I'm new to Golang. I wanted to separate a floating point number into whole and decimal parts. After some research I implemented it but there is a problem in my code. I'm using 5.8 as the input but the result is 5 and 0.79999.
package main
import(
"fmt"
"math"
)
func Round2(val float64) {
intpart, div := math.Modf(val)
fmt.Println(div)
fmt.Println(intpart)
}
func main() {
fmt.Println("Hello, playground")
Round2(5.8)
}
I have tried this, I'm getting an output of:
0.7999999999999998
5
If there is any other way to do this please let me know. I have inserted the go playground with my code in it.
This is an artifact of floating-point arithmetic. A floating-point value is not always exactly the value you expect it to be due to the fact that not all numbers can be represented exactly in the floating-point format used by whatever processor you happen to be using. Obviously, the value is very close to 0.8 and is equal to 15 significant figures. Note that this is not specific to Go.
Instead of
fmt.Println(div)
try
fmt.Printf(".4f
", div) // shows only 4 decimal places
It looks like this article has some good information: https://en.wikipedia.org/wiki/Floating_point#Accuracy_problems
You may want to use a Decimal64p2 type of decimal with fixed .00 precision - https://github.com/strongo/decimal
It is efficient for storing exact amounts of money.
Try this:
f := 5.8
ipart := int64(f)
fmt.Println(ipart)
decpart := fmt.Sprintf("%.7g", f-float64(ipart))[2:]
fmt.Println(decpart)
The complete answer in available in go playground. Basically I used the format specifier %g to achieve the result, moreover, %.6g rounds the decimal part to six places, trailing zeroes are not shown.