How to iterate neighbors of a grid concisely with a for loop.

grid_neighbors.cpp

#include <algorithm>
#include <iostream>

// Suppose we need to iterate over the neighbors of a cell in a 2D grid:
//
// * * *
// * - *
// * * *
//
// This can be done with the following well known pattern:
void all_neighbors() {
    for (int i = -1; i <= 1; ++i) {
        for (int j = -1; j <= 1; ++j) {
            if (i == 0 && j == 0) continue;
            std::cout << i << ", " << j << std::endl;
        }
    }
}

// But what if we just want to iterate the horizontal and vertical neighbors, skipping the diagonal?
// 
//   *
// * - *
//   *
//
// Most programs explicilty repeat the operation 4 times, or use a table.
// But there is a trick to do it in a loop. 
// Think of the neighbor as a vector offset, and then rotate it 90 degrees, 4 times.

template <typename I>
void rotate_2d(I& a, I& b) {
    std::swap(a, b);
    b = -b;
}

void direct_neighbors() {
    int i = 1;
    int j = 0;
    for (int k = 0; k < 4; ++k) {
        std::cout << i << ", " << j << std::endl;
        rotate_2d(i, j);
    }
}

int main() {
    std::cout << "diagonal" << std::endl;
    all_neighbors();
    std::cout << "square" << std::endl;
    direct_neighbors();
}

Since this is coming up in google searches, I think it's worth correcting that the condition i == j yields the wrong output for the diagonal neighbors loop.
You instead want to only skip the iteration where both i and j are 0. By continuing when i == j, you skip -1, -1 and 1, 1, the top left and bottom right.

@nicholaschiasson thanks! 🤦🏻

How does that look?

Great! 😊