GLSL奇偶合並排序

Question

我想了解gpugems網站上的the odd-ever merge sort example，但我無法弄清楚他們傳遞給制服的一些內容。這是整個着色器。

uniform vec3 Param1; 
uniform vec3 Param2; 
uniform sampler2D Data; 

#define OwnPos gl_TexCoord[0] 

// contents of the uniform data fields 
#define TwoStage Param1.x 
#define Pass_mod_Stage Param1.y 
#define TwoStage_PmS_1 Param1.z 
#define Width Param2.x 
#define Height Param2.y 
#define Pass Param2.z 

void main(void){ 
// get self 
vec4 self = texture2D(Data, OwnPos.xy); 

float i = floor(OwnPos.x * Width) + floor(OwnPos.y * Height) * Width; 

// my position within the range to merge 
float j = floor(mod(i, TwoStage)); 

float compare; 

if ((j < Pass_mod_Stage) || (j > TwoStage_PmS_1)) 
    // must copy -> compare with self 
    compare = 0.0; 
else 
    // must sort 
    if (mod((j + Pass_mod_Stage)/Pass, 2.0) < 1.0) 
    // we are on the left side -> compare with partner on the right 
    compare = 1.0; 
    else 
    // we are on the right side -> compare with partner on the left 
    compare = -1.0; 

// get the partner 
float adr = i + compare * Pass; 

vec4 partner = texture2D(Data, vec2(floor(mod(adr, Width))/Width, floor(adr/Width)/Height)); 

// on the left it's a < operation; on the right it's a >= operation 
gl_FragColor = (self.x * compare < partner.x * compare) ? self : partner; 
} 

絆倒我的部分是搞清楚他們分配給Param1和Param2.z。

Param2.x和Param2.y只是圖像的寬度和高度。每次通過循環時，pass變量是否只是一個遞增的數字？

Param1.x，Param1.y和Param1.z讓我完全陷入困境。是否應該在這個程序的CPU端發生什麼事情，他們不在這裏？

任何幫助或清晰度將不勝感激！謝謝

Answer 1

該示例的代碼可用here。 的main.cpp文件的行298是一個你需要的：

glUniform3fARB(oddevenMergeSort.getUniformLocation("Param1"), 
    float(pstage+pstage), 
    float(ppass%pstage), 
    float((pstage+pstage)-(ppass%pstage)-1) 
); 
glUniform3fARB(oddevenMergeSort.getUniformLocation("Param2"), 
    float(width), 
    float(height), 
    float(ppass) 
); 

隨着int ppass = 1<<pass;和int pstage = 1<<stage;