我试图降低侵蚀函数的执行时间,当我尝试用平铺划分问题时,执行实际上更慢,如图所示:
我没有任何调度的代码是:
Halide::Image<uint8_t> erode(Halide::Image<uint8_t> input, int dimension) {
Halide::Var x("x"), y("y");
Halide::Image<uint8_t> output;
Halide::Func limit("limit"), e("e");
limit = Halide::BoundaryConditions::repeat_edge(input);
Halide::RDom r(dimension*-1 / 2, dimension, dimension*-1 / 2, dimension);
e(x, y) = limit(x, y);
e(x, y) = Halide::min(limit(x + r.x, y + r.y), e(x, y));
output = e.realize(input.width(), input.height());
return output;
}
我的代码尝试平铺(我尝试使用示例,如教程中所示):
Halide::Image<uint8_t> erodeTiling(Halide::Image<uint8_t> input, int dimension) {
Halide::Var x("x"), y("y"), x_outer, x_inner, y_outer, y_inner, tile_index;
Halide::Image<uint8_t> output;
Halide::Func limit("limit"), e("e");
limit = Halide::BoundaryConditions::repeat_edge(input);
Halide::RDom r(dimension*-1 / 2, dimension, dimension*-1 / 2, dimension);
e(x, y) = limit(x, y);
e(x, y) = Halide::min(limit(x + r.x, y + r.y), e(x, y));
e.tile(x, y, x_outer, y_outer, x_inner, y_inner, 64,64).fuse(x_outer, y_outer, tile_index).parallel(tile_index);
output = e.realize(input.width(), input.height());
return output;
}
任何有关如何正确安排的提示将不胜感激,因为我对此仍然很陌生。
编辑:用于获取时间的代码:
__int64 ctr1 = 0, ctr2 = 0, freq = 0;
output = erode(input, dimension);
if (QueryPerformanceCounter((LARGE_INTEGER *)&ctr1) != 0) {
// Activity to be timed
output = erode(input, dimension);
QueryPerformanceCounter((LARGE_INTEGER *)&ctr2);
QueryPerformanceFrequency((LARGE_INTEGER *)&freq);
}
std::cout << "\nerosion " << dimension << "x" << dimension << ":" << ((ctr2 - ctr1) *1.0 / freq) << "...";
ctr1 = 0, ctr2 = 0, freq = 0;