我正在研究一个openCL内核，它会加载一些点，决定哪个点最高，然后返回它。一切都很好，但我想在最高评价之前添加一个计算。这将点与一对线进行比较。我有写作和学位，如下：
size_t i = group_id * group_stride + local_id;
while (i < n){
//load up a pair of points using the index to locate them within a massive dataSet
int ia = LOAD_GLOBAL_I1(input, i);
float4 a = LOAD_GLOBAL_F4(dataSet, ia);
int ib = LOAD_GLOBAL_I1(input, i + group_size);
float4 b = LOAD_GLOBAL_F4(dataSet, ib);
//pre-assess the points relative to lines
if(pass == 0){
float px = a.x;
float py = a.y;
int checkAnswer;
//want to write this section as a function
float x1 = tri_input[0].x; float y1 = tri_input[0].y;
float x2 = tri_input[2].x; float y2 = tri_input[2].y;
float check = sign((x1-x2) * (py-y1) - (y2-y1) * (px-x1));
if(check != tri_input[3].x){ //point is outside line 1
checkAnswer = 1;
}
else{
x1 = tri_input[2].x; y1 = tri_input[2].y;
x2 = tri_input[1].x; y2 = tri_input[1].y;
check = sign((x1-x2)*(py-y1) - (y2-y1)*(px-x1));
if(check != tri_input[3].y){ //point is outside line 2
checkAnswer = 2;
}
else{
checkAnswer = 0; //point is within both lines
}}}
//later use the checkAnswer result to change the following
//find the highest of the pair
float4 result;
if(a.z>b.z) result = a;
else result = b;
//load up the previous highest result locally
float4 s = LOAD_LOCAL_F4(shared, local_id);
//if the previous highest beat this, stick, else twist
if(s.z>result.z){ STORE_LOCAL_F4(shared, local_id, s);}
else{ STORE_LOCAL_F4(shared, local_id, result);}
i += local_stride;
}
我想做的是将行检查作为函数调用两次，即代码变为：
size_t i = group_id * group_stride + local_id;
while (i < n){
//load up a pair of points using the index to locate them within a massive dataSet
int ia = LOAD_GLOBAL_I1(input, i);
float4 a = LOAD_GLOBAL_F4(dataSet, ia);
int ib = LOAD_GLOBAL_I1(input, i + group_size);
float4 b = LOAD_GLOBAL_F4(dataSet, ib);
//pre-assess the points relative to lines
if(pass == 0){
float px = a.x;
float py = a.y;
int checkA = pointCheck( px, py, tri_input);
px = b.x;
py = b.y;
int checkB = pointCheck( px, py, tri_input);
}
//later use the checkAnswer result to change the following
//find the highest of the pair
float4 result;
if(a.z>b.z) result = a;
else result = b;
//load up the previous highest result locally
float4 s = LOAD_LOCAL_F4(shared, local_id);
//if the previous highest beat this, stick, else twist
if(s.z>result.z){ STORE_LOCAL_F4(shared, local_id, s);}
else{ STORE_LOCAL_F4(shared, local_id, result);}
i += local_stride;
}
在这个例子中，函数是：
int pointCheck( float *px, float *py, float2 *testLines){
float x1 = testLines[0].x; float y1 = testLines[0].y;
float x2 = testLines[2].x; float y2 = testLines[2].y;
float check = sign((x1-x2) * (py-y1) - (y2-y1) * (px-x1));
if(check != testLines[3].x){ //point is outside line 1
return 1;
}
else{
x1 = testLines[2].x; y1 = testLines[2].y;
x2 = testLines[1].x; y2 = testLines[1].y;
check = sign((x1-x2)*(py-y1) - (y2-y1)*(px-x1));
if(check != testLines[3].y){ //point is outside line 2
return 2;
}
else{
return 0; //point is within both lines
}}}
虽然longhand版本运行正常并返回正常的“最高点”结果，但函数版本返回错误结果（未检测到我隐藏在数据集中的最高点）。即使函数尚未产生整体效果，它也会产生错误的结果。
我做错了什么？
取值
[更新]： 这个修改过的函数可以在注释掉的行中使用，然后挂起来：
int pointCheck(float4 *P, float2 *testLines){
float2 *l0 = &testLines[0];
float2 *l1 = &testLines[1];
float2 *l2 = &testLines[2];
float2 *l3 = &testLines[3];
float x1 = l0->x; float y1 = l0->y;
float x2 = l2->x; float y2 = l2->y;
float pX = P->x; float pY = P->y;
float c1 = l3->x; float c2 = l3->y;
//float check = sign((x1-x2) * (pY-y1) - (y2-y1) * (pX-x1)); //seems to be a problem with sign
// if(check != c1){ //point is outside line 1
// return 1;
// }
// else{
// x1 = l2->x; y1 = l2->y;
// x2 = l1->x; y2 = l1->y;
// check = sign((x1-x2) * (pY-y1) - (y2-y1) * (pX-x1));
// if(check != c2){ //point is outside line 2
// return 2;
// }
// else{
// return 0; //point is within both lines
// }}
}
答案 0 :(得分：1)
一个直接的问题是如何将参数传递给被调用的函数：
/* Styles go here */
.breadcrumb_nav_div{
top:44px;
position:absolute;
width:100%;
background-color:lightgrey;
padding:0.8em 0em 0.5em 1em;
}
.breadcrumb {
list-style: none;
overflow: hidden;
font: 14px Helvetica, Arial, Sans-Serif;
}
.breadcrumb li {
float: left;
margin-left:5px
}
.breadcrumb li a {
color: white;
text-decoration: none;
padding: 10px 0px 10px 55px;
background: grey; /* fallback color */
background: grey;
position: relative;
display: block;
float: left;
}
.breadcrumb li a:after {
content: " ";
display: block;
width: 0;
height: 0;
border-top: 50px solid transparent; /* Go big on the size, and let overflow hide */
border-bottom: 50px solid transparent;
border-left: 30px solid grey;
position: absolute;
top: 50%;
margin-top: -50px;
left: 100%;
z-index: 2;
}
.breadcrumb li a:before {
content: " ";
display: block;
width: 0;
height: 0;
border-top: 50px solid transparent; /* Go big on the size, and let overflow hide */
border-bottom: 50px solid transparent;
border-left: 30px solid white;
position: absolute;
top: 50%;
margin-top: -50px;
margin-left: 1px;
left: 100%;
z-index: 1;
}
.breadcrumb li:first-child a {
padding-left: 10px;
}
.breadcrumb li:nth-child(2) a { background: grey; }
.breadcrumb li:nth-child(2) a:after { border-left-color: grey; }
.breadcrumb li:nth-child(3) a { background: grey; }
.breadcrumb li:nth-child(3) a:after { border-left-color: grey; }
.breadcrumb li:nth-child(4) a { background:grey; }
.breadcrumb li:nth-child(4) a:after { border-left-color: grey; }
而函数本身需要px和py的指针。您应该将该函数称为：
int checkA = pointCheck( px, py, tri_input);
令人惊讶的是，OpenCL不会为此内核提供构建错误。
根据我的经验，一些OpenCL运行时不喜欢单个函数中的多个return语句。尝试将返回值保存到局部变量中，并在函数末尾使用单个return语句。这是因为OpenCL不支持实际函数调用，而是将所有函数直接内联到内核中。因此，最佳做法是将所有非int checkA = pointCheck(&px, &py, tri_input);
函数标记为__kernel，并将其视为这样（即，通过不使用多个return语句，使编译器更容易内联函数）。