Rounding up float point numbers bash

Question

Ok, so I'm trying to round up an input of 17.92857, so that it gets an input of 17.929 in bash.

My code so far is:

read input
echo "scale = 3; $input" | bc -l

However, when I use this, it doesn't round up, it returns 17.928.

Does anyone know any solutions to this?

Solution 1

In case input contains a number, there is no need for an external command like bc. You can just use printf:

printf "%.3f\n" "$input"

Edit: In case the input is a formula, you should however use bc as in one of the following commands:

printf "%.3f\n" $(bc -l <<< "$input")
printf "%.3f\n" $(echo "$input" | bc -l)

Solution 2

To extend Tim's answer, you can write a shell helper function round ${FLOAT} ${PRECISION} for this:

#!/usr/bin/env bash

round() {
  printf "%.${2}f" "${1}"
}

PI=3.14159

round ${PI} 0
echo
round ${PI} 1
echo
round ${PI} 2
echo
round ${PI} 3
echo
round ${PI} 4
echo
round ${PI} 5
echo
round ${PI} 6
echo

# Outputs:
3
3.1
3.14
3.142
3.1416
3.14159
3.141590

# To store in a variable:
ROUND_PI=$(round ${PI} 3)
echo ${ROUND_PI}

# Outputs:
3.142

Comments

[Ludovic Feltz]

Or like @Tim said use printf

[Mark Reed]

If you're rounding to three places, you should actually add .0005, not .005. But why can't you do that?

[Max]

But you have to set a larger scale > 3 first (else 0.0005 = 0.000), and then set scale = 3 and compute the result using, e.g., x / 1. There may also be unexpected results for negative x and you may be required to use if ( x < 0 ) { x = x - 0.0005; } else { x = x + 0.0005 } or similar.

[Gnadelwartz]

try "round xzzzzzzzz" ;-)