CODEWITHSUNDEEP
c
lkemitchll
lkemitchll

Reputation: 2751

C program concept

I'm sorry if this question seems very noobish.

I'm having trouble figuring out the best way to approach this problem.

Here's the basic idea in code:

  int iterator = 0;

  if ((iterator % 2) == 1){
    // do first option
    iterator++;
    return;
  } else if ((iterator % 2) == 0){
    // do second thing
    iterator++;
    return;
  } else if ((iterator % 3) == 0){
    // do third option
    iterator++;
    return;
  } else{
    // error
    return;
  }

Essentially the function will be called once a second (it's a watchapp for Pebble).

I can get the first two options to work but i'm having trouble with the third. I assume it's because of the % 3 being too vague.

How would you guys approach this?

Upvotes: 2

Views: 91

Answers (3)

azz
azz

Reputation: 5940

Is this what you're after?

if ((iterator % 3) == 0){
    // do first option
} else if ((iterator % 3) == 1){
    // do second thing
} else if ((iterator % 3) == 2){
    // do third option
} 
iterator++;

This can be re-written as:

switch (iterator++ % 3) {
case 0:
     // first option
     break;
case 1: 
     // second option
     break;
case 2: 
     // third option
}

Upvotes: 4

AndersK
AndersK

Reputation: 36092

the %2 will give you only two options.

you could also try a different structure

switch ( iterator%3 )
{
  case 0: // option 0
    break;
  case 1: // option 1
    break;
  case 2: // option 2
    break;
  default:
    break; 
}
++iterator;

Upvotes: 1

Ashalynd
Ashalynd

Reputation: 12573

You never get to the third option, because iterator % 2 can be either 0 or 1.

Upvotes: 4

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