Question

E.g.:

for (int i = 0; i < 3; i++) {
  for (int j = 0; j < 4; j++) {
    for (int k = 0; k < 5; k++) {
      ...
    }
  }
}

That’s 3 * 4 * 5 = 60 times executing the innerst code. Now I want to use the values of the indices i, j, and k to generate all numbers from 0 to 59 (not necessarily sorted).

Answer 1

Ok, got it.

Make k count 1, j count k’s limit, i count j’s limit times k’s limit.

for (int i = 0; i < 3; i++) {
  for (int j = 0; j < 4; j++) {
    for (int k = 0; k < 5; k++) {
      int c = k * 1 + j * (5) + i * (5 * 4);
      System.out.println(c);
    }
  }
}

Output: 0..59.