Testing if a number goes evenly into 6?

Question

I am trying to see if two input numbers (integers) including negative numbers go into 6 evenly (remainder is 0). This is the code I was trying.

if((in1)%6 == 0 && (in2)%6 == 0){
    printf("Divisible: both\n");
}
else if((in1)%6 == 0 && (in2)%6 > 0){
    printf("Divisible: only %i\n",in1);
}
else if((in1)%6 > 0 && (in2)%6 == 0){
    printf("Divisible: only %i\n",in2);
}
else{
    printf("Divisible: neither\n");}

This works for all positive integer but for any negatives the printed code is always "Divisible: neither" any help as to how I can show both positive and negative numbers divisible by six with a remainder of 0 would be really helpful

You need to be able to debug your code in order to become a competent software developer - unless you, unlike the rest of us, are always going to write error-free code that runs correctly the first time. I suggest that you add some code before the first `if` statement to print out the values of `in1`, `in2`, `in1 % 6` and `in2 % 6` in order to figure out what's going on. Best of luck. — Bob Jarvis - Слава Україні, Feb 27 '17 at 03:05
@M.M - I think you meant [fizz buzz](http://codegolf.stackexchange.com/questions/58615/1-2-fizz-4-buzz) — Bob Jarvis - Слава Україні, Feb 27 '17 at 03:08
[Modulo of a negative number](http://math.stackexchange.com/q/519845/398889) on the Math Stack Exchange may explain some of what you are observing. Also see [Modulo operation with negative numbers](http://stackoverflow.com/q/11720656/608639) and [modulo operation on negative numbers](http://stackoverflow.com/q/30199474/608639) on Stack Overflow. — jww, Feb 27 '17 at 05:57

score 5 · Accepted Answer · answered Feb 27 '17 at 03:00

You could use != 0 instead of > 0. In C, % of negative number will give a negative result (or zero).

This is because a / b is defined as truncation-towards-zero since C99 (in C90 it was implementation-defined). And a % b is defined as a - (a / b) * b.

Note that you actually do not need this test at all; you can rely on the behaviour of if...else not entering the else case if the if case was satisfied, e.g.:

if ( in1 % 6 == 0 && in2 % 6 == 0 )
{
    // ...
}
else if ( in1 % 6 == 0 )   
{
    // would not reach here if in2 % 6 == 0
}
else if ( in2 % 6 == 0 )
{
    // would not reach here if in1 % 6 == 0
}
else

score 0 · Answer 2 · answered Feb 27 '17 at 05:11

Another consideration, rather than oblige code to test numbers 3 times, re-write to perform only 2 test on the numnbers.

if (in1 % 6) {
  if (in2 % 6) {
    printf("Divisible: both\n");
  } else {
    printf("Divisible: only %i\n",in1);
  }
} else {
  if (in2 % 6) {
    printf("Divisible: only %i\n",in2);
  } else {
    printf("Divisible: neither\n");}
  }
}

Testing if a number goes evenly into 6?

2 Answers2