Rounding up float point numbers bash

问题

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?

回答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)

回答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

回答3:

A little trick is to add 0.0005 to your input, this way you will have your number round up correctly.

回答4:

if you're receiving the round-off error with the number 17.928 try this: read y v=echo "scale = 3; $y" |bc -l if [ $v == 17.928 ] ; then echo "17.929" else echo $v fi

来源：https://stackoverflow.com/questions/26465496/rounding-up-float-point-numbers-bash

标签

bash

shell

floating-point

decimal

rounding