Do something every 1-5 iteration

Question

Need to do somethin for each of every fifth iteration, somehow, this doesn't quite work, can't get the gist of it, what did I do wrong:

for (int i=1; i<1000; i++) 
{
    if (i % 1 == 0) 
    {
        // Do something here for all iterations that end in 1 or 6
        Console.WriteLine("Iteration:" + i.ToString());
    }
    if (i % 2 == 0)
    {
        // Do something here for all iterations that end in 2 or 7
        Console.WriteLine("Iteration:" + i.ToString());
    }
    if (i % 3 == 0)
    {
        // Do something here for all iterations that end in 3 or 8
        Console.WriteLine("Iteration:" + i.ToString());
    }
    if (i % 4 == 0)
    {
        // Do something here for all iterations that end in 4 or 9
        Console.WriteLine("Iteration:" + i.ToString());
    }
    if (i % 5 == 0)
    {
        // Do something here for all iterations that end in 5 or 0
        Console.WriteLine("Iteration:" + i.ToString());
    }
};

Note that `i % 1` is the remainder when dividing by 1, i.e., always 0. — vonbrand, May 23 '21 at 23:56
There are a few answers here trying to improve the code - but it contains just a small error (almost a typo) - all `if`-s should be `if (i % 5 == 2)` instead of `if (i % 2 == 0)`. `i % 2 == 0` check if the number is even (0 reminder when dividing by 2). `i % 5 == 2` checks if the number ends with 2 or 7 - reminder is 2 when divided by 5. — Kobi, May 23 '21 at 23:56
The operand of the remainder operator is the _divisor_, not the expected remainder. See duplicates for more details about how to use it correctly. — Peter Duniho, May 23 '21 at 23:57

score 3 · Accepted Answer · answered May 23 '21 at 23:48

3

You only need to compute the remainder once per iteration, like this:

for (int i=1; i<1000; i++) {
    switch (i % 5) {
        case 0: Console.WriteLine("Remainder 0"); break;
        case 1: Console.WriteLine("Remainder 1"); break;
        case 2: Console.WriteLine("Remainder 2"); break;
        case 3: Console.WriteLine("Remainder 3"); break;
        case 4: Console.WriteLine("Remainder 4"); break;
    }
}

answered May 23 '21 at 23:48

Brian Berns

15,499
2
30
40

Ah, I hadn't thought of it this way. This is better than my answer, good call. – Matt U May 23 '21 at 23:52
Alternative: `switch (i % 10) { case 1: case 6: Console.WriteLine("Action 1/6"); break; ... } }` *Note:* @brianberns solution is shorter, this one might be more readable as it incorporates the text of your comments in the code and humans are used to thinking base 10 ... ;-> – AlWo23 May 24 '21 at 00:13

score 3 · Answer 2 · answered May 23 '21 at 23:49

You're so close. You instead want to do case analysis on i % 5.

for (int i = 1; i < 10000; i++) {
   switch (i % 5) {
      case 1:
         <Do whatever you do for last digit of 1 or 6>;
         break;
      case 2:
         <Do whatever you do for last digit of 2 or 7>; 
         break;
...
    }
}

score 0 · Answer 3 · answered May 23 '21 at 23:52

If I understand the question correctly, you could always have a separate counter that you reset manually:

var iteration = 0;
for (var i = 0; i < 1000; i++)
{
    iteration++;
    if (i == 1)
    {
        // First iteration
    }
    else if (i == 2)
    {
        // Second iteration
    }
    // And so on

    if (iteration >= 5)
    {
        iteration = 0;
    }
}

Do something every 1-5 iteration

3 Answers3