I have a problem that is giving me trouble; my goal is to create an octree.

Actually, it's very cheap (but it's nearly enough for what I want to do with that octree).

My problem is that my std::vector<std::reference_wrapper<Point>> is filled with the same value. So my insertion creates an infinite loop. But here's the code, maybe it'll be more understandable. I put commentaries where my error occurs.

source.cpp

void main(){
    std::random_device rd;
    std::mt19937 rng(rd());
    std::uniform_real_distribution<double> uni(0, 2);
    auto random_integer = uni(rng);
    Point firstCenter = Point(1, 1, 1);
    Point firstHalfDimension = Point(1, 1, 1);
    Octree oct(firstCenter, firstHalfDimension);

    for (int i = 0; i < 3; ++i) {
        double x = uni(rng);
        double y = uni(rng);
        double z = uni(rng);
        Point ptmp = Point(x, y, z);
        std::cout << x << " " << y << " " << z << std::endl;
        auto po = std::ref(ptmp);
        oct.insert(po);
    }
}

Octree.hpp

class Octree {
    Node octree;
    Point firstCenter;
    Point firstHalfDimension;

public: 
    Octree() = default;
    Octree( Point& firstCenter, Point& firstHalfDimension) :
        firstCenter(firstCenter), firstHalfDimension(firstHalfDimension), octree(firstCenter, firstHalfDimension) {}

    void insert(std::reference_wrapper<Point> pt) {
        octree.insert(pt);
    }
};

Node.hpp (where the problem occurs)

#define MAXVAL 2
using Point = gmtl::Vec3d;

class Node {
    Point center;
    Point halfDimension;
    std::vector<std::reference_wrapper<Point>> datas;
    std::array<std::shared_ptr<Node>, 8> children;


    int getOctant(const std::reference_wrapper<Point> p) {
        int oct = 0;
        if (p.get()[0] >= center[0]) oct |= 4;
        if (p.get()[1] >= center[1]) oct |= 2;
        if (p.get()[2] >= center[2]) oct |= 1;
        return oct;
    }

    const bool isLeaf()  {
        return !children[0];
    }

public:

    Node(Point center, Point halfDimension) : center(center), halfDimension(halfDimension){
    }
    void insert(const std::reference_wrapper<Point> p) {
        if (isLeaf()){
            if (datas.size() == MAXVAL) { //Must subdivide
                std::cout << p.get()[0] << " " << p.get()[1] << " " << p.get()[2] << std::endl;
                for (int i = 0; i < datas.size(); ++i) {
                    std::cout << datas[i].get()[0] << " "
                        << datas[i].get()[1] << " "
                        << datas[i].get()[2] << std::endl;
 //The problem is here : the vector is filled with the same values, and it's the same value as p. let's say p = (0.4,0.7,0.8), then the for loop will show to the screen : 
 // 0.4 0.7 0.8
 // 0.4 0.7 0.8
 // 0.4 0.7 0.8
                }
                for (int i = 0; i < 8; ++i) {
                    Point newCenter;
                    newCenter[0] += halfDimension[0] * (i & 4 ? .5f : -.5f);
                    newCenter[1] += halfDimension[1] * (i & 2 ? .5f : -.5f);
                    newCenter[2] += halfDimension[2] * (i & 1 ? .5f : -.5f);
                    children[i] = std::make_shared<Node>(newCenter, halfDimension * .5);
                }
                int octant = getOctant(p);
                children[octant]->insert(p);
                for (int i = 0; i < datas.size(); ++i) {
                    int octant = getOctant(datas[i]);
                    children[octant]->insert(datas[i]);
                }


            }
            else { //Just add
                datas.push_back(p);
            }
        }
        else { //Non-leaf node
            children[getOctant(p)]->insert(p);
        }
    }
};

I really don't understand what I'm doing wrong.

std::vector<std::reference_wrapper<Point>> datas;

Something has to own the objects that this vector contains references to.

You filled your vector with references to objects that no longer exist. It's not clear why you think you need a vector of references, but you shouldn't do something unusual unless you have a very good explanation for why the normal way (a vector of values) won't work for you.