How to efficiently cycle consecutive numbers c++ - c++

Question

I am looking for a solution for cycling through consecutive numbers based on an input value. Similar to modulo, but different for negative numbers. Is there a better solution compared to the inefficient code below? Here is some input/output examples:
Numbers range 0 to 2
-2 -> 1
-1 -> 2
0 -> 0
1 -> 1
2 -> 2
3 -> 0
4 -> 1
//Inefficient Code example
int getConsecutiveVal(int min, int max, int input) //Inclusive in this scenario
{
while (input>max)
input -= (1+max-min);
while (input<min)
input += (1+max-min);
return input;
}
//Incorrect Code example since func(0,2,-1) returns 2
int getConsecutiveVal(int min, int max, int input)
{
return (input % (1+max-min))+min;
}

Answer 1

To be able to increment or decrement, I used the following function. It's more than 1 line, but fewer math operations. It's similar in spirit to the original poster's format. Tested for positive and negative cases.
int16_t cycleIncDec(int16_t x, int16_t dir, int16_t xmin, int16_t xmax) {
// inc/dec with constrained range
// the supplied xmax must be greater than xmin
x += dir;
if (x > xmax) x = xmin;
else if (x < xmin) x = xmax;
return x;
}
Output of cycleIncDec() with various start values and step sizes
x: 11: [phone]
x: 4: -1 3 2 1 0 -1 -2 -3 -4 -5 -6 -7
x: -8: -1 -13 -12 -11 -10 -9 -8 -13 -12 -11 -10 -9
x:-190: -2 -192 -194 -196 -198 -200 -170 -172 -174 -176 -178 -180

Answer 2

In principle, you need the modulo operator. The problem is that in C it doesn't work as expected for negative numbers.
If you know the minimum input value, you can just add a positive number x big enough to transform all negative numbers to positive. It won't affect the result if x % R = 0 (in your example R=3.)
In your example, if you add, say, 3*10 to all inputs and perform the modulo operation you'll get the desired result:
mod(3*10+[-2 -1 0 1 2 3 4], 3)
= 1 2 0 1 2 0 1
(the above is matlab notation and is specialized to the example you have presented. I'll leave it to you to extend it to arbitrary min/max)
A specific formula for the case you have presented:
You have suggested using
((input+abs(input)*(1+max-min)) % (1+max-min))+min
However, this formula does not work. For two reasons:
First, if input=0, the abs() returns 0 and you get the minimum value as output (This is not always what your explicit while-based loop produces)
Second, you forgot to subtract min from the input before the operation.
So the correct formula is the following (using x for input):
(x - xmin + (1+abs(x))*(1+xmax-xmin)) % (1+xmax-xmin) + xmin

Answer 3

You can call % twice to get you the right behaviour, since a%b, for positive b, is guaranteed to lie in [-b+1, b+1].
int getConsecutiveVal(int min, int max, int input)
{
int range_len = (1 + max - min);
input -= min;
return (((input % range_len) + range_len) % range_len) + min;
}