# Reversed Round Function
**URL:** https://discourse.processing.org/t/reversed-round-function/38274
**Category:** Coding Questions
**Created:** [August 8, 2022, 3:08am UTC](https://discourse.processing.org/t/reversed-round-function/38274 "2022-08-08T03:08:14Z")
**Posts on this page:** 4
**Page:** 1
### Author:  [@Mattyjak](https://discourse.processing.org/u/Mattyjak)
#### Post date: [August 8, 2022, 3:08am UTC](https://discourse.processing.org/t/reversed-round-function/38274/1 "2022-08-08T03:08:14Z")
Hello fellow programmers!
I was working on a small code snippet and I ran into an issue.
In my project, I use `floor()` to convert from global coordinates to “chunk” coordinates via a simple mathematic equation: `floor(globalCoordinates / chunkSize)` which gives me a ratio of the two values which I can then use as a coordinate system for the chunks.
The only issue is that certain values will be misrepresented and will give me the chunk right behind them as opposed the chunk ahead. Use `ceil()` does not remedy it, as it only applies it to the opposite values.
I attempted to use `float()` instead but I end up with the worst of both of them. I realize now that if there were some way to switch float from rounding up when it needs to round down (and vise-versa) I wouldn’t have said issue.
If you guys have any ideas as to what I could do to try and make an inverted `round()` function it would be great!
---
### Author:  [@mnse](https://discourse.processing.org/u/mnse)
#### Post date: [August 8, 2022, 5:24am UTC](https://discourse.processing.org/t/reversed-round-function/38274/2 "2022-08-08T05:24:05Z")
Hi @Mattyjak,
see below
Cheers
— mnse
```auto
void setup() {
size(500, 500);
float v1 = 4.49;
float v2 = 4.5;
// floor always rounds down
println("floor("+v1+") = " + floor(v1));
println("floor("+v2+") = " + floor(v2));
// ceil always rounds up
println("ceil("+v1+") = " + ceil(v1));
println("ceil("+v2+") = " + ceil(v2));
// round mathematical round < .5 down >= .5 up
println("round("+v1+") = " + round(v1));
println("round("+v2+") = " + round(v2));
// to see how round works see below
println("myround("+v1+") = " + myround(v1));
println("myround("+v2+") = " + myround(v2));
}
// round function
// rounds float values < 0.5 down and >= 0.5 up
// so 4.49 becomes 4 and 4.50 becomes 5
int myround(float f) {
return int(f+0.5);
}
```
output:
> floor(4.49) = 4
> floor(4.5) = 4
> ceil(4.49) = 5
> ceil(4.5) = 5
> round(4.49) = 4
> round(4.5) = 5
> myround(4.49) = 4
> myround(4.5) = 5
---
### Author:  [@Mattyjak](https://discourse.processing.org/u/Mattyjak)
#### Post date: [August 8, 2022, 7:00am UTC](https://discourse.processing.org/t/reversed-round-function/38274/3 "2022-08-08T07:00:49Z")
Hi there! I already know how to use the round, ceil, and floor methods.
What I needed was an inverse to the round function, as in anything below .5 would be rounded up and anything above .5 would be round down.
I came up with my own solution that seems to work fine:
```auto
int invRound(float val){
if(val+0.5 > 1){
return round(val+0.5)-1;
}
return round(val+0.5);
}
```
It’s pretty simple but it gets the job done.
Thank you anyways for trying \<3
---
### Author:  [@GoToLoop](https://discourse.processing.org/u/GoToLoop)
#### Post date: [August 8, 2022, 7:37am UTC](https://discourse.processing.org/t/reversed-round-function/38274/4 "2022-08-08T07:37:30Z")
Your solution using the ternary operator:
```java
static final int invRound(float val) {
return (val += .5) > 1? round(val) - 1 : round(val);
}
```
> **[Reference](https://processing.org/reference/conditional.html)**
>
> A shortcut for writing an if and else structure. The conditional operator, ?: is sometimes called the ternary operator, an operator that takes three arguments. If the test …