，運行本地代碼，我得到以下編譯錯誤：

着色器無法編譯。頂點着色器 （0）：錯誤C6002：超過指令限制256; 532條指令需要編譯程序 86行，1個錯誤。

我認爲這是由於NVIDIA的gcg編譯器執行某種類型的展開，但我無法弄清楚如何解決這種情況。

現在，我試圖減少if/else語句的類型，而不是if/else，而是一連串簡單的連續ifs（只是爲了查看問題是否已解決），但它不起作用。

我試圖刪除所有的條件，只留下一個和着色器編譯良好（儘管它沒有做我真正需要的）。

這是我的頂點着色器的代碼，片段着色器只包含gl_color信息並編譯良好。

void main(){ 

//Converto direttamente in int, cosi le divisioni dopo sono piu veloci 
int_character_position = int(character_position); 

uniform_reference = (int_character_position/4); 

uniform_reference_sub_item = int(mod(character_position,4.0)); 

if(uniform_reference == 0) {   
    working_float = charsequence_1[uniform_reference][uniform_reference_sub_item];  
} else if(uniform_reference == 1) {  
    working_float = charsequence_1[uniform_reference][uniform_reference_sub_item];    
} else if(uniform_reference == 2) { 
    working_float = charsequence_1[uniform_reference][uniform_reference_sub_item]; 
} else if(uniform_reference == 3) { 
    working_float = charsequence_1[uniform_reference][uniform_reference_sub_item]; 
} else if(uniform_reference == 4) { 
    working_float = charsequence_2[uniform_reference - 4][uniform_reference_sub_item]; 
} else if(uniform_reference == 5) { 
    working_float = charsequence_2[uniform_reference - 4][uniform_reference_sub_item]; 
} else if(uniform_reference == 6) { 
    working_float = charsequence_2[uniform_reference - 4][uniform_reference_sub_item]; 
} else if(uniform_reference == 7) { 
    working_float = charsequence_2[uniform_reference - 4][uniform_reference_sub_item]; 
} else if(uniform_reference == 8) { 
    working_float = charsequence_3[uniform_reference - 8][uniform_reference_sub_item]; 
} else if(uniform_reference == 9) { 
    working_float = charsequence_3[uniform_reference - 8][uniform_reference_sub_item]; 
} else if(uniform_reference == 10) { 
    working_float = charsequence_3[uniform_reference - 8][uniform_reference_sub_item]; 
} else if(uniform_reference == 11) { 
    working_float = charsequence_3[uniform_reference - 8][uniform_reference_sub_item]; 
} else if(uniform_reference == 12) { 
    working_float = charsequence_4[uniform_reference -12][uniform_reference_sub_item]; 
} else if(uniform_reference == 13) { 
    working_float = charsequence_4[uniform_reference -12][uniform_reference_sub_item]; 
} else if(uniform_reference == 14) { 
    working_float = charsequence_4[uniform_reference -12][uniform_reference_sub_item]; 
} else if(uniform_reference == 15) { 
    working_float = charsequence_4[uniform_reference -12][uniform_reference_sub_item]; 
} else if(uniform_reference == 16) { 
    working_float = charsequence_5[uniform_reference -16][uniform_reference_sub_item]; 
} else if(uniform_reference == 17) { 
    working_float = charsequence_5[uniform_reference -16][uniform_reference_sub_item]; 
} else if(uniform_reference == 18) { 
    working_float = charsequence_5[uniform_reference -16][uniform_reference_sub_item]; 
} else if(uniform_reference == 19) { 
    working_float = charsequence_5[uniform_reference -16][uniform_reference_sub_item]; 
} else if(uniform_reference == 20) { 
    working_float = charsequence_6[uniform_reference -20][uniform_reference_sub_item]; 
} else if(uniform_reference == 21) { 
    working_float = charsequence_6[uniform_reference -20][uniform_reference_sub_item]; 
} else if(uniform_reference == 22) { 
    working_float = charsequence_6[uniform_reference -20][uniform_reference_sub_item]; 
} else if(uniform_reference == 23) { 
    working_float = charsequence_6[uniform_reference -20][uniform_reference_sub_item]; 
}     
v_texCoord.y = a_texture.y; 
v_texCoord.x = a_texture.x + (0.0105 * float(working_float)); 
gl_Position = myMVPMatrix * myVertex; 

Answer 1

OpenGL ES 2.0的實施方案中通常具有它們支持的指令的最大數量，但由於它是不切實際的最小數目不被授權的API。 Nvidia Tegra實現將源代碼編譯爲532條指令，該指令超出了其最大限制。除了修改OpenGL ES 2.0實現之外，唯一可能的方法是重寫着色器，以便將其編譯爲更少的指令。

如果不知道Nvidia實現的細節，像working_float = charsequence_6[uniform_reference -20][uniform_reference_sub_item];這樣的語句與分支結合在一起會產生多條指令（您有減法，加載，存儲和數組索引）。一般來說分支通常是有問題的，因爲由於不同的代碼流，編譯器可能無法編譯多條指令並行運行。

你如果 - 否則，如果 - 否則，如果代碼是很容易改寫，以便它可以被編譯成更小的指令，同時仍具有相同的含義：

if(uniform_reference < 4) {   
    working_float = charsequence_1[uniform_reference][uniform_reference_sub_item]; 
} else if(uniform_reference < 8) { 
    working_float = charsequence_2[uniform_reference - 4][uniform_reference_sub_item]; 
} else if(uniform_reference < 12) { 
    working_float = charsequence_3[uniform_reference - 8][uniform_reference_sub_item]; 
} else if(uniform_reference < 16) { 
    working_float = charsequence_4[uniform_reference -12][uniform_reference_sub_item]; 
} else if(uniform_reference < 20) { 
    working_float = charsequence_5[uniform_reference -16][uniform_reference_sub_item]; 
} else if(uniform_reference < 24) { 
    working_float = charsequence_6[uniform_reference -20][uniform_reference_sub_item]; 
} 

可能已裝配到256指令限制。 （在着色器代碼中，規則的大拇指通常只是GLSL代碼的大小，粗略定義了指令的數量。）如果不是這樣，看起來您可以將uniform_reference - {4,8,etc}替換爲uniform_reference_sub_item，因爲它似乎是相同的值;這也會節省一些說明。最終，你可以把charsequence_ [123456]值到一個數組和索引它的基礎上uniform_reference一個常數，還有的CharSequence劃分，而不是，所以，到底你有這樣的事情：

working_float = charsequence[uniform_reference/4][uniform_reference - uniform_reference_sub_item][uniform_reference_sub_item]; 

也許你也可以合併由uniform_reference和uniform_reference_sub_item索引的兩個數組以排除一個索引級別。