编程形态学开放返回输入

Question

我在编程形态学打开，它返回的结果，这是相同的输入图像。我是不是误解了对它的解释？我的代码是：

Var x, y;
Func limit, erosion, dilation;

ImageParam input(type_of<uint8_t>(), 2);
Param<int> dimension;

RDom r(-1 * dimension / 2, dimension, -1 * dimension / 2, dimension);
limit = BoundaryConditions::repeat_edge(input);

erosion(x, y) = argmin(r, limit(x + r.x, y + r.y), "erosion")[2];
dilation(x, y) = argmax(r, erosion(x + r.x, y + r.y), "dilation")[2];

erosion.compute_root();
dilation.vectorize(x, 4).parallel(y);

Target target = get_host_target();

target.set_feature(Target::NoRuntime, true);

dilation.compile_to_static_library(path, { input,dimension }, target);

Answer 1

下面的结果似乎与我的OpenCV教程映像相似：

#include "Halide.h"
#include "../../tools/halide_image_io.h"

using namespace Halide;

int main(int argc, char **argv) {
    Var x, y;
    Func limit, erosion, dilation;

    ImageParam input(type_of<uint8_t>(), 2);
    Param<int> dimension;

    RDom r(-1 * dimension / 2, dimension, -1 * dimension / 2, dimension);
    limit = BoundaryConditions::repeat_edge(input);

#if 0
    erosion(x, y) = minimum(limit(x + r.x, y + r.y), "erosion");
    dilation(x, y) = maximum(erosion(x + r.x, y + r.y), "dilation");
#else
    erosion(x, y) = argmin(r, limit(x + r.x, y + r.y), "erosion")[2];
    dilation(x, y) = argmax(r, erosion(x + r.x, y + r.y), "dilation")[2];
#endif

    erosion.compute_root();
    dilation.vectorize(x, 4).parallel(y);

    Image<uint8_t> in = Halide::Tools::load_image("/some/path/morphology.png");
    input.set(in);
    dimension.set(10);

    Image<uint8_t> result = dilation.realize(in.width(), in.height());

    Halide::Tools::save_image(result, "/some/path/morphology_out.png");

    return 0;
}

#if 0块只显示使用最小和最大运算符，而不是argmin/argmax。对argmin/argmax的r参数也是可选的。

我会问，是在如何将图像传递给Halide编译函数，还是在如何处理返回的输出方面出现了错误。您还可以使用debug_to_file或Halide代码中的打印来调试管道的中间阶段。