如何访问指向指向对象的指针列表的指针向量的向量数组中的前向列表？

Question

Here's the relevant piece of code:

std::vector<std::vector<std::forward_list<Person*>*>*> grid;

for (int i = 0; i < criticalValuex; i++) {
    std::vector<std::forward_list<Person*>*>* list;
    for (int j = 0; j < criticalValuey; j++) {
        std::forward_list<Person*>* flist;
        list->push_back(flist);
    }
    grid.push_back(list);
}
for (Person *person : people) {
    int i = (int)((person -> getPosition().x - city.getPosition().x + citySize.x / 2) / cellSize);
    int j = (int)((person -> getPosition().y - city.getPosition().y + citySize.y / 2) / cellSize);
    if (i < citySize.x / cellSize && j < citySize.y / cellSize) {
        grid[i][j]->push_front(person);
    }
}

Ignore the Person or any city parameters. The error isn't due to the variables or classes. The error occurs when I write grid[i][j]->push_front(person); even though it should work.

I think that grid[i][j] should be a pointer to a forward list so if I deference it, then I get back the forward list right? But it tells that it has no member "push_front".

Answer 1

Attention! The element in the grid is pointer.

As declared std::vector<std::vector<std::forward_list<Person*>*>*> grid; , type of grid[i] is std::vector<std::forward_list<Person*>*>* . It is not an array but a pointer of array. So grid[i][j] wont compile, while (*(grid[i]))[j] does, and its type is std::forward_list<Person*>* .

If want to perate to the forwardlist, the right code is (*(grid[i]))[j]->push_front(...) .

Much more, declare the grid as type std::vector<std::vector<std::forward_list<Person*>>> is more easier way.